. , ... w... ......n.....:......r....3..... "a. g
“......wbéxewhé $39.3”...

..QA' ...:AA A}? I}
.. {Lyn‘lpuflncvoh ...; SM. ...-tumuA
A .... .
1

9 Ant )l 0A J
wwwwfiwufimnmww.“ . ..m...»km.%ku.fifi
I O ‘.V‘ A
... ..m. ... .5535 35......

.L. ~ ... .
gflwumm. wfifimflfififisfi ,. «Mu . ..

     

              

   
    

a... a: .
A5 lfiahtrtfia a"!
t n I A A ..«P

3.2.... ......2...i<.msii L
a... gixl Int. ‘vuuxsx

pr:-n...V...:ona-.vsa ...-IvKJsABWNA

     

                
  

             

 
    

      

.. .../7r: «xvu. Sq.
_ HRH...” rerr tuna {Vtrfiers luv-A

:13}: at...
v‘ .
pr r..t..r.t..1, {if *vvvw‘r Jr r.

w
. . .. ..m.u.u.n.u.n...u.y... ......L... a .. «...... ,
. . . tenor-n: (e: . . . ...
. 39......“ .... a. ”flvmrrm r»... rt: 9 x c ... 1r: ..qu .. - nrvrxvvakur . t x .. = n x." Bah...- ) «a:

. P I11. El .
z 3 c.....x.....tcv1zv.wt r v r ..Jd. imm‘: «v.53. .... v4.9.xfiwflutflfl$~flflfidrtfl-nuéctg-fifi‘wflfiv-a “HAW-“kflfitacnfikflh. .

J...

   

        

              

 

        
       

        

 

         

 

.
...er...rp..x:..n .- .... air... . 1 A e .— ‘ Al.
rr tvrzvac-KZ} 13...: £333.! a m: A :39”?ng ...... a. .v: (“at J. A .. .
......"fifi..U.U.H.u.n.n..§.m...... ...“...v......wh.$.uawwm.ww ..wfi .... . . 2.444%... ..ék.ar......m.........;n...._xu.-
.r;x>t .... .1. 5.1.. .. .. A k... n t. A. . a , t... A u» . . ALA .. (A s... .
...?acrwnuuhunmunrxAJanerAsh? chums”: . Any. Mex-3.1%... «a. many-HY AxAki‘A :«HAIAK H8» AAA . l “Aggy ”9&- A... It). A .3: AL flan... ..
v. fr... z, .. 111;... m a A ..r. R... r v.1”: at NI-tcfififia .fisy Aka-0011‘“? % A‘WJH ...-WM.- IA «MIL. 3.9.. h. ..11 6 Al‘st‘flttsmfl 33“....” VAL—’4'
. .: . .. . .. . . pm. 4. A A ... .9 A AU: MW.»

             

has}... .. x16. ._ I: turn}.

1:: :6. . a . . .

:rw.........y..... ..r u... o

vv»c..vv.'rsux
...

        
               

   

. 3&5... ...u. ..n. . . .. . . .. . ....
....uu...«.,..n. $3.... ....mm 4.4%.... ...»..«mw. fink? .. ...... 33....

lit-{AA

...... A. nu 0.. A" . , a . trAuitnltle
:54... .....Fufi: 4%.: N «a A 5‘ «a... A. 7.2.51 .. . A u Thur fist» gduuhnun.
.. g%.u§a.fl..fi¥ figufiwfi: .wflnflfiwmmv Mafia“? t¢Wn¢fy$. . . A. . 3:... t M
a... A on. on A a o A... A AA. 05161 t . . . ugfifiaux
. u. “unwang. w.%aummm..&mmfi. ......r. a... ..

         

       

          

 

  
  

                             
                            
                                       

   

 
 
 

 

l A .. . .fflnuf. . 7.3.11.1}
A 7 “Ant t 93. ”3M. ) fithh Aolw - A 0 As. A8 ....A at $0.15!. AAA? I.v.l....
...rnr ,«AVAJHNJ‘AJVVVHxWIflA‘JflEMH‘”scjsnanveoirfl tamnbufmwaw: “mg-Pain A: \AOAVAI. flog-AA rho- da tibtoflrg ...ot-“slibiwwgk‘zu .Humu4Hl. . :13...”qu {Am-...
93:... sAnJE-“nizu . 5‘04. t A935... A c .- t I An. ”AA A)! .- .VAAAJ. A1: A O£ltnsA§AiAAIA¢u it! .05 t3;
tkta‘ A a 2.th if ..AAi: o A AfiffinA b t b 55‘ A: .. i.¢.v.ohlt..§|t.rvl.:1. .{ttf-x.“ (A;
940 ..." A A «MA ‘9... A ALI ... AtA qsgté‘Ait6§.i.-A (L.XILDAI;IPA
.. Ain't .. , 9‘39.“ 0 SVAFtAgs AA églltldcnh so! ‘54 A AOAI goz All.-o.f..3uift.itt6$.a§lt.bi .. vb
...-Av 911.... A .. v A 0.30.. A A ...—AAA... n 2.50 1... 0.10.3“ gi A JALAAAS.AAA$AASIAA$AA\A£$§31:3.
9.01.3? Alba: A kt... IO. J‘utfltAhcn oMfi-boilchv all) 0.: AA. , «Ad... Atbtaklsi {At}. 1 in?!) A:
i do}... .. A121. c Audra“. 5dr... ...: A7..¢A¢...o A AA AA A AAfihAi-Alli‘, DA t..t.aiAAzAtAA. A $.35. $1.315}...
v5.02; ‘7”. $1.... ...!- “11 mm. ”d, c acfuvpos nomaamvrnflnflot Htifiihfigruuauum A 5¥ANAA it? Artivv-lfiauw. 1' ”mum-T “Viulamflshtnfihfidfi‘ A “gnfibuAtAfimefi? ?.Ar.u¢..%h.m.hnwthc Ahufihhfirhutr
.. c 1. c a ... AA AA A A Ina .3 A: A path .835 it... e u CAL. P) A. 5‘ (I; O» ..ot..(....ix.. (7... 11.17.5531,
Ann .2... ......rs $51: 3......th A on“ oleAtlfivt-vlzaP u foquNs-fio tut “AA! «- $43 if... tux A A3AA.‘|LG§A$¢A¢ASA$§EAZA&APAA 9:31.... '1
:a at) u! f A a 1.3. A A. a no... it: n. . I. A I... 3 A 5.503;. Ari-Ag... AAAAtAAl.Ab‘OAzfiAN.o...AJAA;AAAif}??? ..iit’tahl!
...: A... Wag: .. :35: 4 A -..-.53.: Qahfi a $2504 A . AA... ...AaAotoAAA A ...-dogging. A AA. AitAast‘A‘Atiliiklx .
Agni-AA. a». ..., day Acflti‘og 52$.“ 4.. AAAAA: fol-.5». :85: n6: .«AJAA A A Aza>ulss AfNAAAAAfga: 3.4.1113): ...?!
......:.f.... 2...: .. 3:...5... . .5. z Aureafi .....f... ..t......%: fig}: . t.:.9.¢¥3u.i.d..uhzmfi119:1§e. .3...

I
i).
6.

   

   

 

‘ I
z .... . .
J A At ..I. A :- 3.4 to A AAA A. AZ)... A. . A . T (A17 .
an"... ”NARA. Jana): 2AA." v AANapMWMAAAA new; a... hfifiafla yfifzfiflb AAA gfisflflrfl33.54..anx...vsn..i.ux‘bz$t).. ..LXFAM.
m to ,

 

    

    

          

 

          

     

 

. Jmava 9.9.0.93.- .. . A OLA-AA .
A n. A. .1! t .aflc 13.“- g I 3 I C. V (A ufltthzn¢$il 4161!} Alinaghz
.. .....nu...»fi$...n«..Vkmfimwmp:¥m anymw. ......1... ......mm. m... 3... .w. .. ”Waflm..neh..2:uunzr§z$z. a. .3... Want. :11}:
.... honing-$4 59.5555“ “A... 43159,: liaNAnAaosatt no. A A AA A ADJ-VIA In A.» AVAIL than“! ...-AAA 511A ...-It:
"At )3 cc 54...: n 1 WA} A «tad-both A: Jurv .....(iCA-d‘hXVoicha v t A AQA 3g MAAA..A£. 1.97:}! A haltlftgat

Atty-Ar.- brfiéaa.‘ , n tzAAA twang: Cl A .3: AAA at”... 5A A.!‘AA A’flAOAIOAA‘AItV z...
fa... «......ux... @9311... «$.53 .... ......P..u...:...::. ..m. ....mmmmyqawzkt infinwflmmfi ...?»uéfififiu .... :3...» u”.

q... x. u. 10... . .10 0: AA . a... . . . . A4 A
AflbhflbcuguufiWAI fiMfiuflfleewfimanmIM$oh~ae “4:9..an AAA .n a “Mnflusfifihfiflmftflz.fléi Ilbigsvvflqpfiflu‘A
sokqutcAu “...-Pl a, oiffl a... It? m.% 5!. s A}... lath-fl . ..A :13: .A‘AO}. ... . . (rig. 35$)..-
icA 5:1,... I i A II: ,I AOOA‘A n t Aft...“ 4. . gs? A“ ’AIA AA is“; FLA:
vasuooralliA: Auflvnq I. Otaood 6 c A 2: 523 h. I A Mi. 64...... 9a.: . b A)“ t DAL-{A I
A... is A 05:11:...“ . «AAA A no 950.“: AAA 1. AA! A Aoc¢xbnbt (I A

055.1. “AH .r . A Ilia; A: if 0.. c A A $9,312.. 91.. A’Akaflg ... -.v§.z..9p|
'AIAlutOt- .3 .. it‘QA-U I93! 5 IHfiA-m . ... . t cattfltf Ate—cva‘Iltb 5.9.0:...- 335A;
...... $4....3uufivva .... , ...... 83.3... ... ,i.}fu....!u1.§.y3:..m.¢.uz..rfimma strumvnwuw... 2 :qWMmmnatuns~isa.t.irifitmvflabiun
IA. .0 A wclt- .0 0,- . . 5501.!Oil9- . it: Oh! . fl IA 5).. At
....“anfinnamrcmt .....vthAAAtiLP . . . . . ,QIA‘WIAAU‘AAAIJAAHQAAmlhflhnWAbAA30.9.69”?Atfihxtyg .-
...: :1...$.B.:w%.§ .. . .. marthuwgésxntishfi$s
. . u ‘ '«u

n
U
A
a"?
o;-
'4‘
5‘:
‘1.
.3
“I
:b
,0
I
‘0
. ‘ J
:I
‘,

 

.
5 A
.2:
A
.‘a
5'
, A
i. t
2%:
F‘
{i
1;:
‘ :s
f'
2%
.3
h

r
I
I
a
A
a
l
J
a
:1
‘x
'2
'a
'5;
‘I
'u
. st:
’ D
b
C
b

v
7
i
..b
k

.

O

L

a

A

9

 

'Afo
41.1..
‘41:

O Q
.
4‘5
x. r
Hi
uh
Ag.
[5‘
.l
’2
f;
j;

J
{L
I
3 3
A
:5.
‘5

 

- a
9
a
A
A

2;.
v
a
A
A
O
E
,
A
A
é
3
.

 

: 3
.
:4: l ‘
Ag: ‘
1‘7 1
5‘ O
.43;
'0 I
it:
.0
I‘,
if:
I
A
i
D
I

£
1
S.
I
i

Q! s
.....fixgmfiafibfiidtz
. . ....wmv ...,mwmmflvua:
%fl§..hdfinfi. 3... $6.93....
A A fill: .
u.-

.. C
fig...”

a

r

t
1
h
t

a
to
3.
E.
C

i
i
' 5+3
I
2
:

 

.

A’s
)
’
,
’c
a
.
a
O
A
A1
I
A
A
h
r
J
.
t

I
A
I

7
i
b
;
v
e
A
A
2
I
E
l
r

I
e
3
¥
.
f
V
v
A
O
5
‘9
I
A
c
if»?
'2

 

       

3
i:
z i
“i?

C
-.x
‘
U
u

a;

   

flow:

09”! ..
azénrag‘ flt‘vlfllqhufloi (2-,!)

      

.
"i
i I
‘i
4 i
v}?!
w
é
. ‘,

 

            

 

I
5%
3
3:

$1..

a

5
i:
"i
,t ~
Ii

 

I
J
a .
ta“-
’QE
7‘ ‘
:n}
,1
mi
I I
k‘:'
A ;

 

€35:
Q:
W
H
M:

 

2
as
' - g9.

  
                      

1A.. A «3.101.320
A 09:10::V
{~55 .IAAi‘t QdA. A

      

(«A. ,At
. .vn..1\. ...A .
1...\AI\?II€

            

                  

... . )‘b...« #:8\ \1

 

        
  

                    

   
    

        

 

        

   

   

...... :x.u....v..«.n>...( a ~ Iv 3...: . 1,v....u\‘.n&11A.u....¢A.,...w.. .x \ . ...)...figt.
. . .. . 3.31;va
. . . .. , . . . . ; .1. . . . . .. . . . . . . . \. ..\.... .. . I... . .... ..t‘ltK..: .
:- ....~<us...........t..a.. 0- .. . .. .«4.:...nflnLsz_.5\m....¢)2|..iA.m$u.A—‘ttuli.av 1.x..1,,\...A|¢.‘.;.......,. ... ,. , 5.3.4: AAAdIJAA.4t.~IUAoA.I.pA»..«A«.\....14»N.A.4J-O<\.A.!A$
1 pol..;.;«qx...n.;1y.f:. , aiopjfiiudwut: .RdAtvvCuAA u. 1 .. 1.1005921 . . . . . . r A , t . u . ”whit‘sfiiv.‘
......nnu........r... ..s.................. 65.3.. an. it... . . 5.54». A. .335 . can... gwgufitnk. wfifi aw .Qnrufisiv 5v¢uuxxb
94..) ... 1 $95 ......rxl. , . ...:OAIHJJKJA). thy! DAWAAII).).AA4Y.3 . fig. bio 5.... in“: ...:AJKAAfiiA fluid; a...“ 0145.134. 7.911. ..}\I . ... Ah. .. A, . Ax. .\.r ...... ....
. . .\ . . . .. .l . ... . . . . .. .. 1. ‘IA ...-... §...v ... .

 
 

 

 

 

ABSTRACT
PROGRAMED INSTRUCTION FOR
REMEDIAL ELECTRONICS STUDENTS

by HerbertrPaul Froehlich

Programed instruction began as an application of principles from the
laboratory to the technology of education. The two major forms of pro—
gramed instruction are the linear and non—linear, introduced by Skinner
and Crowder, respectively. While linear programs are designed to elicit
correct learner responses by means of a fixed sequence of small instruc—
tional steps, non—linear programs provide a flexible sequence of large in-
structional steps. Because of these and other differences between linear
and non—linear programs, related to their degree of organization and read—
ability, it was hypothesized that the relative effectiveness of these two
tYPeS of programs in teaching electronics would depend on the academic,
intellectual] and reading abilities of the students who used them.
Furthermore , since the results of previous investigations suggested
that the sequéncing of instructional steps provided an organization of sub—
ject matter for the low-achieving student which he was not able to provide
fer himself, it was hypothesized that the two forms of programed instruc—
tion Would assist a group of slow learners undertaking remedial training
to aChieve at a higher level than a conventional lecture. This hypothesis

was tested againSt criteria which were designed to measure (a) achievement

Herbert Paul Froehlich

on electronics subject matter specifically taught in lecture, linear, and
non-linear presentations, and (b) later academic performance on electronics
concepts which depended on an understanding of the more elementary prin—
ciples developed in the remedial presentations.

The fact that responses are constructed in a linear program rather than
selected from among several alternatives, as they are in a non-linear pro-
gram, suggested the hypothesis that there was an interaction between the
form of response required by the program and the form of response required
by a criterion test (multiple choice and construction), the criterion test
measuring electronics ability. A second hypothesis stemming from the
difference in program response form was that the student would need more
time to complete the linear program than the non—linear. Obviously, then,
linear programs would prOVide less efficient learning than non—linear pro—
grams, from a time point of view only, everything else being equal.

The subjects of this study were 494 naval trainees undertaking reme-
dial instruction. These students had failed an examination given at the
end of the second week of a 19—week basic electronics course and were
required to repeat the second week of instruction at night while continuing
the third week of instruction during the day.

Remedial classes , with students repeating the second week of instruc-
tion, met for two hours each of three nights. Each class received compa—
fable lecture , linear! or non-linear instructional presentations using the

Second week subjeCt matter. These three instructional treatments were

random}, assigned to classes.

Herbert Paul Froehlich

There were two studies due to the fact that the two forms of the crite-
rion test (81 and 82) , used alternately as pre-treatment and post-treatment
measures of electronics ability, were not parallel. One study was con-
cerned with the analysis of data from classes provided with the 81—82 or-
der of presentation, while the second study considered classes given the
82-81 order. Each form of the criterion test contained 15 multiple choice
and 15 construction items.

Because of unequal treatment sample sizes, and for greater precision,
regression analyses were conducted. These analyses were, in effect,

analyses of covariance using two control variables within each of two
regression models. One regression model included scores on intelligence
and prior academic achievement as independent variables, and the other
used reading comprehension and pretest scores. Each of several regres-
sion equations provided an error sum of squares which was used in an F—
statistic designed to test hypotheses concerning interactions between
learner and method variables, and treatment main effects. Intercorrela—
tion matrices included all the variables for each treatment group.

The results from one set of data showed that the linear program was
best for students with low prior achievement scores , and that the lecture
was best for students with high scores. An examination of the mean and
the distribution of the posttest criterion scores suggested that this inter-

aCtion and the lack of a significant treatment main effect, was due to

Criterion tests which were too difficult. High ability individuals were

Herbert Paul Froehlich
academically superior to those of low ability, although neither form of
programed instruction, nor the lecture, permitted the high ability students
to reach the academic level usually attained by non—remedial students.
Although it took students using the linear program on the average 10 percent
longer to complete instruction than students using the non—linear program,
based on the total number of minutes to complete instruction, the average
student using either form of programed instruction completed remedial
training better than 30 percent sooner than those in the lecture.

It was concluded that either the principles of learning developed in
the laboratory were not directly applicable to human verbal learning as
represented by programed instruction, or designers of programed learning
were far from providing faithful implementation of these principles of

learning,

PROGRAMED INSTRUCTION FOR

REMEDIAL ELECTRONICS STUDENTS

By

Herbert Paul Froehlich

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOC TOR OF PHILOSOPHY

Department of Psychology

1964

Sir
AyM/v’“

.'I / I‘

/

15m

-4

ACKNOWLEDGMENTS

Many persons contributed directly and indirectly to the conduct of
this study and deserve my sincere thanks for their cooperation and patience:
the Chief of Naval Air Technical Training, Memphis, Tennessee, who made
this study possible and the Commanding Officer, Naval Air Technical
Training Center, Memphis, Tennessee, in whose schools the investiga—
tion was conducted; the Naval Air Station, Memphis, Tennessee, the
School of Aviation Medicine, Pensacola , Florida , and the Computer Center,
Michigan State University provided their computer facilities for processing
the data.

Personal thanks are extended to my Dissertation Committee which in—
cluded Dr. James S. Karslake (Chairman), Dr. Eugene Jacobson, Dr.
Frederic Wickert, and Dr. David Krathwohl. Dr. Martin Fox, of the Mich—
igan State University Statistics Department, suggested the regression
analyses used in this study. I am most grateful for his assistance. Dr.

G. D. Mayo, of the Naval Air Technical Training Command was instru—
mental in mobilizingthe initial interest and manpower needed to conduct

the research.

ii

 

 

TABLE OF CONTENTS

Page
ACKNOWLEDGMENTS ......................... ii
LIST OF TABLES ............................ v
LIST OF FIGURES ........................... viii
LIST OF APPENDICES ......................... ix
Chapter
I. INTRODUCTION ....................... 1
Purpose ......................... 3
Literature Review .................... 7
Theoretical ...................... 7
Practical ....................... 11
II. HYPOTHESES ........................ 15
III. SETTING .......................... 17
Subjects ......................... l7
Variables ........................ 17
Intelligence ..................... l7
Criterion Tests .................... 18
Davis Reading Test .................. 19
Weekly Examinations ................. 22
Time (T) ....................... 23
Development of the Instructional Presentations ..... 23
Design .......................... 25
IV. ANALYSIS OF THE DATA ................... 28
Regression Analysis ................... 28
Correlational Analysis .................. 38

iii

 

TABLE OF CONTENTS —— Continued

Chapter Page

V. RESULTS ................ . ........ 39

Main Analyses .................... 45

Hypothesis la ................... 45

Hypothesis lb .................... 50

Hypothesis 2a ................... 52

Hypothesis 2b ................... 54

Hypothesis 3a . . . . . .............. 55

Hypothesis 3b .............. . . . . . 55

Hypothesis 4a ................... 57

Hypothesis 4b ................... 57

Hypothesis 5 ................... 58

Hypothesis 6 ................... 59

VI. DISCUSSION ....................... 60

VII. SUMMARY AND CONCLUSIONS .............. 72

The Learning Process .......... . ...... 72

The Measurement Process ............. . . 75

APPENDICES ............................ 76

BIBLIOGRAPHY ........................... 6 91
iv

Table

10.

LIST OF TABLES

Page

Differences Between Skinnerian and Crowderian
Approaches to Programing Instruction ......... 4

Independent and Dependent Variables Used in
Testing Hypotheses of Interactions Between
Ability and Method Variables, and Method
Main Effects ..................... 6

Item Content, Discrimination, and Difficulty Indices
for the Final Form of the Criterion Tests Based on
Two Samples of Regular Students ........... 20

Reliabilities of Forms 81 and 82 When Used as Pretests . 21

Intercorrelations, Means, and Standard Deviations for
Intellectual, Reading, and Academic Variables
Based on a Sample of 500 Regular Students . . . . . . 22

Estimates of the Kuder—Richardson Reliabilities for
Three Weekly Tests Administered During the
19—Week Basic Electronics Course .......... 23

Breakdown of Subjects by Class, Treatment, and Form
of Criterion Test Administered First (81 or 82) . . . . 27

Variables Used in the Regression Analysis ........ 33

Correlations Within Each Treatment Group Between
Part and Total Scores of the Criterion Tests ..... 40

Summary Table for Differences in Means of Pre—
Treatment Variables for Subjects in the Lecture,
Skinner, and Crowder Treatments (Set 1) ....... 41

Summary Table for Differences in Means of Pre—
Treatment Variables for Subjects in the Lecture,
Skinner, and Crowder Treatments (Set 2) ....... 42

LIST OF TABLES -- Continued

Table Pa ge

12. Summary Table for Differences in Means of Post-
Treatment Variables for Subjects in the Lecture,
Skinner, and Crowder Treatments (Set 1) ....... 43

13. Summary Table for Differences in Means of Post-
Treatment Variables for Subjects in the Lecture,
Skinner, and Crowder Treatments (Set 2) ....... 44

14. Tests of Significance for Interactions Between Ability
and Method with Posttest and Retest Dependent
Variables (Hypothesis la) ............... 46

15. Estimates of the PTT tot Means for High, Average,
and Low WK 1 Scores for the Lecture, Skinner,

and Crowder Treatments ................ 50
16. Tests of Significance for Method Main Effects with

Posttest and Retest Dependent Variables

(Hypothesis lb) .................... 51
17. Tests of Significance for IQ Main Effects Across

Programed Instruction Treatments (Hypothesis 2a) . . 52

18. Estimates of the PTT tot Means for High, Average,
and Low IQ Scores for the Skinner and Crowder
Methods ....................... 53

19. Estimates of the PRT tot Means for High, Average,
and Low IQ Scores for the Skinner and Crowder
Methods ....................... 54

20. Tests of Significance for RC Main Effects Across
Programed Instruction Treatments (Hypothesis 2b) . . 54

21. Tests of Significance for Interaction Between Ability
and Form of Programed Instruction with Time as
the Dependent Variable (Hypothesis 3a) ........ 55

22. Tests of Significance for Programed Instruction Main

Effects with Time as the Dependent Variable
(Hypothesis 3b) . . . ................. 56

vi

 

 

Table

23.

24.

25.

26.

LIST OF TABLES -— Continued

Tests of Interaction Between IQ and Method with
Achievement on Related Subject Matter as the
Dependent Variable (Hypothesis 4a) ........

Tests of Significance for Method Main Effects with
Achievement on Related Subject Matter as the
Dependent Variable (Hypothesis 4b) .........

Tests of Significance for Method Main Effects Using
PTT mo and PTT fi as Dependent Variables
(Hypothesis 5) ....................

Estimates of the PTT tot Means for High, Average,

and Low RC Scores for the Skinner and
Crowder Mahods ...................

Vii

Page

57

58

62

LIST OF FIGURES

Figure Page

1. Representation of the Interaction Between WK 1
and Method by Plots of the Treatment Regression
Coefficients with PTT tot as the Dependent
Variable (Set 2) .................... 48

2. Representation of the Interaction Between WK 1
and Method by Plots of the Treatment Regression
Coefficients with PTT tot as the Dependent
Variable (Set 1) .................... 49

viii

Appendix

A

B

LIST OF APPENDICES

Supplementary Tables A — I ...............
Skinner Program .....................
Crowder Program .....................
Answer Sheets Used with Skinner Program ........
Answer Sheets Used with Crowder Program ........

Criterion Tests ......................

Raw Data: Scores on All Variables for All Subjects

Page
77
100
207
635
660
663

680

CHAPTER I

INTRODUCTION

While learning theorists have agreed upon certain principles estab—
lished in the psychological laboratory, they have not, in the past, always
been able to institute such principles in the classroom. Glaser (32) has
pointed out the unfortunate lack of influence which education and the de—
veloping science of learning have had on each other. Major obstacles to
the fruitful interaction between laboratory findings and learning in the
classroom may be attributed to differences between the experimental lab—
oratory and the classroom, namely, differences in species of the subject,
conditions of learning, and subject matter. Skinner (63, 64) and Crowder
(16), however, have sought to close the gap between the laboratory and
the classroom, assisted by recent advances in educational technology.

Kendler (40) cautioned against making an automatic connection be—
tween theory and its successful application to a practical problem, while
Glaser (32) argued that applied research was the meeting ground of applied
and basic endeavors. It was at the applied level that solutions to both
theoretical and practical issues were attempted, often in the same exper-
imental program. Indeed, the work of Skinner and Crowder, on verbal
learning by means of self-instructional devices, stimulated both theoret—

ical and practical aspects of research in learning. Through the use of

self-instructional devices - those concerned with the presentation, activity,

1

 

and feedback phases of the teaching cycle (23) — researchers functioning
in an educational context were able to manipulate and study parameters of
learning previously confined to laboratory analysis. Thus, "schedules of
reinforcement (42, 57), " "contingencies of reinforcement (65) , " and
”shaping" became terms applied to human as well as non-human learning.
On the practical side, educational specialists gave more attention to
an examination of instruction, putting aside for a time attempts at patching
up activities which supported instruction, such as the manipulation of
teacher-student ratios, the. lengthening of courses, or the introduction of
training aids to instruction. Because self—instructional devices provided
frequent and immediate reinforcement, allowed for active student response,
and took into account the fact and the nature of individual differences,
psychologists felt that such learning devices were often more efficient
sources of communication and control of the learning process than were
classroom instructors (65). Those concerned with education attempted to
capitalize on this potential. They saw in programed instruction a way to
relieve shortages of teachers and educational facilities (66) , a means of
providing students with "individualized instruction (65) , " and an approach
to assisting those with limited intellectual capabilities (ll). ‘
Thus , learning principles , translated by an educational technology,
allowed theoretical and practical research to blossom within a few short

years (25 , 31).

 

Purpose

The present research, conducted in a military training environment,
had both theoretical and practical implications. The theoretical implica-
tions stemmed from characteristics of Skinnerian and Crowderian approaches
to programing. These characteristics, outlined in Table 1, suggested to
the present investigator that differences in a learner's intellectual, reading,
and academic abilities might condition the degree to which different ap-
proaches to implementing knowledge of results, active responding, and
program step size affect student performance. The chief purpose of the
present study, therefore, was to investigate the relationship between
learner and/instructional variables.

Table 1 shows that responses are constructed in a linear program
rather than selected from among several alternatives as they are in a non-
linear program. Hedberg (34) pointed out that a criterion test of only mul—
tiple choice items would favor Crowder's non—linear method of programing
if the criterion test was for that kind of program. The present study, there-
fore, investigated the possible interaction between the response form of
the criterion (multiple choice and fill-in) and the form of programed instruc—
tion.

Practical implications were based on several empirically supported
assumptions: first, that students using programed instruction could attain
a common, but high level, of academic achievement. Skinner (64) , for
example, has pointed out that low ability students were capable of sub-

stantial, even extraordinary, achievement if permitted to move at their own

 

 

 

4

TABLE 1

DIFFERENCES BETWEEN SKINNERIAN AND CROWDERIAN
APPROACHES TO PROGRAMING INSTRUCTION

 

Skinnerian

Laboratory
Principle

Programing
Principle

Crowderian

Programing
Principle

 

The relation between
behavior and the
consequences of be—
havior is brought about
by "contingencies" of
reinforcement

Complex performance is
shaped by changing con—
tingencies of reinforce—
ment progressively in
the direction of the re—
quired behavior

Behavior is brought
under control of the en—
vironment by reinforcing
only when particular
stimuli are present

Before behavior is rein—
forced the response
must be omitted

Appropriate stimulus—
response connections
must be made

Immediate and frequent
reinforcement of behav—
ior appropriate to the
desired goal

Gradual progression
from simple to complex
concepts in successive
steps. Instruction is
presented in a fixed
"logical" sequence of
small steps

The stimuli to be dis-
criminated are present
and reinforced in a
variety of stimulus
contexts

Students respond by
constructing the re-
sponse before receiv-
ing confirmation

Correct responses are
assured through
prompting and cueing.
Errors leave traces of
erroneous responses

Students proceed at
their own rate through
the sequence

Teaching is a communi-
cation process. Feed-
back to the student de—
termines whether the
communication was suc—
cessful or not

If communication is not
effective the student
may be shifted to sub—
ject matter not mastered
(branching). Steps are
large and allow flexi—
bility of instruction

Students respond by
selecting a response
from among alternatives

Error is used to guide
the student who is
learning

Individual differences
are accounted for
through branching

 

 

pace, if frequently and immediately reinforced, and if allowed to follow a
coherent sequence of instructional frames. Porter (53), using a linear
spelling program showed that there was almost no correlation between
aptitude and achievement following learning by programed instruction.
Such a finding was predictable if, in fact, programed instruction allowed
students, progressing at their own pace, to attain a common level of per—
formance, since achievement test scores following programed instruction
would then show little variation. Second, it was assumed that the time
devoted to conventional training could be shortened, even when training
low ability or remedial students, as in the present study. Previous studies
showed a savings of 13 percent for a high ability group using an elec-
tronics program (10) and almost a 50 percent savings in training time for
a program in German (24).

There were, thus, two other major purposes to the present study.
Since programed instruction provided the organization which low ability
or "remedial" students could not independently develop from texts and
lectures , the present research investigated the extent to which programed
instruction could actually assist a group of slow learners to achieve at a
high level. The present study considered as well whether programed in—
struction reduced conventional training time, and whether the Skinnerian
or the Crowderian program was more effective in such reduction.

Table 2 presents the instructional and learner variables used in the

Present study for making tests related to the interactions between ability

 

 

 

6
and method, and method main effects. The following section examines in
greater detail the theoretical and practical foundations on which the present

study was based.

TAB LE 2

INDEPENDENT AND DEPENDENT VARIABLES USED IN TESTING
HYPOTHESES OF INTERACTIONS BETWEEN ABILITY AND
METHOD VARIABLES, AND METHOD MAIN EFFECTS

 

 

Class of Variables
Variable in Class
Independent
Variable
Instructional (method) Lecture
Skinner
Crowder
Learner (ability) Intelligence (IQ)
Reading Comprehension (RC)
Week 1 grade (WK 1)
Dependent
Variable
Immediate test of Total posttest score (PTT tot)
retention Multiple choice posttest score (PTT mc)
Fill-in posttest score (PTT fi)
Delayed test of Total retest score (RT tot)
retention Multiple choice retest score (RT mc)
Fill—in retest score (RT fi)
Related subject Week 3 grade (WK 3)
matter Week 5 grade (WK 5)
Time to program Total time (T)

completion

 

.-- ..
"- .

Literature Review
Theoretical

There are at least two assumptions about the way people learn which
have various implications for programing. One such assumption regards
the learner as a receptive mechanism for whom associative connections
take form so as to mirror experience. Luchins (46) has called this concep—
tion the ”homo mechanicus" model of man. Another assumption is that the
learner is a selective self—organizing mechanism which selects and extracts
information from his environment. Luchins (46) referred to this conception
as a "homo sapiens" model of the learner.

Stolurow (66) indicated that one implication of the first assumption for
a method of programing is that the degree of freedom in response be reduced.
The program should consider the learner as a responsive system which has
to be channeled into the formation of "correct" responses and has to be
prevented from making wrong responses. An example of this viewpoint is
presented in Appendix B as the Skinnerian or linear approach.

A programer acting on the "homo sapiens" model of the learner would
allow the learner greater freedom of response through branching and the use
of multiple choice items containing a number of wrong response terms. The
Crowderian or non—linear approach to programing represented this view-
point; an example of it is presented in Appendix C.

A linear fOrm of programing would seem to be advantageous for those

of limited intellectual ability. In a study of concept formation, for example,

~-~

...- ...__

8

Detambel and Stolurow (19) found that at low ability levels , an ordered
rather than a random sequence of frames allowed for more students to learn
the concept. At high ability levels there were no significant differences.
These investigators concluded that the best sequence did for the poorest
ability group what the highest ability group could do for itself regardless
of sequence. This result suggests that the linear program might be effec—
tive for the low ability remedial student in the present study. Evidence
presented on page 10, however, shows that results conflict as to whether
high or low ability individuals benefit more from programed instruction.

Because the Skinner program provided small linear steps, while the
Crowder program used large instructional steps, the linear approach might
be more adequate for remedial students who have reading difficulties.
According to Skinner (65), the Crowderian approach left most of the
learning to reading rather than to responding. The linear or "small step"
format, on the other hand, placed more emphasis on responding and less
on reading. With respect to these differences in programing Hedberg (34)
asked, "How important are differences in reading level in the use of a
Crowder or Skinner program? " With this question in mind, the present in—
vestigator considered that there might be an interaction between reading
comprehension and the method of programing.

Trittipoe, Trittipoe, and Hahn (68) recently tested a similar proposi-
tion using fourth and sixth grade students and three different approaches

to designing linear programs. These investigators believed that prompting

required the most cueing and provided the most voluminous reading task;
conversational chaining had less cueing, and was, therefore, the most
complex reading task; confirmation provided the least cueing and the
easiest reading task. The findings indicated that reading was especially
important for the fourth grade group. This was supported by a reading—
achievement test correlation of . 95 in the conversational chaining treat—
ment. The correlation between reading ability and achievement was . 43
for sixth grade students in the conversational chaining group. The inves—
tigators concluded that in a program which was difficult to read, achieve—
ment would be closely related to reading ability.

The conceptual scheme considered in the present study included
learner and instructional variables in view of the possible interaction be-
tween them. Past research using such a framework has been limited (66),
although a few recent studies have considered interactions. Homme,
Willey, and McMahan (35) found interactions between intellectual ability
level and methods of instruction such as television, text, machine, and
lecture. The teaching machine was poorest for low ability students and
best for high ability. This result was found in programs of alternating cur-
rent but not those of direct current.

Roe (56) could not detect any aptitude—method interaction on an ele—
mentary probability criterion test or time to complete instruction. Lambert
(42) found no relation between ability and schedules of reinforcement, and

Silberman, Coulson, Gunn, and Melaragno (61) did not demonstrate an

10
interaction between form of program and achievement characteristics of
students. The latter researchers went so far as to say that there was no
sound basis for exploring relations between task, method, and student
variables as these determined the efficiency of student learning under
programed instruction.

Although research on the relationship between learner and instruc—
tional variables has been limited, some studies have considered the vari—
ables of intelligence and reading ability apart from their possible inter—
action with methods of programing. Wong and Douglass (71) found that
students of high intellectual ability proceeded through a program faster
than those of low ability. Faltz (22) and Sharf's (57) low ability students
made more errors during learning than those of high ability. Skinner (64),
Fry (29), and Shay (58) indicated that programed learning was applicable
to individuals within a wide intellectual range and that special programs
based upon ability were unnecessary. Keisler (38) suggested that pro—
gramed instruction was more appropriate for brighter students , whereas
Little (45), Freeman (27), and Porter (54) showed that programed instruc—
tion benefited low ability students more than those of high ability. High
ability students using programed instruction were superior to high ability
control students in several studies (3, 10). Silberman, Lal. (61) con—
cluded, "once an overachiever, always an overachiever. "

As to reading ability, Wong and Douglass (71) recognized the pos—

sible influence of this variable upon the outcome of research in programed

 

ll

instruction. They therefore eliminated subjects from their study who did
not comprehend the written word. Dite, Woodruff, and Faltz (20) also
recognized that reading ability was related to success in programed instruc—
tion. Using the Gates Reading Test these researchers found more correct
responses to a program for students of high reading ability than for those
of low reading ability. It remained, however, for these and other investi—
gators to show that these groups actually differed on a subsequent test of
achievement.

The evidence relating to intellectual and reading ability is inconclu—
sive. The sample of remedial students in the present study included both
low and high ability students. Therefore, in addition to an investigation
of interactions between ability and methods of instruction, some emphasis
was placed on the academic consequences of programed instruction for low

and high ability individuals.

Practical

The practical purposes of the present study grew out of the nature of
the military training environment. A previous study (28) done in the same
l9—week basic electronics school within which the present study was con—
ducted, indicated that the school had an overall drop rate of 14 percent.
About 6, 000 students went through the school in a year. When compared
to drop rates of under 5 percent for other technical schools, the present
investigator realized the magnitude of the problem. Since there was al—
ready a shortage of qualified men for technical jobs in the armed services,
theloss of almost 1, 000 electronics technicians a year further strained

our military defense capabilities.

 
 

 

12

Our national defense is dependent to a large extent on our technical
readiness. The military services , therefore, must turn out as many qual—
ified men as quickly as possible. Bryan and Rigney (6) have noted the
special problems facing the military, in this connection, such as the
short—term availability of manpower and the scarcity of high grade talent.
As already noted, some research evidence suggested that programed in—
struction might reduce the length of time required for training (24). Re-
sults were not conclusive, however. Silberman (60) observed that, of
seven studies he had seen, only three showed that linear programs took
longer than non—linear, while three studies showed no significant differ—
ences. Again, previous work also indicated that low ability students
might benefit from programed instruction (54) , that is , attain a higher level
of academic excellence than would be possible without programed instruc—
tion. Bryan and Rigney (6) have supported the use of teaching machines
for purposes of training low ability individuals and for reducing the length
of time required for training. Military managers, themselves, have viewed
the teaching machine as holding some promise for solving such training
problems (50).

A previous study by Froehlich (28) was directed primarily toward a
solution of the time—in-training problem to the neglect of the problem posed
by the lack of ability of many of the students. The present study, on the
other hand, evaluated the influence of programed instruction on the low

ability and "remedial" student, who, as Mowrey and Webb (52) have

13
shown, were inadequately prepared, poorly motivated, and unable to com—
prehend the principles of electronics.
Where there are differences in levels of student ability and motiva—
tion, some flexibility in training seems to be necessary. As Crawford

(15) noted

. . . Heterogeneity of input places particular burdens on a training
system which has a rigid output quota to meet on a fixed schedule.
Adaptation of training to varieties of aptitude and background
characteristics of trainees constitutes a major research problem.

Both Skinner (65) and Crowder (16) have stated that programed instruction

allowed an adaptation of training. The question raised in the present

study was whether such adaptation gave practical results, that is whether
remedial students would benefit more, academically, from programed in—
struction than from the conventional lecture.

Past research on this question has been inconclusive. Out of fifteen
studies comparing programed instruction with the conventional lecture,
six showed no differences; nine favored programed instruction on an
achievement dimension, while all fifteen showed that programed instruc—
tion on the basis of time to completion was superior (60).

Hughes and McNamara (36) were able to show that programed text—
books were superior to conventional classroom instruction. Wong and
Douglass (71) felt that one of the reasons many past studies showed sig-
nificant differences , especially between programed instruction and the

conventional lecture, was because conventional instruction was conducted

under unfavorable conditions and so almost had to show significant

g

 

 

\-'-‘-

.-_
‘v-

 

14
differences. Using twelve pairs of subjects and a branching program on
direct current electricity, these researchers supported their position by
demonstrating that there were no significant differences between an auto—
mated programed instruction group and a conventional lecture group given
optimal conditions for learning.

Programed instruction has several advantages over the conventional
lecture apart from those attributed to the implementation of learning theory.
Four of the ten offered by Briggs (4) were of especial interest in the present
context. Among these advantages were the fact that (a) an expert program
writer could reach a number of students, (b) misconceptions held by min—
imally qualified instructors would not be passed on to students, (c) slow
students would not be embarrassed by their lower rate in learning, and
(d) good instructors could use their time to better advantage than providing

rote drill for students.

CHAPTER II

HYPOTHESES

Theoretical and practical research directions were outlined in Chap—
ter I. Differences between Skinnerian and Crowderian methods of pro-
graming subject matter, as summarized in Table 1, indicated that achieve-
ment following either form of programed instruction might depend on the
intellectual and reading abilities of the learners.

Practical consequences of programed instruction centered around
bringing low achievers to a higher level of achievement than was possible
with the conventional lecture. The present investigator indicated that
little work had been done in relating learner and instructional variables,
and that studies comparing several forms of instruction on achievement
and time criteria had been inconclusive.

The present study, therefore, tested several hypotheses concerned
with the relation between learner and instructional variables. Table 2
outlines the variables referred to in these hypotheses. In testing these
hypotheses, an attempt was made to reduce some of the limitations in
previous research. These limitations related to the lack of a sufficient
number of subjects, the shortness of time of many of the experiments, and
the shortness of many of the programs used. It was because of such

shortcomings that Gilpin (33) and Firm and Perrin (25) felt that success in

I”!

16

the field of programed instruction had been limited.

The hypotheses, in null form, are:

la. On both immediate and delayed tests of retention there will be no
interactions between ability and instructional method.

lb. There will be no immediate or delayed posttest differences for
remedial students who have been subjected to either the programed pre—
sentations or the conventional lecture.

2a. Within programed instruction treatments, there will be no imme-
diate posttest differences for persons of varying intelligence.

2b. Within programed instruction treatments, there will be no imme-
diate posttest differences for persons of varying reading comprehension
levels.

3a. With time to program completion as the criterion, there will be
no ability and method interaction within programed presentations.

3b. With time to program completion as the criterion, there will be
no differences between the Skinner and the Crowder presentations.

4a. There will be no interaction between ability and method with
achievement on related subject matter as the criterion.

4b. The academic achievement of students who learned under pro-
gramed instruction will not differ from students who learned under the
conventional lecture on related subject matter.

5. There will be no interaction between the form of response required
by the program and the form of response required by the criterion test.

6. There will be no relationship between time to complete the pro-

gramed instruction and academic achievement.

CHAPTER III

SETTING
Subjects

The subjects of this study were 494 naval trainees undergoing reme—
dial instruction. These students had failed an examination given at the
end of the second week of the l9—week basic electronics course. They
were required to repeat the second week of instruction at night while con—
tinuing the third week of instruction during the day. The remedial class
met for two hours each of three nights.

In terms of a combination of verbal intelligence (GCT) and arithmetical
reasoning (ARI) the remedial students had a mean of 118. 38 with a stand-
ard deviation of 7. 53. This compared to a mean of 122. 71 and a standard
deviation of 9. 61 for a sample of 500 basic electronics students taken
from day school classes. Since this latter sample included some of the
remedial students in the present study, an appropriate statistical test of
the difference between means could not be made. When the means were
considered as independent samples, a "t“ of 7. 59 was significant be—

yond the . 01 level.

Variables
Intelligence
Intelligence in this study was based on a combination of a verbal
(GCT) and an arithmetical reasoning (ARI) test. The GCT consisted of

17

\.

...- -.._

18
sentence completion items, opposites, and verbal analogies. The ARI

contained both arithmetical reasoning and computational items.

Criterion Tests

In order to evaluate the influence of the instructional procedures on
achievement, two 30-item criterion tests were constructed. Each test
was used alternately as either pretest or posttest. Appendix F reproduced
the final two forms of these criterion tests. Each form contained fifteen
construction or fill—in items and fifteen multiple choice items. Item anal—
yses were conducted on experimental forms of the criterion tests and con—
sisted of administering one form (81) to 323 regular basic electronics stu—
dents and the other form (82) to the next 345 students. Item statistics
were developed on the thirty-six items contained in each experimental
form.

The item analysis was based on one suggested by Mosier and
McQuitty (51) for getting difficulty and discrimination indices, and was
the method used by the testing section of the basic electronics school.
The analysis assumed that knowledge of the item and total score on the
test were normally distributed. Data were collected on Pu and P1, the
proportion from the upper and lower quarters, respectively, of the total
test score distribution. An abac was used which allowed the reading of
the difficulty index of the item and the item—test correlation. The item
test correlation was based on Chesire, Saffir, and Thurstone's (12) com—

puting diagrams for the tetrachoric correlation coefficient, and used a

19
graph for which the division of the test score was at the quartile. This
gave the proportion of the upper or lower quarter who would pass an item
of any given difficulty with any particular r. These diagrams provided the
proportions for Pu and P1 for various values of r.

The investigator selected items for the final form of the criterion tests
based on the difficulty and discrimination indices, The difficulty index
ranged from O to 9 indicating the proportion, low to high, that passed the
item. The discrimination index had the same range, although an item with
a low index indicated poor discrimination. Table 3 presents the data on
the items selected.

Data for Kuder—Richardson reliabilities (KR #20) were collected from
the remedial students and involved an analysis of the two criterion test
forms when used as pretests. Night classes were used for each form.
Table 4 summarizes the data for forms 81 and 82. The means of the two
forms are also presented and indicate that 81 was a more difficult test
than 82. These means were basically a percent transformation of the raw

scores.

. Davis Reading Test
The Davis Reading Test (18) was chosen because of favorable critical
comments in Euros (7) related to high reliability, high correlations with
other reading scores and English grades, and ease of administration. Two
scores , reading comprehension and reading speed, were obtained within an

hour, without separate timing of the parts. These scores were converted

20
TABLE 3
ITEM CONTENT, DISCRIMINATION, AND DIFFICULTY INDICES

FOR THE FINAL FORM OF THE CRITERION TESTS
BASED ON TWO SAMPLES OF REGULAR STUDENTS

 

 

 

 

 

Form Sl Form S2
Fill-in Fill-in
Item Content.- Discrimination Difficulty Discrimination Difficulty
1 Ohm's Law 4 7 2 9
2 Ohm's Law 3 8 4 8
3 Powers of Ten 5 6 4 5
4 Powers of Ten 6 6 5 6
5 Powers of Ten 6 6 5 6
6 Series Circuits 6 6 6 5
7 Series Circuits 7 5 4 8
8 Series Circuits 5 6 6 4
9 Series Circuits 6 4 7 5
10 Series Circuits 4 7 4 8
11 Parallel Circuits 5 7 5 5
12 Parallel Circuits 6 6 5 6
13 Parallel Circuits 4 5 6 6
14 Parallel Circuits 6 6 4 4
15 Parallel Circuits 5 7 6 6
Subscore mean 5.20 6. 13 4. 87 6.07
Multiple choice Multiple choice
16 Ohm's Law 5 7 4 7
l7 Ohm's Law 5 7 4 8
18 Powers of Ten 4 8 5 7
19 Powers of Ten 6 4 5 4
20 Powers of Ten 4 7 6 7
21 Series Circuits 4 8 5 7
22 Series Circuits 3 8 4 7
23 Series Circuits 3 8 4 7
24 Series Circuits 5 6 5 7
25 Parallel Circuits 5 6 5 6
26 Parallel Circuits 3 5 6 6
27 Parallel Circuits 5 6 5 7
28 Parallel Circuits 4 7 6 7
29 Parallel Circuits 5 6 6 7
30 Parallel Circuits 6 5 5 6
Subscore mean 4 46 6 53 5.00 6. 67
Total score mean 4.83 6.33 4.93 6.37

 

21
TAB LE 4

RELIABILITIES OF FORMS 81 AND 82 WHEN USED AS PRETESTS

 

 

 

Pretest Form N Reliability1 Mean2
S1
Fill—in 180 . 69 37. 02
Multiple choice 180 . 46 51. 74
Total 180 .73 45.48
82
Fill—in 184 . 59 46. 88
Multiple choice 184 . 61 56. 23
Total 184 .74 51.64

 

lReliability based on Kuder—Richardson formula #20.

2
Means based on the percent of total items correct.

by the smoothed equi—percentile method described by Flanagan (26) and
allowed their direct comparison. The average parallel form reliabilities
for 1,096 grade 12 students as given in the manual were . 77 for reading
comprehension and . 85 for reading speed. Form A was used throughout
this study.

Means and standard deviations were computed for 500 regular basic
electronics students and were similar to those reported in the manual. The
reading comprehension (RC) mean and standard deviation were 73. 26 and
6. 37 respectively; for the reading speed score (RS) these statistics were,
respectively, 74. 00 and 6. 33. Table 5 shows the intercorrelations among
the GCT, ARI, IQ, RC, and RS variables. The scores on the M were
more closely related to GCT than ARI and they were themselves fairly well

related. Other aspects of Table 5 were examined in Chapter 4. '

22
TABLE 5
INTERCORRELATIONS, MEANS, AND STANDARD DEVIATIONS

FOR INTELLECTUAL, READING, AND ACADEMIC VARIABLES
BASED ON A SAMPLE OF 500 REGULAR STUDENTS1

 

 

ARI IQ WK 3 WK 5 RC RS Mean SD
GCT 21 84 34 31 59 63 61.87 5.80
ARI 83 39 38 32 26 60.85 5.71
IO 44 41 54 53 122.71 9.61
WK 3 74 28 24 78.19 10.75
WK5 23 18 74.20 13.12
RC 69 73.26 6.37
RS 74.00 6.33

 

lDecimal points omitted.

Weekly Examinations

Table 6 presents the reliabilities of tests given at the end of the first,
third, and fifth weeks of the 19-week basic electronics training program.
Extimates of the Kuder-Richardson reliabilities (KR #20) were made using
400 regular basic electronics students and a standard deviation based, in
each instance, on about the first 100 students. Data for estimating these
reliabilities were furnished by the testing section of the 19—week basic
electronics school. The week one (WK 1) test contained twenty-five items
on elementary mathematical manipulations and introductory electronics
principles. The week 3 (WK 3) test of fifty items covered direct current
electricity. The week 5 (WK 5) test also had fifty items and dealt with
alternating current electricity. The tests were designed for use in regular
classes and had difficulty indices of 7. 4 and discrimination indices of

3. 4, as defined above.

23
TABLE 6
ESTIMATES OF THE KUDER-RICHARDSON RELIABILITIES

FOR THREE WEEKLY TESTS ADMINISTERED DURING
THE 19-WEEK BASIC ELECTRONICS COURSE1

 

 

Week
1 3 5
.52 .69 (.60) .79 (.66)

 

1Figures in parentheses are reliabilities as esti—
mated for the remedial school sample (67).
Time (T)

The time required to complete the programed instruction was recorded
in minutes by the instructor as the student handed in his answer sheet.
The highest time score possible in a two=~hour evening was 120 minutes.
About 5 percent of the students did not finish the booklet on any one night.
Therefore, minutes were added to the time score in direct proportion to the
time it took a student to complete the number of frames already completed.
Since students who did not finish the program nevertheless completed bet»
ter than 90 percent of the frames, the number of minutes added was quite
small. This procedure provided an estimate of the time it would have

taken a student to finish had he been given more time.

Development of the Instructional Presentations

The linear program used in this study (Appendix B) was a revision of
a navy program originally designed for use in a teaching machine (35).

The Crowder program (Appendix C) was adapted from an electronics program

#1.?th 4

 

24
used in the Air Force by Crowder and Zachert (17). Both programs covered
the same electronics subject matter as was covered in the 19—week navy
basic electronics school including Direct Current and Alternating Current
electricity. From this material Ohm's Law, Powers of Ten, Series Cir—
cuits, and Parallel Circuits were selected for night school adaptation.
These were the concepts presented during the second week of instruction
which students in the past had found most difficult to learn according to
educational specialists of the basic elecrronics school.

The main ways of insuring that these programs and the conventional
lecture were parallel in Ligpih and extent 9;" coverage were by (a) a master
schedule which gave an hourly outline of the curriculum, (b) student work
books which clearly defined the subject matter under investigation, (0)
manuals of instruction which outlined in detail the subject matter covered
in class, and (d) familiarity with the existing electronics curriculum pos-
sessed by educational specialists.

The completed programs are presented in the appendix. The linear
program has 348 frames included in ninty—six pages. The non—linear pro—
gram contains a total of 167 ”correct" frames. Two to five choices are
provided for each instructional frame bringing the total number of pages to
423. Answer sheets designed for use with these programs are reproduced
in Appendix D and Appendix E.

These programs were given a trial run on men who were waiting to
enter the basic electronics school. No additional work was done on the

programs since these men experienced no difficulty in going through them.

25

In attempting to see how "good” the frames were in practice - how
many errors were made per frame - the incorrect answers, as taken from
the answer sheets of remedial students were tallied for each frame. For
each of the three booklets, Skinner and Crowder versions, the average
number of errors per frame was estimated and divided by the number of
students in each treatment. When this latter result was multiplied by 100
an estimate of the average error rate for the average student was obtained.
This error rate, in percent, for nights one, two, and three was 3. 74, 3. 49,
and 4. 54, respectively, for the linear program: it was 7.99, 10. 80, and
12. 29, respectively, for the scrambled booklets. Apparently, then, the
Crowder programs were somewhat more difficult from the first to the last
booklet.

The differences in difficulty between Skinner and Crowder programs ,
for the most part, do not reflect differences in the number of pages or the
number of frames contained in each program, but differences in consequence
of step size. One of the criteria for designing linear programs was low

frame error rates (5 6).

Design
The basic treatment groups were the lecture, Skinner, and Crowder
presentations. These treatment groups were randomly assigned to incoming
remedial classes using a table of random numbers though there were some
restrictions to the assignment process. An equal number of classes was

assigned to each treatment group and each treatment group was exposed

26
an equal number of times to the criterion tests , which were used alternately
as pretest and posttest. One form of the test was designated as 81 and
the other as S2. A total of twenty—four classes was used.

Table A in Appendix A indicates the schedule of data collection. (a)
Scores on the GCT and ARI were collected while the students were in basic
training or "boot" camp. (b) The Davis Reading Test was administered
three days before students entered the basic electronics day school. (c)
The criterion test which was used as the pretest was given at the first
meeting of night school just before instruction began. (d) The criterion
test used as posttest was administered immediately after the third night
of instruction. (e) Two weeks after remedial instruction all subjects were
retested with the form of the criterion test which had been used as the
posttest.

A preliminary examination of the total criterion test means of the first
six classes indicated that these tests were not parallel. A difference be-
tween total score means was significant at the .05 level. The investigator
therefore decided to consider the 81—82 order of presentation as one study
and the SZ—Sl order as a second study. Table 7 summarizes the data on

class assignment with respect to the various forms of instruction.

27

TAB LE 7

BREAKDOWN OF SUBJECTS BY CLASS, TREATMENT, AND
FORM OF CRITERION TEST ADMINISTERED FIRST (81 OR 82)

 

 

 

 

fl
Lecture Skinner Crowder
Class N Class N Class N
1 23 3 l9 5 7
10 18 12 23 7 13
22 26 18 25 14 22
24 26 20 22 16 22
Total 93 89 64
S2
Lecture Skinner Crowder
Class N Class N Class N
4 10 6 l4 2 15
8 20 9 13 11 3O
13 26 15 28 19 27
17 20 21 27 23 18
Total 76 82 9O

 

CHAPTER IV

ANALYSIS OF THE DATA

This study, as indicated earlier, involved the analysis of two sets of
data. Set I was concerned with the analysis of data on students who took
the criterion tests in the order of Sl—SZ. Set 2 considered those students
who took the 82 criterion test as the pretest and the SI criterion test as the
posttest. The analysis described in this chapter applied to both sets of

data.

Regres sion Analysis

The basic analyses centered around regression and correlational anal—
yses. Hypotheses one to five might have been considered in an analysis
of variance or analysis of covariance framework (70). That is, we might

have considered a 3 x 3 fixed treatment design in which the lecture, Skinner,

 

and Crowder presentations were one factor, and "high,' medium, " and "low"

I the levels on the second factor. This second factor might have been any

i one of the independent variables associated with the five hypotheses , e. g. ,

l intellectual ability, reading ability, academic ability — either pretest score
or Week 1 score. The dependent variables might have included posttest,
retention, Week 3, Week 5, and Time scores. Any one, or some combina-

tion of independent variables might have served as covariates.

By means of computers, such as the Data Systems 3600 Computer at

28

29
Michigan State University, it was possible to consider an approach which
was not before practical. This approach, basically one of regression, was
more sensitive than the analysis of variance just described in that it used
more of the information contained in the data. The approach was also able
to handle the problem of unequal sample sizes which in the past has been
handled less satisfactorily through the unweighted means analysis sug—
gested by Bendig (2) and Winer (70).

The first problem encountered in handling the data within the regres—
sion framework was to decide what independent variables to include. In
the analysis of variance or covariance single variables might have been
investigated for their effects on the dependent variable, or any number of
variables might have been considered in a factorial design for the purpose
of analyzing the interactions produced by the single variables. Winer (70)
has described the fixed linear model underlying a 3 x 3 factorial design:

Yijk = 14.. +71 + 51‘ +‘f61j + Eijk
where, Yijk= an observation on element under treatment combination ab.
+1.. = the overall mean.
Ti= factor a.
Bj = factor b.
761]: interaction, ab.
eijk= error.

In the regression analysis followed in the present study another model

was used which made it desirable to include more than one independent

variable, but only as many variables as would not result in a singular mat-

rix. To include orthogonal independent variables in the model would, on

 

 

30

the one hand, have provided a more sensitive test of a null hypothesis by
creating a smaller error term; though on the other hand, it would not have
been possible to calculate a regression if the independent variables ex-
plained much of the same variation. A description of the regression pro—
gram provided by the Michigan State University Computer Center (Appendix
C of ABS Program Description 4) indicated that under conditions when an
independent variable was linearly dependent on other independent variables
of the same model, possible solutions to the regression were infinite. It
would then be necessary to drop at least one of the independent variables
from the regression or to combine them into a smaller number of variables.

It was necessary to decide which independent variables would be in—
cluded in the model. To include intellectual ability, reading comprehen—
sion, pretest, and Week 1 grade (which was the measure of prior academic
ability) among the independent variables treated here , would not only have
been impractical in terms of handling the statistical decisions required in
this procedure, but would also have included variables with much common
variance. It was therefore decided to select the two best variables from
the above group; "best" in terms of predicting the dependent variable, pro—
viding the least variance in common with the other independent variable,
and having practical significance for training.

Several two variable models were used. One included IQ (GCT + ARI)

and the Week 1 grade (WK 1). Another model included reading comprehen—

sion (RC) and the pretest (PRT). The first model met the criteria for inclusion

31
best. The second model was considered because of the hypothesis related
to reading comprehension. However, as can be seen from Table 5, RC
added little predictive variance over IQ when considering the prediction of
WK 3 and WK 5. The second model was less practical than the first since
PRT and RC scores were not usually available in the training situation.

The above discussion suggested that IQ and RC should not be included
in the same model. When these variables were included, in a trial run, a
singular matrix resulted and the computer provided no meaningful output.

The model and the strategy described below held for any two independ-
ent variables which were included in it. Thus, reading comprehension
(RC), pretest (PRT) or WK 1 scores were substituted as independent vari—
ables in the model. The dependent variable, where appropriate, was the
total, multiple choice, or fill-in scores of the PTT and retention test (RT)
(hypotheses 1, 2, and 5); the dependent variable was also time (hypothesis
3), WK 3 or WK 5 grade (hypothesis 4).

The general model and the decision strategy employed is outlined be-
low with IQ and WK 1 as the independent variables and total PTT as the
dependent variable. Table 8 lists the variables used in this and other re—
gression analyses. The fixed linear model was written as follows:

Yij =41 + Bixmij +V1X11U + Eu

where , i = method.
j = subject.
Y = dependent variable.

X101]: WK 1 score for the jth subject in the ith treatment.

‘hkA .

 

32
X111]. = IQ score for the jth subject in the ith treatment.
<1 = effect due to treatment.
B = effect due to WK 1.
~r= effect due to IQ.
e = error
The hypotheses tested with this model were
H1271 = 72 = 73
H2381 = 82 = 83
Hstdi = c12 = C‘3
In other words, H1 stated that with respect to IQ the slopes of the three
treatment groups were equal (IQ plotted against posttest) , or that there was
no interaction between IQ and method. H2 indicated that there was no
interaction between WK 1 and method (WK 1 plotted against posttest) , and
H3 that there was no main effect due to the methods of instruction. The
main effect for treatments represented a three plane plot (one for each
treatment) of IQ and WK 1 against posttest. Where the first two hypotheses
considered the slopes of the regression lines, H3 considered their inter-
cepts.
Table 8 lists the lecture, Skinner, and Crowder presentations as vari-
ables. These were "dummy" variables, that is , they were included in the
regression as either 1 or O, in other words, included in a calculation or

not included. With this in mind, the above model was written
Yij = <11 (1,0) +42 (1,0) +03 (1.0)
+ 61(1,O)X101j+82(1,O)X101j+ 83 (l/O)X10ij
+ Y1 (1,0)X11 +‘Y2 (1.,O)X111j+‘(3(l,O)X11U+ 61]-

 

33
to indicate the existence in the model of the interactions between IQ and

method, and WK 1 and method.

TAB LE 8

VARIABLES USED IN THE REGRESSION ANALYSIS

 

Variable Variable
Number Name

 

1 Multiple choice posttest
2 Fill—in posttest

3 Total posttest

4 Multiple choice retest
5 Fill—in retest

6 Total retest

7 Lecture1

8 Skinner

9 Crowder1

10 Week 1 score

11 Intelligence

12 Reading comprehension
13 Reading speed

14 Multiple choice pretest
15 Fill-in pretest

16 Total pretest

17 Week 3 Score

18 Week 5 score

19 Time

 

lVariables 7, 8, and 9 were coded 100,
010, and 001, respectively, for purposes of
analysis.

It can be seen that since interactions were built into the model the
assumption of additivity could be directly tested. The assumption of lin—
earity of regression was tested by plotting the scores for the independent
and dependent variables of each model used in the study, taken two at a

time, against each other. The resulting plots were examined for their lack

-..—..‘

 

34

of apparent linearity. There was no indication as to a lack of linearity.

Some caution was used in the interpretation of results, especially for
just significant F—ratios. The above model assumed that the factors, in
addition to the treatments , were fixed by the experimenter prior to the con—
duct of the study. This, of course, could not be accomplished where the
factors were test scores, since the test scores were not under the control
of the investigator. In addition, no test for homogeneity of variance was
made as this would have proven difficult to do; too many tests were al—
ready being made on the same data, and there was little that could have
been done if the variance were heterogeneous. The investigator assumed
that this condition was satisfied through the random assignment of treat—
ments to subjects. 1 There was no physical evidence, such as suggested
in Lindquist (44), that a bias in assignment had been involved.

To the variables listed in Table 8 were added six new or "transformed"
variables reflecting the interaction between IQ and method and the inter—

action between WK 1 and method. In computer terms

X(21) = x(7) * x(10)
X(Z 2) = X(8) * X(10)
X(2 3) = X(9) * X(10)

x(24) = x(7) * X(11)

X(25) = X(8) * X(11)
X(26 6=) X(9) * X(11)

Where, for example, variable (21) was the product of variable (7) and variable

 

1The analyses related to interactions, main effects, and tests of as—
sumptions were suggested by Dr. Martin Fox of the Michigan State Univer—
Sity Statistics Department. Mr. William Rubel, Dr. Walter Stellwagen,
and Dr. Einar Hardin were also consulted.

 

35
(10). Variables (21) to (23) reflected the WK 1 and method interaction,
and variables (24) to (26) the IQ and method interaction.
In computer terminology, the following regression equations , and
those similar to these, were developed to test H1, H2, and H3 which re-

lated to hypotheses l, 3, 4, and 5 of this study:

[1] x(3) = Q(X(7). . .x(9), X(21). . .X(26))* [n]
[2] x(3) = Q(X(7)...X(9), X(11), X(21)...X(23))* [ml]
[3] x(3) = Q(X(7). . .x(10), x(24). . .X(26))* [(62]
[4] x(3) = Q(X(7). . .x(11))* [w3l
[5] x(3) = P(X(10), X(11))* [04]

where the variable to the left of the equation was the dependent variable,
in this example PTT tot, and those to the right the independent variables. 1
The Q in the equation indicated that the regression was calculated with—
out a constant term, while the P in equation [5] indicated that a constant
was calculated.

AES Program Description 4 indicated the test for the effects of two or
mOre variables in a regression. For this test two regressions were calcu—
lated, the overall regression and a regression that contained the same vari—
ables, except that the independent variables were omitted. This was, in
effect, what was done in equations [1] to [5] . Thus, to test for the in—
teraction between IQ and method equations [1] and [2] were used. Equa—
tion [2] did not contain the terms related to the interaction between IQ
and method, i.e. , variables (24) to (26). The first equation provided an

error sum of squares (SSn), as did the second (SSW). In other words, SSn

 

The dots in the equation represent "inclusive. "

 

36
indicated how well the full model fit the data, and SSUl indicated how well
the hypothesis fit the data. The F statistic took the form

(sswl — SSn)/df
SSn/df

F =
and indicated how much better the model was than the hypothesis.

If the first null hypothesis, H1, was accepted, equation [2] was
tested against equation [4]; that is, one behaved as if there was no inter—
action between IQ and method in testing for the interaction between WK 1
and method. If there was a significant interaction between IQ and method
the terms related to this interaction were retained. The interaction between
WK 1 and method, in the latter instance, was tested by using equations
[1] and [3].

If neither interaction was significant a test was made for treatment
main effects. Equations [4] and [5] provided the appropriate error terms.
With a significant interaction between WK 1 and method a test for treatment
main effects would not be meaningful as the difference in treatment effects
WOuld obviously be dependent on the WK 1 grade of the student. In addi—
tion, the test for main effects assumed homogeneity of regression, which,
With significant interactions, would not be supported.

The above strategy was used for two as well as three treatment groups.
HYpOthesis 2 examined the IQ and RC main effects across programed in—
Struction treatments. Where no interactions existed these main effects

Were tested by equations of the form

x(3) = P(X(10),X(ll))*
x(3) = P(X(10))*

 

37
for the IQ main effects, and

x(3) = P(x(12),x(16))*
x(3) = P(X(16))*

for the RC main effects. Total PTT was the dependent variable in the above
examples. Tests for interaction effects were made again, especially where
they were found in the larger model. Where interactions existed, the test

for IQ main effects was made with an equation of the form

C

The variables included in the above equations are identified in Table 8.

In summary, the first five hypotheses were tested using a linear regres-
sion model. Since this model had the interaction between ability and
method built in, a test of hypotheses 1a, 3a, and 4a was possible. Since
the same model also provided for the testing of main effects, hypotheses
1b, 2a, 2b, 3b, and 4b were also testable.

Hypothesis 5 suggested that students using the construction or the
multiple choice programs would do as well on the fill—in part of the post—
test as on the multiple choice part of the posttest. If the null hypothesis
did not hold one might expect the Skinner subjects to be superior to the
Crowder subjects on the fill—in criterion. The Skinner subjects could con-—
ceivably be as good as the Crowder subjects on the multiple choice post—
test_ These possibilities suggested an analysis of treatment main effects
With both the fill-in multiple choice parts of the posttest as the dependent

Variable in a regression analysis. With parallel fill—in and multiple choice

 

 

38
parts of the criterion test a factorial design would have been feasible,
with instructional methods as one factor and mode of response as the

other factor.

Correlational Analysis

 

Hypothesis 6 sought the relationship between time and achievement
within each treatment group. An International Business Machines 1620
was used to obtain intercorrelational matrices for each treatment group as
well as the means and standard deviations for the variables used in the
present study. Except for variables 7, 8, and 9 , Table 8 lists these vari-
ables. All students had scores on these sixteen variables except the con-

ventional lecture group which had no time scores.

 

 

CHAPTER V

RESULTS

There are at least two reasons for the inconclusive results found in
past studies of programed instruction. One concerns the inadequacy of
measurement and design, the other concerns the effectiveness with which
the principles of learning were incorporated within the structure of the
program. The latter includes the question of whether investigators could
duplicate these principles in the classroom with the same meaning as in
the laboratory.

Tables 9 to 13 present data related to measurement and design char—
GCteristics of the present study. The principles of learning as incorporated
in programed sequences are discussed in Chapter VI. Table 9 confirms
the unreliability of the criterion tests noted in Chapter III; Tables 10 and
11 indicate significant differences in the WK 1 and WK 2 means for the
Lecture, Skinner, and Crowder Treatments. Tables 12 and 13 provide only
a suggestion of possible differences in post—treatment means for the three
treatment groups, analyses of variance indicating differences between
Time (T) means and Retest (RT) means. Tables B to E in Appendix A show
the actual means of the pre— and post—treatment variables. These pre—
liminary results provide the context within which the main results were

interpreted .

39

 

40

Table 9 was taken from the intercorrelation matrices of Tables H, I,
and J of Appendix A, and provides correlational data on the criterion tests.
Since the posttest and retest scores were based on the same form of the
criterion test, the (PTT) (RT) correlation might serve as a form of test—
retest reliability. The total, multiple choice, and fill-in re-test correla—
tions were somewhat lower than the reliabilities shown in Table 4. The
correlations between pretest and posttest were also of the same magnitude,
except for the very low (PRT mc) (PTT mc) correlation in set 1. The cor—
relations in Table 9, thus, give additional evidence for low reliabilities.
Low reliabilities might be one reason why the components of the pretest
(mc and fi) did not correlate well with each other, even though they meas—

ured the same content.

TAB LE 9

CORRELATIONS WITHIN EACH TREATMENT GROUP BETWEEN
PART AND TOTAL SCORES OF THE CRITERION TESTS

 

 

Lecture Skinner Crowder

Set 1 Set 2 . Set 1 Set 2 Set 1 Set 2

(PRT mc )(PRT fi) .44 .55 .39 .48 .44 .56
(PRT)(PTT) .55 .75 .49 .61 .71 .69
(PRT f1)(PTT £1) .38 .48 .47 .43 .62 .54
(PRT mc)(PTT mc) .41 .55 .13 .48 .46 .57
(PTT mc)(PTT n) .60 .64 .51 .54 .55 .71
(PTT)(RT) .58 .72 .50 .65 .62 .68
(PTT f1)(RT fi) .47 .59 .45 .47 .59 .55
(PTT mc)(RT mc) .47 .66 .38 .56 .47 .59

 

 

41

Tables 10 and 11 indicate that there were significant differences be—
tween the Lecture, Skinner, and Crowder means for WK 1 and WK 2 of set
1, and for WK 1 of set 2. Tables 12 and 13 show significant differences
for RT fi and T of set 1, and for RT tot, RT mc, RT fi, WK 3, and T of set
2. An F max (69) of 2. 37 for the RT fi variances of set 1 was significant
at the . 02 level; other variances were homogeneous. The two differences
in pre-treatment means were probably due to the method of assigning
classes to treatments; the remaining means, however, were equal.

TABLE 1 0

SUMMARY TABLE FOR DIFFERENCES IN MEANS
OF PRE—TREATMENT VARIABLES FOR SUBJECTS IN THE
LECTURE, SKINNER, AND CROWDER TREATMENTS (SET 1)

 

 

Variable Source of Variation Df SS MS F

IQ Treatment 2 217.98 108.99 2.52
Within 243 10511.02 43. 25
Total 245

RS Treatment 2 128.92 64.46 2.26
Within 243 6945. 26 28. 58
Total 245

RC Treatment 2 133.05 66.58 2.06
Within 243 7845.96 32. 29
Total 245

WK 1 Treatment 2 1104.38 552.19 3.06*
Within 243 43904.05 180. 68
Total 245

WKZ Treatment 2 665.76 332.88 4.63**
Within 243 17480. 57 71. 94
Total 245

PRT tot Treatment 2 411.12 205.56 1.00
Within 243 49947. 21 205. 54

Total 245

 

 

 

L

42

TABLE 10——Continued

 

 

 

Variable Source of Variation Df SS MS F
PRT mc Treatment 2 562.84 286.42 1.19
Within 243 58627.91 241.27
Total 245
PRT fi Treatment 2 402.80 201.40 .60
Within 243 82222.99 338.37
Total 245
* **
p<.05. p<.025.
TABLE 11

SUMMARY TABLE FOR DIFFERENCES IN MEANS
OF PRE-TREATMENT VARIABLES FOR SUBJECTS IN THE
LECTURE, SKINNER, AND CROWDER TREATMENTS (SET 2)

 

 

 

Variable Source of Variation Df SS MS F

IQ Treatment 2 89. 22 44. 61 . 64
Within 245 17001.28 69. 39
Total 247

RS Treatment 2 81.83 40.91 1.57
Within 245 6394.01 26.10
Total 247

RC Treatment 2 125. 49 62. 75 2. 42
Within 245 6363. 98 25. 98
Total 247

WK 1 Treatment 2 1688.45 844.22 4.89***
Within 245 42295. 64 172. 64
Total 247

WK 2 Treatment 2 90.91 45.45 .54
Within 245 20517. 93 83. 75
Total 247

PRT tot Treatment 2 266.65 133.33 .54
Within 245 61005.03 249. 00
Total 247

PRT mc Treatment 2 53. 84 26. 92 . 09
Within 245 75965.48 310. 06
Total 247

PRT fi Treatment 2 1170.92 585.46 1.74
Within 245 82406. 55 336. 35
Total 247

***p< . 01

 

43

TABLE 12

SUMMARY TABLE FOR DIFFERENCES IN MEANS
OF POST-TREATMENT VARIABLES FOR SUBJECTS IN THE
LECTURE, SKINNER, AND CROWDER TREATMENTS (SET 1)

 

 

 

Variable Source of Variation DF SS MS F

PTT tot Treatment 2 143.90 71.95 . 33
Within 243 52447.93 215.84
Total 245

PTT mc Treatment 2 180. 77 95. 39 . 33
Within 243 71283.17 293. 35
Total

PTT fi Treatment 2 306.86 153.43 .55
Within 243 68296.45 281. 06
Total 245

RT tot Treatment 2 403.55 201.78 1.07
Within 243 45965.80 189.16
Total 245

RT mc Treatment 2 54.56 27.28 .10
Within 243 64415.43 265.08
Total 245

RTfi Treatment 2 2293.92 1146.96 4.62**
Within 243 60375. 34 248.46
Total 245

WK 3 Treatment 2 543.21 271.60 2.99
Within 243 22065.69 90.81
Total 245

WK5 Treatment 2 572.74 286.37 2.46
Within 243 28324.47 116.56
Total 245

T Treatment 1 44241.20 44241.20 21.14****
Within 151 315965. 10 2092.48
Total 152

**
p< . 025.
****

p<.005.

 

44

TAB LE 13

SUMMARY TABLE FOR DIFFERENCES IN MEANS
OF POST—TREATMENT VARIABLES FOR SUBJECTS IN THE
LECTURE, SKINNER, AND CROWDER TREATMENTS (SET 2)

 

 

 

Variable Source of Variation Df SS MS F
PTT tot Treatment 2 1214.56 607.28 2.43
Within 245 61340 250. 37
Total 247
PTT mc Treatment 2 1261.51 630.75 2.05
Within 245 75326.72 307. 46
Total 247
PTT fi Treatment 2 1389. 25 694. 63 2.18
Within 245 77902. 23 317.97
Total 247
RT tot Treatment 2 1667.77 833.88 3.51*
Within 245 58263.57 237.81
Total 247
RT mc Treatment 2 2867.69 1433.85 5.02***
Within 245 69929. 80 285.43
Total 247
RT fi Treatment 2 3637.43 1818.72 5.94****
Within 245 74974.25 306.02
Total 247
WK 3 Treatment 2 656.09 328. 05 3. 23*
Within 245 24913. 90 101. 69
Total 247
WK5 Treatment 2 45.15 22.57 .17
Within 245 32068. 71 130. 89
Total 247
T Treatment 1 21638.00 21638.00 8.17***
Within 170 450225. 30 2648. 38
Total 171
*
p< . 05.
***
p< . 01.
****

p<.005.

 

45
Main Analyses
The first five hypotheses were tested by the regression analysis out-
lined in Chapter IV. In essence, that analysis was an analysis of covar—

iance, and used an F statistic of the form

r _ (8sz — SSn)/df
‘ SSA/df

As was indicated in Table 2, hypotheses of interaction and main effects
were tested against several dependent variables (PTT, RT, WK 3, WK 5 , T)
within two main regression models. One model included IQ and WK 1 as
independent variables, the other used PRT and RC. Tables F and G in Ap—
pendix A indicate the appropriate sum of squares for the F tests used.
Appendix G presents the raw data available on all individuals included in

the study.

Hypothesis 1a

On both immediate and delayed tests of retention there will be no
interactions between ability and instructional method.

Table 14 shows that the only interaction between ability and instruc—
tiOnal method was that between WK 1 and method (PTT tot and PTT mc as
dependent variables) for set 2. In other words, the advantage attributed
to one method or the other depended on the subject's WK 1 score.

Figure 1 represents the relationship between WK 1 and instructional
method. It shows a plot of the regression coefficients of variables in—
Cluded in the regression equation which was used to test for the interaction

between WK 1 and method. For the PTT tot dependent variable, this equation,

 

46

TABLE 14

TESTS OF SIGNIFICANCE FOR INTERACTIONS BETWEEN
ABILITY AND METHOD WITH POSTTEST AND
RETEST DEPENDENT VARIABLES (HYPOTHESIS 1a)

 

 

 

 

Dependent IQ-Method1 WK 1-Method2 PRT—Methodl RC—Method2
Variable Interaction Interaction Interaction Interaction
Set 1 Set 2 Set 1 Set 2 Set 1 Set 2 Set 1 Set 2
PTTtot 1.93 .98 .58 3.81** 1.05a .93a 2.18 .25
PTT mc 2.33 1.44 .51 3.60* 2.622 2.34b 1.23 .19
PTT £1 .72 .57 .28 2.64 1.05C .61C .94 .21
RTtOt 1.26 1.94 1.55 1.47
RTmC 1.87 1.75 .46 1.44
RTfi .33 1.24 3.01 .71
lF-ratios with 2/237 df for set 1 and 2/239 df for set 2.
2

F-ratios with 2/239 df for set 1 and 2/241 df for set 2.

a

Total pretest—method interaction.

b

Multiple choice pretest—method interaction.

c
Fill-in prete st—method interaction.

*
p<.05.
**
p< . 025.

in computer terminology, was

x(3) = Q (X(7)..

and is the same as equation [21 on page 35.

Of the variables included in this equation were

- 7.54579
16.54162
—l3.81762
.49373
.08428
.53467

(Lecture)
(Skinner)
(Crowder)

(WK 1)(Lecture)
(WK 1)(Skinner)
(WK 1)(Crowder). (

.x(9) ,x(11) ,X(21). . .x(23))*,

The regression coefficients

_ _._ ————a-'-i

 

47

To make the plot for each treatment group, two points were needed:
the estimated PTT score for the lowest WK 1 score (taken as l), and the
estimated PTT score for the highest WK 1 score (taken as 100). The first
three regression coefficients provided the first point of the Lecture, Skinner,
and Crowder methods, respectively. The last three regression coefficients,
multiplied by 100 and added to the value of the lowest point, provided the
second point. Figure 1, thus, presents the WK 1 - PTT tot plots for the
three methods using set 2 data. This figure shows that the Skinner pro-
gram was best for individuals with low WK 1 scores, while the lecture
approach was best for individuals with high WK 1 scores. Figure 2 pre—
sents the WK 1 — PTT tot plots for the set 1 data.

A very rough estimate of the same result was obtained by dividing
each treatment group into a "high, " "average, " and "low" WK 1 subgroup.
The division was based on the average mean of the three treatments and a
range which approximated i 1/2 standard deviation. The high subgroup
had scores of 74 and above; the low subgroup had scores of 60 and below,
and the average subgroup ranged between 61 and 73. Table 15 presents

the results and shows the same relationship among treatments as Figures

land 2.

48

N Hum

HMHdemdS HZmQH/Hmmmfl mHHH m4. Hoe. .H..H.nH HHHH>> mHZmHOHanHmOO ZOHmmmmwmm HZm2H<mmH mHIHH. .“HO

mHOHnH ME QOHHHMHZ Q24. H vH.>> meEmm ZOHHOHEPHZH mHHH nHO ZOHH<Hmemmmmm
H mmDOHnH

.OHQMHHMN/ uflmegflwtfl. £00m 1H0.“ muCOHOMMHOOO

conmoeoon one So moEOOm Ho“. Hem Have one. eon mew oncHHflo one. ocon Bonanza oHHHH

mmOOm H vaHm>>

 

00 H 0
one :6:
lo H I
Ovo l I
v0.0? w
10
1w
IwH
nocchHm
IVN
QVA/Pog IWM

 

 

Posttest Total

50

TAB LE 15

ESTIMATES OF THE PTT tot MEANS FOR HIGH, AVERAGE, AND LOW
WK 1 SCORES FOR THE LECTURE, SKINNER, AND CROWDER TREATMENTS

 

 

Lecture Skinner Crowder

Set 1 Set 2 Set 1 Set 2 Set 1 Set 2
High

(74 and above) 63.95 63.00 63.27 53.90 65.89 63.62
Average

(61 — 73) 58.21 60.30 54.81 52.09 53.96 48.85
Low

(60 and below) 51.81 45.88 50.50 51.42 42.40 44.47

 

Hypothesis lb

There will be no immediate or delayed posttest differences for remedial

students who have been subjected to either the programed presentations or

the conventional lecture.

A main effect for method, with the posttest as criterion, was not es-

tablished in set 2 because of the significant interaction between WK 1 and

method that was just noted. There was a treatment main effect in set 2

with the retest criterion. Table 16 shows these results.

Table E in Appen—

dix A indicates that on the retest, the Lecture mean was greater than the

means for the programed instruction treatments.

 

 

51

TAB LE 16

TESTS OF SIGNIFICANCE FOR METHOD MAIN EFFECTS WITH
POSTTEST AND RETEST DEPENDENT VARIABLES (HYPOTHESIS 1b)

 

 

 

Dependent
Variable IQ-WK 1 Model PRT—RC Model
Set 1a Set 2b Set 1a Set 2b

PTT tot 1. 05 _f 2. 07C 2.41C
PTT mc 1. 39 ___f 1.19“1 2.98d
PTT f1 .50 1.67 .99e 1.11e
RTtot 1.51 3.91**
RT mc .19 7.18****
RT 11 5. 22*** 4. 39**

 

aF-ratios with 2/241 df.

bF—ratios with 2/243 df.

cTotal pretest—RC model.

dMultiple choice pretest—RC model.
eFill—in pretest-RC model.

fWith a significant WK 1 method interaction the test for main effects

is not meaningful.

**
p<.025.

***
p<.01.

****
p<.005.

52
Hypothesis 2a
Within programed instruction treatments , there will be no immediate
pOsttest differences for persons of varying intelligence.

This null hypothesis was rejected. The data in Table 17 suggested

 

that a difference in PTT score following programed instruction was depend-

ent on intellectual ability, especially with respect to the more difficult

PTT fi subtest.

TABLE 17

TESTS OF SIGNIFICANCE FOR IQ MAIN EFFECTS ACROSS
PROGRAMED INSTRUCTION TREATMENTS (HYPOTHESIS 2a)

 

 

Dependent a b
Variable Set 1 Set 2
PTT tot 9. 85**** 3. 84C
PTT mc 3. 38 . 73
PTT fi 11.99**** 6.64**

 

aF—ratios with 1/150 df.

bF—ratios with 1/167 df for PTT tot and PTT mc, and
1/169 df for PTT fi.

CIf the interaction between WK 1 and method were dis-
regarded the main effect would be significant at the . 05
level.

**

p < . 025 .

****

p < . 005 .

Table 18 indicates that high ability individuals had higher PTT tot

SCOres than those of lower ability. The data in Table 18, like those in

 

53
Table 15, are only suggestive, and were not suitable for statistical anal-

ysis.

TABLE 1 8

ESTIMATES OF THE PTT tot MEANS FOR HIGH, AVERAGE,
AND LOW IQ SCORES FOR THE SKINNER AND CROWDER METHODS

 

Skinner Crowder Total
Set 1 Set 2 Set 1 Set 2 Set 1 Set 2

 

High
(122andabove) 61.32 57.15 63.05 59.24 62.12 58.39

Average
(115—121) 56.11 52.37 56.87 54.82 56.48 53.54

Low
(114andbelow) 52.00 47.21 50.31 49.77 51.51 48.59

 

The PRT tot means were grouped in the same manner as the PTT tot
means. Table 19 shows that there was only a slight difference in PRT tot
for the three ability levels in set 1. This fact, and the greater (and sig—
nificant) PTT tot differences presented in Table 18, indicated that the high
ability subgroup gained more from programed instruction than the low abil—
ity Subgroup.

The data in set 2 did not reveal similar findings and suggests that
performance was related to the form of the criterion test which was admin—
iStered first. It will be recalled that the less difficult criterion test (82)
was the pretest in set 2, and the more difficult test (S1) the posttest.

Thus, When used as either pretest (set 1) or posttest (set 2) , the more dif-

1cult criterion test did not permit the expreSSion of ind1v1dual differences.

54

TAB LE 19

ESTIMATES OF THE PRT tot MEANS FOR HIGH, AVERAGE,
AND LOW IQ SCORES FOR THE SKINNER AND CROWDER METHODS

 

Skinner Crowder Total
Set 1 Set 2 Set 1 Set 2 Set 1 Set 2

 

High
(122andabove) 48.23 56.80 52.74 56.07 50.31 56.37

Average
(115—121) 48.46 49.12 48.75 52.77 48.60 50.85

I Low
(ll4andbelow) 50.00 45.47 44.77 46.14 48.40 45.83

 

Hypothesis 2b

Within programed instruction treatments , there will be no immediate
posttest differences for persons of varying reading comprehension levels.

The main RC effects in set 1 were highly significant as Table 20 points
out. Thus, while level of reading comprehension did not have a bearing
on which form of instruction was superior (Table 14), there was some ev—
idence that achievement was related to level of reading comprehension as
measured by the Davis Reading Test.

TABLE 2 0

TESTS OF SIGNIFICANCE FOR RC MAIN EFFECTS
ACROSS PROGRAMED INSTRUCTION TREATMENTS (HYPOTHESIS 2b)

 

 

 

Dependent a b
Variable Set 1 Set 2
PTT tot 12.39**** .51
PTT mc 5.56** .25
PTT fi 11.47**** .50

aF—ratios with 1/150 df. bF—ratios with 1/169 df.
** *‘k**
p<.025. p<.005.

55
Hypothesis 3a
With time to program completion as the criterion, there will be no
ability and method interaction within programed presentations.
Table 21 indicates that, with time as the dependent variable, there

was no interaction between IQ and method or between WK 1 and method.

TAB LE 21

TESTS OF SIGNIFICANCE FOR INTERACTION BETWEEN
ABILITY AND FORM OF PROGRAMED INSTRUCTION WITH
TIME AS THE DEPENDENT VARIABLE (HYPOTHESIS 3a)

 

 

IQ--Method1 WK l-Method2
Interaction Interaction
Set 1 Set 2 Set 1 Set 2
. 22 . 55 . 22 . 09

 

1
set 2.

F—ratios with 1/147 df for set 1 and 1/166 df for

2F—ratios with 1/148 df for set 1 and 1/167 df for
set 2.

 

 

 

Hypothesis 3b
With time to program completion as the criterion, there will be no dif-
ferences between the Skinner and the Crowder presentations.
As anticipated in Tables 12 and 13, Table 22 shows a significant
treatment main effect with respect to time. Tables D and E of Appendix A
Showed that it took a student longer to proceed through the Skinner than

the Crowder program .

 

56

TAB LE 2 2

TESTS OF SIGNIFICANCE FOR PROGRAMED
INSTRUCTION MAIN EFFECTS WITH TIME
AS THE DEPENDENT VARIABLE (HYPOTHESIS 3b)

b

 

Set la Set 2
17.10*** 5.39*

 

aF—ratio with 1/149 df.
bF—ratio with 1/168 df.
*
p<.05.
***
p<. 01.

It might also be mentioned that there was no IQ main effect, with the

time criterion. This was indicated by an F of 2. 32 (with 1/150 df) in set

1, and an F of 1.56 (with 1/169 df) in set 2.

Hypothesis 4a
There will be no interaction between ability and method, with achieve-
ment on related subject matter as the criterion.
Based on the non—significant F tests in Table 23, the above null hy—

pothesis was accepted.

57

TAB LE 2 3

TESTS OF INTERACTION BETWEEN IQ AND METHOD
WITH ACHIEVEMENT ON RELATED SUBJECT MATTER
AS THE DEPENDENT VARIABLE (HYPOTHESIS 4a)

 

 

Dependent IQ—Method1 WK l—Method2
Variable Interaction Interaction

Set 1 Set 2 Set 1 Set 2

WK3 .64 .35 1.82 1.92

WK5 1.57 0.00 1.77 1.88

 

lF-ratios with 2/237 df for set 1 and 2/239 df
for set 2.

2F—ratios with 2/239 df for set 1 and 2/241 df
for set 2.
Hypothesis 4b
The academic achievement of students who learned under programed

instruction will not differ from students who learned under the conventional

 

lecture on related subject matter.
There were no significant differences among either WK 3 or WK 5 means

for the three methods of instruction, as indicated in Table 24.

TABLE 24

TESTS OF SIGNIFICANCE FOR METHOD MAIN EFFECTS
WITH ACHIEVEMENT ON RELATED SUBJECT MATTER
AS THE DEPENDENT VARIABLE (HYPOTHESIS 4b)

 

 

Dependent a b
Variable Set 1 Set 2
WK 3 . 93 l. 37
WK 5 . 79 . 99

 

aF—ratios with 2/241 df.
bF—ratios with 2/243 df.

 

 

58
Hypothesis 5

There will be no interaction between the measurement form of the

criterion and the form of programed instruction used.

The hypothesis of no interaction between the measurement form of the
criterion (mc and fi) and the two forms of programed instruction was not
supported under the present technique of testing. The mc and fi means
were not equal prior to the period of remedial training (Tables B and C in
Appendix A) and could, therefore, not be appropriately analyzed subsequent
to training. However, it seemed clear from the insignificant method main
effect shown in Table 25 that no interaction existed. Table 25 shows that
neither the lecture, Skinner, nor Crowder methods of instruction proved
Superior when either PTT mc or PTT fi were the dependent variables. The
Separate analyses of variance conducted on these two dependent variables

(Tables 12 and 13) suggested the same result.

TABLE 25

TESTS OF SIGNIFICANCE FOR METHOD MAIN EFFECTS
USING PTT mc AND PTT fi AS DEPENDENT VARIABLES (HYPOTHESIS 5)

 

 

 

Dependent Variable Set 1a Set 2b
PTT mcl 1.19 2.98
PTT £12 .99 1. 11
aF-ratios with 2/241 df. bF—ratios with 2/243 df.

lA dependent variable for a model including RC and PRT mc as
independent variables.

. 2A dependent variable for a model including RC and PRT fi as ,
1ndependent variables. I

 

 

 

59
Hypothesis 6
There will be no relationship between time to complete programed in—
struction and subsequent academic achievement.
The correlations between time to complete programed instruction and
achievement on the posttest were negative and non-significant. Tables I

and J in Appendix A provide the correlation coefficients.

 

W

CHAPTER VI
DISCUSSION

The challenge which some researchers have been trying to meet,
seems, has been to discover how closely they could mold human learr
to fit sub-human requirements. With the attainment of this goal, the
perimenter or teacher could more easily manipulate the behavioral asp
Of learning than was previously possible. The present research was,
part , directed toward an investigation of success in attaining this goa
and assumed that linear and non-linear programs of instruction incorp<
rated the principles of learning developed in the laboratory.

While results of the present investigation indicated that the learr
process could not yet be controlled to the extent envisioned by Skinne
and Crowder, the results did point to the fact that students were able
learn from programed instruction. Unfortunately, neither the high nor
l0W ability remedial students using programed instruction, and those .
the lecture presentation, achieved at an acceptable level (scores of 8
althOugh the high ability individuals had a greater posttest mean than
thOSe of low ability.

It was difficult to reconcile the differences in results between se

and Set 2, with respect to the RC main effects and the interaction bet‘
WK 1 and method. They might have been due to differences in difficu.

be
1“Ween the criterion tests, differences between academic performanc

60

L

 

 

 

61

developed by the instructional methods and that required by the criterion
tests, or differences as a function of pre—treatment inequalities in WK 1
score. There was no systematic attempt to determine which factor explained
the results , although the first factor stood out. A position consonant with
that proposed by Skinner (65) would suggest that the interaction between
WK 1 and method was primarily due to the fact that the students using the
linear program advanced to a common, but low level, of attainment.

The lack of significant results in general, might be viewed in the
light of low reliabilities, in addition to the limitations of student assign-

ment and criterion test development. In the present study reliabilities of

the total posttest scores were about . 74, with the 15-item components
even lower. The total reliabilities were somewhat lower than those of . 85
reported for typical tests containing between 30 and 50 items (29, 41, 58,
61 , 62). As estimated by the Spearman-Brown formula, in this study a 60=-
item criterion test might have resulted in a test with a reliability of . 85.
There were some difficulties in interpretation as a consequence of
r1On-parallel criterion tests and nonmparallel subscores within each test.
The latter shortcoming found in past studies (13, 21) also restricted the
fun exploration of the interaction between the measurement form of the
Criterion’and the form of the program. Data suggested that the difference
in the RC main effect between set 1 and set 2 was due to the different
leVels of difficulty of the criterion tests. For example, Table 26 indicates

t
hat the more difficult posttest of set 2 placed an upper limit on the academic

loeI‘fOI‘rnance of students within each of the RC levels. On the other hand,

g

     

62
the less difficult posttest of set 1 allowed the expression of individual
differences in achievement related to reading ability. Such differences in

main effects obtained between set 1 and set 2 pointed to the possibility

 

that the inconsistency in conclusions found among previous studies , was
in part, due to the use of criterion tests of unknown but varying levels of

difficulty.

TAB LE 2 6

ESTIMATES OF THE PTT tot MEANS FOR HIGH, AVERAGE,
AND LOW RC SCORES FOR THE SKINNER AND CROWDER METHODS

 

Skinner Crowder Total
Setl Set2 Setl Set2 Setl Set2

 

High
(75 andabove) 62.27 55.00 60.91 52.42 61.58 53.49

Average
(69—74) 54.91 52.46 53.82 55.80 54.54 54.19
1 LOW

(73 andbelow) 51.92 50.57 57.21 56.19 54.26 53.28

 

Differences between PRT tot and PTT tot for set 1, found in an examina- -
tion of Tables 18 and 19, suggested that the intellectually able students
gained more than the less able. The question raised, in this connection,
" Waswhy the "high ability" students , and others, did not perform even
better than they did. One reason just described, was the difficult posttest,
especially in set 2. It was also more likely than not that remedial stu—
dents reached close to their peak academic performance, regardless of

measured intellectual performance. Thus ,. rather than considering higher

 

 

63 ,
ability remedial students as "underachievers (29)" an equally reasonable 1'
proposition, especially in the present study, was that measures of intel— ' I

ligence “overpredicted” students' ability. While achieving somewhat be~

yond the low ability student, the high ability individual functioned closer

to his "real" ability than to his measured ability.

Another question regarding the criterion tasks and their relationship
to the research results was raised by the findings. Did the criterion tests
measure behavior which the programs were designed to develop? This
question did not imply that the content of the criterion and the program
differed, but that there was no great transfer of training, as measured by
the criterion. 1

With respect to Skinner—type programs, Briggs, Goldbeck, Campbell,
and Nichols (5) indicated that an investigator should not expect even a
mOderate level of learning of program content which was not elicited during
training. Indeed, if the Skinner and Crowder programs emphasized rote
learning, and the criterion tests were designed to measure conceptual,
al'lalytical, and applied tasks. transfer and measured improvement in per—
f0I‘mance would be minimal.
As Table 16 showed, there were no method main effects on the post-
test criterion. Rather than conclude that there were no treatment differences ,
it WOuld be more appropriate to conclude that a fair evaluation of possible
’ differences was not made. What was needed for a fair evaluation of method .
2

l . . .
m This pomt of View was suggested by Dr. M. Ray Denny of the Depart- !
ent of Psychology, Michigan State University. '

g —#

64
main effects were criterion tests with a higher ceiling and those which
measured the type of academic performance which the programs attempted
to develop.

Wong and Douglass (71) suggested one additional reason for a finding
of no significant differences among treatment means: the lecture treatment
was well presented and organized. This interpretation might have validity
in the present study, in as much as the remedial curriculum was highly
organized and, according to instructor evaluation reports well presented.

An interpretation of the interaction between WK 1 and method {set 2.)
follows the above arguments closely. The interaction may be the conse-
quences of one of several phenomena already mentioned. Reference is
made to Figure l and Table 15 for another interpretation. The low ability
individuals could not organize the subject matter on their own, To the
extent that the Skinner program provided this organization, low ability in»
dividuals in the Skinner program surpassed those in either the Lecture or
Crowder presentations. At the high ability level, individuals in the
Crowder and Lecture groups were able to assimilate generalizations pro-
vided by the presentations. High abilizy Skinner students on the other
hand, learned fewer generalizations and relationships and so could not
apply their rote knowledge to the test of transfer which followed.

Because of some superiority in ability, the high ability Skinner stu—
dents out performed the low ability Leotzxe and Crowd r studenzs. Note

that these high ability Skinner ind‘vi uals in set 1, Table 1.3, were able

 

65
to perform even better on a less difficult posttest; that is , they were not
as much restricted by the posttest as the subjects in set 2 were.

Thus, the interaction between WK 1 and method was some function of
the low ceiling of the posttest, the applied nature of the posttest, and the
rote characteristics of the Skinner program. The minimal Skinnerutreatment
slope for set 2 in Figure 1, appeared to support a Skinnerian assumption.
The assumption was that students using programed instruction would achieve
at a common level; in this research, because of the nature of the criterion
tests, it was a common but low level of achievement. The set 1 data,
however, did not verify this assumptiono

The finding of an interaction might also have been a function of pre—
treatment inequalities in WK 1 score° The method of assignment did not
provide for the level of sophistication which would assure equal means on
all significant dimensions. In terms of WK 1, the Skinner group had a sig—
nificantly lower mean score than the other instructional groups (Table C of
Appendix A). The individuals in set 1 also had low WK 1 scores, especially
the Lecture and Skinner treatments , but showed no interactions.

No analysis was undertaken to ascertain which set of factors accounted
for the interaction between WK 1 and method. The investigator believes that
the differences in results between set 1 and set 2 may best be explained
by the characteristics of the criterion tests just discussed, i.er _. their
low ceiling and applied nature.

It was difficult to explain the significant main effect favoring the lec—

ture method on the retest in set 2. in set 1 the only significant difference

 

66

occurred on the RT fi criterion; the F test was conducted with heterogeneity
of variances (. 02 level) and no consideration of a possible interaction be—
tween WK 1 and method which was just short of significance at the . 05
level. The differences in results were hard to substantiate since there was
much uncontrolled variation during the two weeks between the first and
second administration of the posttest, All students received the same
training between posttest and retesto There was no doubt, though, that
differences favoring the Lecture method appeared two weeks after remedial
training. These differences were greatest for the fi criteriono

There were clear-cut findings with respect to time (T). It took an
average of about 10 percent longer to complete the linear program than the
non-linear. One reason for this result is that students had to read all the
frames in the linear version, but the brighter students using the non-linear
program could skip much of the materialo it also took longer to go through
the linear program because more writing was involved; this was the dif_
ference between constructing an answer and selecting one. This was , of
course, not a definitive study of the time dimension and its relationship
to programed instructiono Rather, the results show that programed instruc—
tion reduced the time usually required for the six hours of remedial instruc=~
tion.

Tables I and I in Appendix A provide the correlations between time to
complete programed instruction and other variables. The negative correla-

tions, of low magnitude, confirmed the findings of past research: higher

67
ability students take less time to progress through a program than lower
ability students. Ruling out an interaction between ability and method,
and a significant treatment main effect, the implication of these findings
is clear. Educational specialists could design a remedial training program,
or any other course of instruction, such that most students would be as-
signed to programed instruction groups (instead of being required to sit
through 360 minutes of the lecture presentation (two hours each of three
nights). Each student would progress at his own rate.

The average student, as shown in Tables D and E of Appendix A, com-
pleted instruction better than 30 percent quicker than the average student
in a lecture group. Some students completed the course even sooner, and
were available that much sooner for other activities. It took the fastest
student only one hour to complete the six hours of instruction. Students
whose estimated completion time for programed instruction was longer than
the conventional lecture would be assigned to the conventional lecture, at
no loss in academic proficiency.

This research provided little support for a system of differential assign—
ment of students to treatments, such as may be used in making personnel
decisions (15). The lack of evidence ~=— for interactions between ability and
method— was a consequence not only of the difficult posttest, but also the
more limited ability range of remedial students. In fact, the range of meas—
ured intellectual ability- IQs of 90to 149, in this study— may be a phenomenon

of poor measuring devices, as already noted. It is hypothesized that the

 

68
"real" range of ability in a remedial group is narrower than specified by
ability tests.

Perhaps an even larger sample size would have produced method main
effects and interactions. Possibly a longer program would have allowed
the full potential of programed instruction to be realized. On these points,
a study by Campbell (8) which compared programs of three levels of adap—
tiveness on multiple choice and fill-in criterion tests, showed that each
program was equal in teaching effect. His sample included 2,000 students
in the fourth through the twelfth grades. Crowder and Zachert (l7) eval—
uated an electronics program of 1600 frames. The study lasted four weeks
and showed no differences from a lecture control group.

Most of the studies reported in the literature have used small sample
sizes and short programs. While the two studies just mentioned had some
favorable characteristics, there were shortcomings. The experimental and
control groups in Crowder and Zachert's study contained only 28 subjects.
In other words, it does not appear feasible to eliminate all shortcomings
in any one study. Thus, there will probably always be some factor, espe—
cially in a field experiment, which places a limit on what the investigator
can say, or should say, about his data.

In addition to the measurement problem in programed instruction, Roe
(56) felt that ". . . present theories and auto-instructional techniques are

inadequate to achieve individualized instruction, and that a usable auto-

mated teaching system will require further analytical and hardware development. "

69
The next few paragraphs consider Roe's comment and the theoretical
strength of programed instruction, as it relates to practical consequences.

These results invited a serious questioning of the direct application
of principles developed in the psychological laboratory to human learning
and confirmed past difficulties in assessing this application.

Table 1 indicated that the Skinnerian and Crowderian forms of pro-
gramed instruction differed, although the principles of learning around
which differences in implementation centered were similar. These prin—
ciples were consistently verified in the laboratory and were demonstrable
in rats , pigeons, and dogs by a number of different experimenters. It was
shown in the present and previous studies that a direct translation of
learning principles to the classroom setting was difficult, in a practical
sense, Porter (53) has commented on the difficulty of generalizing results
from animal studies to studies concerned with humans. The research re-
sults have, thus, been discouraging because of over-conceptualization,
improper implementation of learning principles , or, as implied throughout
this chapter, poor measurement.

The purpose of the remainder of this section is to review several
studies that suggested over-conceptualization; this was certainly one fac-
tor which could explain the findings of the present investigation. Since
better than 80 percent of the research in programed instruction has used
linear programs, these studies deal with the Skinnerian approach.

Although knowledge of results seems to have been particularly advan—

tageous in psychomotor performance (47), investigators reached no clear—

 

70
cut conclusions concerning its effectiveness in recent studies in programed
instruction. Early studies using the Pressey punch board found that learning
was enhanced by immediate knowledge of results (1, 37). There are sev-
eral reasons why knowledge of results was not effective in studies utilizing
linear programs. These reasons relate to the low error rate of linear frames,
and the source of reinforcement.

Possibly knowledge of results was not as important to the learner as
a "reinforcement" in linear programs as it might have been with a more dif—
ficult program sequence. In other words, the satisfying effects were min-
imal when the students were correct 95 percent of the time. The student
knew the answer before confirmation was provided.

A difficulty in studying knowledge of results was that, even without
an external source of knowledge of results , there usually was an intrinsic
source. As Skinner pointed out (64), completing the task itself was a sig—
nificant reinforcement. In laboratory studies the intrinsic source was not
a confounding variable as in human studies.

In addition to reinforcement and low error rates , Skinner attempted to
shape behavior by means of a fixed and logical sequence of instructional
frames directed toward the criterion behavior. Recent evidence suggested
that the sequence did not have to be fixed, logical, or ordered (43, 48,

49, 55). Mager found that students generated sequences different from
those developed by their instructors. It should be pointed out that studies
in this area have been limited by the very small number of items included

in the program sequence.

71

Where learning was measured in terms of motor performance, bar
pressing, for example, active responding on the part of the subject was
reasonable; lower animals are not equipped with other means of communica-
tion, and bar-pressing was the behavior being taught. In the laboratory,
conditions are such that in order to receive a reinforcement the animal has
to trigger the food magazine by responding. In verbal learning by means
of programed instruction, the value of active responding seems to be to
involve the student in his subject and to draw attention to the appropriate
response, rather than to build up the response probability of a single re-
sponse. But it is begging the question to assume that a one word response
is anything more than that; it is certainly a very limited aspect of the ter-
minal behavior being developed.

Since reinforcing, sequencing, and responding have a significance
in the laboratory different from that of the classroom, direct application of
principles from one environment to the other is limited. Even if this trans-—
lation were meaningful, present apparatus and programs do not allow faith-
ful duplication of the laboratory principles. And if both translation and
duplication were readily effected, investigators would still have to sharpen
measurement and design characteristics, especially under practical op—

erating conditions .

CHAPTER VII

SUMMARY AND CONCLUSIONS

The present investigator agrees with Silberman, Coulson, Gunn, and

Melaragno (61) that
. . .the appropriate and critical dimensions for specifying the learning
task, the teaching method, and the student have not yet been iden—
tified. Consequently, there is not a sound basis for exploring re—
lationships between task, method, and student variables as they
determine the efficienty of student learning under programed in—
struction.

There were several reasons for difficulty in identifying the appropriate
and critical dimensions of task, method, and student variables in the pres—
ent study. As indicated in Chapter VI, these reasons related to aspects of
measurement and design, the nature of remedial students with respect to
measured ability, and limitations in applying laboratory developed princi-
ples of learning to students in the classroom. For investigators to define
significant aspects of the "instructional gestalt (59)" and make predictions
about academic achievement within it, they must provide more meaningful

conceptualizations of the learning process and devise more valid instru—

ments for evaluating the outcome of that process.

The Learning Process
The present and previous studies have provided little support for the
"homo mechanicus" model of man, developed from insights gained in the
animal laboratory, and represented by linear programing. Evidence for the

72

Va,

73
"homo sapiens" model, as represented by non—linear programing was equally

discouraging. In the present study, neither the linear nor non-linear forms

of programing provided remedial students with the kind of subject matter

organization or development which would allow them to achieve at a level

equivalent to their non—remedial fellows. Even those students with high

measured intelligence did not perform well, although they performed better

than those of low ability. Prior academic ability, intelligence, and reading

ability did not interact with the form of subject matter presentation — Lecture,

linear, non-linear. The only change brought about by programed instruc-

tion was to shorten the time required for training when compared to a con-

ventional lecture. Since little was learned from programed instruction,

there was no transfer of advantages to subject matter which followed reme=-

dial training.

In spite of the negative results found in the present study, the homo
sapiens model seems to have more potential than the homo mechanicus for
dealing effectively with learners characterized by a wide range of individ-
ual differences in academic achievement, intelligence, and modes of

learning. The more flexible branching programs are designed to take into

account the initial differences among individuals. Non—linear programs

may also be designed to respond more realistically to the needs of the

learner as he progresses through the program. A student might find linear

programs useful where his reading and verbal skills have not yet matured

and where his study habits have not been completely formed.

\

74
While the programer incorporates stimulus differentiation and general-

ization into linear programs to some degree, however unsuccessfully, he

neglects other concepts involved in human learning and retention. The

programer presumably recognizes and accounts for the effects of proactive

and retroactive inhibition by an emphasis on low error rates and repetition

of concepts throughout the program. Low error rates, however, do not in—

sure against the accumulation of interferences generated as the student

advances through the program. Non—linear programs, on the other hand,

provides for some form of branching in recognition of the fact that errors

are cumulative throughout learning. Whatever the cause of the errors ,

either poor communication or inability of the student to discriminate or
generalize, the non-linear procedure capitalizes on, rather than overlooks,
the kinds of behaviors characteristic of human learning.

In the present study, neither program provided much flexibility in sub-

iect matter presentation. In the future investigators should direct more

attention toward investigating the various patterns involved in human
learning, i.e. , the dynamics of acquisition, integration, and retention of

verbal material, than emphasizing the application of principles of behavioral

control developed in the animal laboratory. Computers might assist in an

attempt to integrate past findings in the area of human learning for the
purpose of application to program construction and machine presentation.
Additional light on human learning might be gained through the use of an

artificial subject matter, with computers presenting, modifying, and

rec ording behavior.

75
The Measurement Process
It is obvious that the outcome of research into the learning process,

and its interpretability, depends on the proper application of principles of

measurement and design. In the present study, the criterion variables

were inappropriate to the tasks being taught by the instructional methods.
An adequate evaluation of a theoretical position also depends on the de-

gree to which experimental implementation faithfully reproduces elements

of the theoretical model.

Measurement sophistication, as far as programed instruction is con—
cerned, includes (a) adequately developed criterion tests reflecting the
tasks taught by the program, (b) adequately designed programs reflecting
the process of learning hypothesized by the investigator , and (c) a detailed

description of the subjects used in a study on personality, motivational,

and ability variables. In order to allow more meaningful interpretation

and comparison among several investigations, there should be standard-

ized programs and criterion tests, developed on subjects possessing dif—

ferent patterns of personality, motivation, and ability. The gain in knowl—

edge about human learning stemming from work in programed instruction

Will certainly be limited, and less meaningful, until a comprehensive and

systematic research program is developed. This basic idea is, of course,

fundamental to most research, although not always implemented.

APPENDICES

76

APPENDIX A

SUPPLEMENTARY TABLES A — I

77

 

78

TABLE A

SCHEDULE FOR ADMINISTRATION OF TESTS
AND INSTRUCTIONAL PRESENTATIONS

 

 

Class Treatment Pretest Posttest Retest
No. N Form Date—Form Date-Form Date-Form
1 23 Lecture Mar 27 SI Mar 29 82 Apr 13 82
2 15 Crowder Apr 3 82 Apr 5 81 Apr 20 SI
3 l9 Skinner Apr 10 81 Apr 12 82 Apr 27 82
4 10 Lecture Apr 17 82 Apr 19 81 May 4 SI
5 7 Crowder Apr 24 S1 Apr 26 82 May 11 82
6 14 Skinner May 1 SZ May 3 81 May 18 SI
7 13 Crowder May 8 SI May 10 82 May 25 82
8 20 Lecture May 15 82 Mayl7 SI jun ISI
9 13 Skinner May28 82 May 31 SI jun 15 SI
10 18 Lecture jun 5 SI jun 7 82 jun 22 82
ll 30 Crowder jun 1‘2 82 jun 14 SI jun 29 SI
12 23 Skinner jun 19 81 jun 21 S2. jul 6 82
13 26 Lecture jun 26 82 jun 28 SI jul 13 SI
14 22 Crowder jul 2 SI jul 5 82 jul 20 82
15 28 Skinner jul 10 82 jul 12 SI jul 27 SI
16 22 Crowder jul 17 SI jul 19 82 Aug 3 SZ
17 20 Lecture jul 24 82 jul 26 SI Aug 10 SI
18 25 Skinner jul 31 81 Aug 2 82 Aug 17 82
19 27 Crowder Aug 7 82 Aug 9 81 Aug 24 SI
20 22 Skinner Aug 14 31 Aug 16 82 Aug 31 82
21 27 Skinner Aug 21 32 Aug 23 81 Sep 7 SI
22 26 Lecture Aug 28 81 Aug 30 82 Sep 14 82
2.3 18 Crowder Sep 4 82 Sep 6 81 Sep 21 SI

24 26 Lecture Sep 11 SI Sep 13 82 Sep 28 82

 

 

79

TAB LE B

MEANS AND STANDARD DEVIATIONS OF PRE—TREATMENT
VARIABLES FOR SUBjECTS IN THE LECTURE,
SKINNER, AND CROWDER TREATMENTS (SET 1)

 

 

 

 

 

Lecture Skinner Crowder
Mean S D Mean S D Mean S D
IQ 117.60 6.17 117.11 6.49 119.45 7.08
RS 71.80 5.55 72.22 5.12 73.60 5.21
RC 70.83 5.40 71.44 5.53 72.70 6.13
WKl 64.70* 12.60 66.59 13 51 70.09 14.18
WK 2 52.88** 7.45 54.82 9.04 57.06 8.86
PRT tot 46.58 14.42 49.32 13.05 49.12 15.51
PRT mc 50.89 15.66 54.37 14.30 53.10 16.58
PRT fi 41.12 18.19 43.69 16.77 43.85 20.31
*
Difference between means significant at . 05 level.
**
Difference between means significant at . 025.
TABLE C
MEANS AND STANDARD DEVIATIONS OF PRE—TREATMENT
VARIABLES FOR SUBjECTS IN THE LECTURE,
SKINNER, AND CROWDER TREATMENTS (SET 2)
Lecture Skinner Crowder
Mean S D Mean S D Mean S D
IQ 119.19 7.84 117.81 8.07 119.51 8.78
RS 73.46 4.70 72.02 5.00 72.88 5.51
RC 72.71 4.91 70.97 4.45 71.83 5.77
WKI 70.44*** 13.31 65.52 11.92 71.28 13.75
WKZ 55.34 8.19 54.01 8.14 54.14 10.49
PRT tot 52.56 15.59 50.14 15.94 52.21 15.62
PRT mc 55.65 19.10 55.29 17.39 57.08 16.71
PRT fi 48.65 17.10 43.30 18.20 46.34 18.80

 

**
Difference between

means significant at

. 01 level.

80

TAB LE D

MEANS AND STANDARD DEVIATIONS OF POST-TREATMENT
VARIABLES FOR SUBjECTS IN THE LECTURE,
SKINNER, AND CROWDER TREATMENTS (SET 1)

 

 

Lecture Skinner Crowder

Mean S D Mean S D Mean S D
PTTtot 57.59 15.11 55.92 13.31 57.37 15.53
PTT mc 63.06 17.02 61.15 16.11 61.56 18.21
PTT fi 50.78 17.12 49.19 15.53 52.01 17.47
RT tot 64.65 13.16 61.75 11.65 63.89 16.66
RT mc 68.08 16.15 69.14 14.49 68.87 18.29
RT fi 60.29** 15.23 53.25 12.68 57.82 19.54
WK3 67.83 8.51 68.39 9.24 71.45 10.98
WK5 62.07 9.88 61.99 10.09 65.51 12.61
T ——a —a 251.91W44.31 217.43 46.96

 

**
Difference between means significant at . 025 level.

****
Difference between means significant at . 005 level.

TAB LE E

MEANS AND STANDARD DEVIATIONS OF POST-TREATMENT
VARIABLES FOR SUBjECTS IN THE LECTURE,
SKINNER, AND CROWDER TREATMENTS (SET 2)

 

 

 

 

\ Lecture Skinner Crowder
) Mean S D Mean S D Mean S D
' PTT tot 57.86 15.31 52.34 14.06 55.01 17.39
1‘ PTT mc 60.86 17.77 55.25 15.40 59.19 19.24
t PTTfi 54.06 16.77 48.42 16.79 49.87 18.92
“ RT tot 63.89* 15.38 59.35 14.44 58.16 15.98
‘ RT mc 64.84*** 15.82 65.74 16.67 58.93 17.93
RT fi 61.39 17.55 51.81 17.29 56.14 17.01
WK 3 71.07* 10.34 67.69 9.43 71.52 10.29
WK 5 62.93 12.17 63.57 9.72 64.51 12.37
'I' a a 267.51***44.38 247.67 56.19

 

a
. Time to complete programed instruction was not applicable to the
lecture group.
*

Difference between means significant at . 05 level.
***
Difference between means significant at . 01 level.
** *
Difference between means significant at . 005 level.

‘ _+

 

-..--.."

81

 

NNMMN .02 N 423.03 3mm - .38
:N 393 .NmmS mamm c2828.:
3. NSNTSN 8N 2.84.828 3% 89:82; Es.
NNmNm.$m N 3393: awm - 3%
8N 213 .Nomz 3% c9888.:
8 .N 58 .mmN NNN 82% .288 qmm 8582-9 05 NE
288.2: N $8938. 8% - 43mm
EN EN: .932. 43mm 885
mo .2 8.0.3.02 SN 88... .432. .33 522 8522
$88 .3: N $8M. .NON 3% - 3%
SN 38m .392. «3% 83028.:
mm: ONO? .2: SN N33448:. 3% 8:52; VS,
33.8 NS N 33543 dmm - 3%
mmN N N H m .XWNQ 3% 830285
34 @NNEN: NMN $82 .8va dmm 8522-9 #8 Em 0: .2
m 898m Eocomum mmumsvm wo 53m uobm “~me oBch> mummfioaxwm
552 m0 moouomm “coccmdofl

 

HQ Emmy mHOmmmm 23.2 QOEHME 02¢ SHAME/N. OZHQ<mm .mOmeHquHZH 02.4.
.QOEHMSH 072 3354. meEmm ZOHHOENHZH mg OH. Qmefimm mmmmmHOmwm
mam Emma mmm. mOm MOZ<OHmHZOHm LO memme MOM mmm<DOm MO Ebm mOmmm

..m mqmdfi

82

 

818.8.” N .38ng 3% - 3%
SN NNNmN .mmmmm 8% 8:8qu
3 .N «$2.52 NNN 239303 3% 89:82-2
8:: .mi N 38.8. .omN 3% .. 3%
NNN 2 m3 .30?” H3% 83285
mo .2 SEN .NNN NNN N88 .mmNNN dmm aofimzéma so“ Ea
EomogNN N 3:: .NVN 9% ... N3mm
EN :03, 483 43mm 385
om. ONO”: .NeN SN ENS .mmmmm 8% 822 85.22
483 .3 N N338? 3% u 9.3
SN ENE .mmmmm 92% 830885
NN. 8N3 .EVN NNN NONNeNmeNm Hamm 89:22; V3,.
$43.82 N NSNvam 4% n 3%
NNN NONNN .Novwm 3% 830985
NN. 8mm: .meN NNN NNNVN 52% 4% 89:32-9 a E
SooNSmN N NONQNK 9% - 43mm
3N $EN.mNNN© 3% 9.685
mm .H @2353 SN 238.828 Namm :82 89:22
h 39.6w Eocoofi moumsvw Ho 83m Noam “mom. m~nmflm> mflmofioaxnm
c002 yo momemmm “259$de

 

835600 .. a 3mg

 

83

 

@3868 N NENNNS dmm - 3%
NNN ENE .NNvom H3% 8:883
mm. emNS .:N NNN N23 .Nooom 9% 8562-8 a NE
SENSE, N 25$.on 8% - 3%
2N «EmNNONNo 3% 88mm
2 4 £va .NmN :VN mmNNSNNONo 3% :82 8522
$34.88 N 339N314? - 3%
:N 3:... 58$ 3% 8828.:
NM; $63.43 NNN 29.5638 3% eofimzbsbt
42.3.2.” N 3:3.sz dmm - 3%
NNN 53 .826 3% 830983
N; 25383 NNN 843N308 cmm 89:22-8 9: Ea
moomNNNN N NNNSNS 3% - 43mm
2N 848 .32}. 43mm 385
S .N SNNNNZ SN NNNoN .235... N33 582 8522
m oumscm Eocmofi mohmzvm Ho Esm 5km “woe. @3353, mflmmfioqxnm
smog wo moohmofl unoccoaom

 

85:19:00 i ..N mums“

 

84

 

808.4ch N 22:58 3% .. 3%
mvN $me .3va vamm muommm
34 3.0365 SN 33meva mamm 322 @0522
Hmmmm .EN N monomNVm 3mm 1. m3mm
:NN Hmmmm .5va m3mm coflomumg
mm; wNmmodZ mmN mzmwémw; 3mm cofibEi M3
ommmdeN N 326.03» dmm u Tumm
mmN mime 6mm; H3mm qofibmngfi
MN J «3% .EH \LmN Soon Nam; dmm cofimfiug How am
mmmnmNoN N :02 .m? msmm ... vsmm
2N 3N8 .8sz 43mm 868mm
mm. mfiowoJHN :NN mmfimnémmom msmm 522 @0562
N38 .NNN N 230.3% 3mm u 3%
EN 933 .Emom mamm cofiomumfi:
mo .H owvmmdaN mmN AmmonNNvom H3mm cofioEIEHmm
m mumsum Eocomfi mmuwscm mo Esm .Hobm “mom. manmtm> mflmwfiomfim
G032 Ho mowumom “coccoqmm

 

coscficoo II m mam/5H.

 

‘_

85

 

ONNNNNNNH N SEN-.NNN 9% - 3%
2N 83.4.83... 2% 88%
NN .m 42.24.53 SN 03$.2NNS 9% 582 8.82
828.2: N N888: 5% - 9.8
:N 08.3 .HNNE, 8% 88882
86 $2323 NNN $284.33 23mm 8.82; 25
NNNwo .2. N 22.2 .2: 88 - 3%
NNN N23 .428 22mm 88882
mm. 2wa .va RN 328 .38 GS 8282-02 a. .E
223:2. N 883.3 N38 - 43%
MN 283 .322 M5% 888
2. £08 .NNN SN 82: .328 N38 282 8.82
@886: N NBSNNN 23mm. - 88
SN 32: .328 9% 88882
8.. NS: .omN NNN mom: .HNNS 23% 8282-2 0:5
88482. N SSNNNN dmm - 23%
NNN mom: .HNmmm 3% 88885
NN .2 223 .NNN NNN Nommm .NNN?“ amm 8282-9 08 E
m 89.6w Eocoopm mobmscm mo 85m 52m ”.me bang-20> mfimofioxflwm
C002 MO wmmnmwm HGOUGOQQQ

 

voscfisoo II m Ema;

 

6
8

 

ES: SEN : 0:2 .2:N 4% - :38
:2 @8358? 23mm. 38:2
3:9,: 433.8: om: NNSNNNNNN dmm 582 cm a NE
38.8: 2 3er.8: dmm - 23mm
3: 823.08% 23mm 388
382 omooo .EN om: 2.2:: .2:NN 8.8 58.2 cm as NE
NNNNNNN: : NNNmNNNm: qmm - 3mm
:2 N88222:: 3% 38:2
.3183 .N: NSSNN: O: 83.23: dmm 582 cm :8 NE EN
2:2 .mSN : 2:: .mSN 4% - 23mm
1: «.88 .83.” 23mm 888
4448?: OOSNNN 9: 3.9: .3on 4% 582 0: :2 NE
BN3 :2 2 ENE .SN .8 - :38
2: 223.03ch 23mm 888
mm .N 8N2 .NNN 0.: 889$:me dmm 582 c: 08 NE
NNNSNNE : NNNSNS: 4% - :38
E: 830 .memvN 3% 38:2
.143 .m 2:8 .8: 0.: N88 .8va dmm 582 c: :8 EL 8N
m 895m Eocooum moumSUm Mo Esm .Hotm “mom. o328> mfimofiognm
cwmz wo moo-2.85 “coccodofl

 

88:80 - .2 59¢

 

N $222.32 88 - 3mm

 

285$ 2N NNNNTNNN: 243mm. 8822
mm. N288 .NN 22.N 022:3: Namm 22282 8.82
$82.22 N 222 .mmN 23mm - Namm
22.N 32% N8: 238 8888222
NN .2 ENE .NN NNN $28 .NNN: 23mm 8282-2 V2.3
2:8 .222. N 23.8.22 dmm - 23mm
NNN £228 .NNN: 23mm 822882222
8. 2822me NNN 2483.222: dmm 8282-02 N V222 22V .82.
NNNN 2:22 2 NNNN 2:22 8% - 2.3mm
02 38-288..” 243mm 88222
070 ......02 N2 N2: SON N22 832N222 9% 22282 8282
SNNNNV 2 SNNNEV 23mm - 8%
N22 @2222. .2N NNNN 9% 822882222
NN. 222: SEN N22 NQNNNNNNNN 23mm 8282-2 V22:
22;: .32. 2 £2.82. 4mm - 23mm
222 mONN .NNNNNN 23mm 822828222
NN. NmNNNNoN N22 omNmNNommN 4% 8282-02 2 2m .8
.2 396m Eocoobm moan-56m mo Esm 2022-2. Ewe @3328, "2892209222
2.22822 Ho woeoom 2223:2930

 

 

 

8252280 - ..2 22mg

88

.moo .. VQ

«2&2
.222 . v22
2;...»
.mNo . Va
232..

.m 332222.093 20.2 mm2msvm 20 Sam. 20.2.5 @222 Om? 3 322.226

Km 0.2 om mmmmq 220 63223922 @222 mommUSESm 20
3mg 9: @222st moflcmnomfi @222. 95 mm2msvm we 85m 202.8 @222 m2m $9: 22033 202, .53. am. @222. 20 22220.2 @229

 

 

2
3252.? N 2822.32 mama .. 2.3mm
3N NNNNN .vmmNN 3% 382.22
NN.. 2.25222 .3 22VN Nommm .222me Nsmm 332 2522622
822222.022 N $213.23 23% ... 9%
22...N Nmmmw .222va Ngmm 2522988222
2 .2 2NNNm.©m NNN NNN22 .NNONN 3% 26226222 22>
$322.82 N $225.28 4% n 23mm
NNN NNN22 .NoomN 23mm 2522028222
3 .2 NNEO .2; .NNN 238 .NQNNN cmm 252282-02 m 22>
“2 895m Eovmofi mewsvm 20 Sam 2022m 2mg. $33232 m2mmfioa>m
22.822 20 $522.50 2.289.5me

 

6922232200 II m mam/Nd.

 

282.22

 

was. 22222 2.92222
8322.322 N 82:22.22 23mm Nsmm
N2.N momma .222 28 mgmm 2822822222
.25 .N 3222.228 22.... NSNN 322.2 22mm 2.926222 22>
2N2.N2 .NNN N NNNNN .NE dmm 3 23mm
SN NOSN 532.22 23% 2822..-.28222
2.2. .2 NNQNN .2mN NNN 32% .N2o2.© amm 2522532302 2: 2.22
mpUmwwm
car... 222022 2.822222
9
8 SEN 2.2: N 332.252 23$ Ngmm
EN 25va .3222 msmm 2822298222
......222 .N 22.2 .NON 22.N 9.2223, .2382 22mm 2522822 22>
2:28 .NON N $80 .82. dmm a 23mm
22.N 83m .2283 23mm 822882222
NN. £32228 NNN 283.832. cmm. 252262-02 20.2 222 22 62
n2 998m Eovmmfi mewsvm 20 Sam 2022m 2mg. wEmCm> mHmmfioEwm
2522 20 mmm2mmm 2222225me

 

 

 

NAN Hmmv mHOmmmm 23.22 QOEHME 02d. 5.55:3. UZHQ<mm .mOZmUHAquZH 02.2.2.
QOmHmz 024. E554. meEmm ZOHHOémHZH Mme OH DmHfimm mmmmmHOmwm
mam Emma mm; mOm mOZ<OEHZUHm m0 mHmmH mOm mmm<DOm m0 23m mOmmm

0 mg mflw

2"!

90

 

$2.2m .mNN N 223N903 Namm - 2.3%
EN 22.8.8212». 2.3mm $82222
22..N 2.2.8 .82 N2.N om2.NN .N2NNN Namm 222222 2.22222
$222.8 N 022% .3 23mm - mgmm
N2.N omvNN .N2NNN Ngmm 2522882222
N. 32.22 .82 22.N 02.2mm .om2NN 23mm 28222222622
22282.32 N NNNNNNE dmm - 23mm
22.N 222.23 .22 m2NN 23mm 2822882222
mm. 282 2.2 NNN $82.2 .SONN amm 25222222222 28 .222
Noo2N .32. N 882.2222 Namm .. 3%
m2.N 822.22 22$ 2.3mm 88222
222 .2 2228 .EN m2.N 28$ .Noooo Namm 22222 2822222
2822...: N NNNNN .NN2.2 23mm. - Ngmm
m2.N 28$ .Nmmow Ngmm 2822822222
2.22 .N N852 .EN 22.N 2.282 .NNNmo 23% 28522-2 22>
8322.82 N £82 .oom qmm - 23%
22.N 2.222222 .NNNQ 23% 822828222
3. N88 .EN mNN 222222823212 dmm 252222-02 22 2.22
m 983% Eovmem mm2msvm 20 Saw 2025 22me @3323, mHmmnqu>m
c0022 20 $2.52me 2222225an

 

Umscficoo I: U mamfiw

91

 

2223 .22N N 22222.32 9:22 a 2.322
EN @2222 22.222 2.322 38222
22 .2 2822 .mmN N2.N 22222.2 2.2222 .2322 222222 2.922222
2.22N2N 62.2 N 222.8 .mmN 23% .. 322.
EN 2222.2 2.232 3% 822828222
22. 222222 .222.N 22.N 22222 .oNomm 23% 262222222 2.22
£32..Nm N 0322.252 3mm .. 23.22
22.N 22222 .2822 23mm 2828822222
2N. 2.vam .N2.N mmN 2822 .2222 2.22 822222-02 22 2.22
22222.22 N 3222 .2NN2 N322 ... 2.322
m2.N 2228 .mNNS 2.322 302222
22 .N 8222 .moN N2.N A282.22. .382 2322 222222 2522222
8222.222. N 2222.2 .222 2322 u N322
N2.N 82.2.2. .8252 N322 2522828222
2.N .N 2.2Nmm.NoN 22.N 2228 .2822. 23.22 822222-05 .222
22823 N 22222.2..22 4mm - 232m
SN @2822 .2822. 23mm 2522822222
22. 22NN2.2.22N 2NN 222.8 2:222. 2.22 2522222221022 922 2.22
.2 895m 2220226222 mewsvm .20 Saw 2022m 5.3. 22322222; 3.2939232
222232 20 mmm2mmfl2 222022ch220

 

 

6922232200 I: U mam/2H

92

 

82.22.2222 N 222.22.222.22 23222 .. 2.322

m2.N 2228 .22N22 2.322 228.222

2:23.22 .2 222222 .22N m2.N 22222 222.8 2322 2222.22 220222222
2N2: 2.22., N 3222.222 2322 .. 23%

m2.N 22222 222.22 2322 8228222222

2.2..2 22.22.N .ooN 22.N 22222 22222 2322 282222222 22>
222.2622. N 2.82.2602 dmm ... 2322

22.N m2 222 222% 2322 2828222222

22 .2 22.28 .22N 22N 22Nmm 22222 2.22 252222-02 9: .22

2.282.222 N 222202 .2222 2322 .. 2.322

m2.N ...SNm .2222 2.322 22.8222

2222.2 222.22 .moN EN 2232 .822. 2322 222222 2.622222
822.23 N 32222.22 2222 .. 2322

EN 2282 53.22. 2322 8228222222

22.2 82.22.NON 22.N N2N22 .3222. 2322 25222222 2-2>>
2.22: .222 N 8222.222 .222 - 23%

22.N N232 .2322. 232m 25228222222

2.22 .2 2.2228 .20N 22N mNmom .2322. dmm 2822222262 202 22
..2. 896m 222025822 mm2mzvm 20 Sam 2022m 22mg. @3323, m2mm222oa>m
22me 2o mmm2omm Emuvcmamm

 

 

 

220222232200 ..I O mama?

93

 

©NN©2N222 2 oNNomNmZ dmm. a N322.
E2 3222 .2222, N322 3.02222
2.1.2.2 $22.2 .omN 222 22.23.3222. 3.22 22.222 02 22 2.22
20.22.22.222 2 2222.2 .222 2.222 .. N33
2222 22.2.22. 2222.2. N322 228.222
2.... S22N SN 22 022222 2.22.2. cmm 2822 02 us. .222
N222.N .222 2 N222.N .222 .qmm .. N322,
222 2222.22 .2822 N322. 282.222.
2.2 .2. 22.222 .22N $2 022.82 .2222. 2.2.2 5.22.2 02 28 2.22 2.2
2N2222 .2222 N mvaN .22NN 23mm 2.322
22.N mmNN2 .0228 2.3mm 2202222
3.22 .2. 2.222 .NON EN @282 2.32.2 232.2 22222.2 2.922222
22.222 .222 N 22222 .222 23mm a 2322
m2.N 2282 2.82.2 .2322 822882222
22. 222.202 .EN 22.N 8222 .2222 232.2 26222222 22>
22222 .on N 2222.2.N8 dmm - 23%
22.N 8222 .2228 2322 2828822222
2.N .2 02222 .22N 22N 2.22.2.2 2.2.23 2222 82222.02 22 .22
...2 39.5w Eovmmfi mm2msvm 20 Sam 2022m 2mg. $33232 m2mmfioa>m
22me 20 mmemQ 2252225me

 

229222222200 ..I U mam”?

94

 

2222.22222 2 2222.22222 2322 - 2322
222 2222 .222222 2322 2202222
2.22 .2 2222.2222 222 2222 .222222 2322 22222 2922222
2222 .222 2 2222.222 2322 - 2322
222 2222 .222222 2322 252825222
22. 2222 .2222 222 2222 .222222 2322 29222222 22>
2222 .2222 2 2222 .2222 G22 .. 2322
222 2222 .222222 2322 2822222222
22. 2222.2222 222 2222.222222 d22 285222-02 2 22 .22
22222.2: 2 22222.222 q22 - 2322
222 22222.22222 2322 228.222
22. 22222 .222 222 22222 .22222 C22 22222 02 22 222
22222.22 2 22222.22 d22 -2322
222 22222 .22222 2322 2202222
22. 22222.222 222 22222 .22222 2222 22222 02 92 222
22222.22 2 22222.22 G22 .. 2322
222 2222222222 2322 228222
22. 22222 .222 222 22222 .22222 q22 c2222 02 28 222 22
h QHMSUW EOUQQPM mQHmSUW MO 85% hoghm “mmH @HQMflH@> mflmOSHOnrnm
2202.222 20 $5325 222902202225

 

229222222200 1: O mam/5H

95

2223 $2.22 6222an 2022222209: @222 @2222 292222.52 20 2252 20.2.5 @222 9222 22222222 22022.23 .202 22222 :22 @222 20 8.202 92.2.

Km 02 om mwmmq :0 2522222223 9222 mocwoflggm 20

2

 

 

22222 .222 2 22222 .222 2322 .. 2322
222 2222222222 2322 228222
22. 22222 .222 222 2222222222 2322 22222 2922222
22222 .222 2 22222 .222 2322 .. 2322
222 22222 .22222 2322 822022222
22.2 22222.222 222 22222.22222 2322 282222222 22>
22222 .2 2 22222.2 2222 a 2322
222 22222 .22222 2322 2022028222
22 .2 22222 .222 222 22222.22222 d22 222222-02 2 22>
22222.222 2 22222 .222 2322 .. 2322
222 22222 .22222 2322 228222
22 .2 22222 .22 222 22222 .22222 2322 222222 2822222
22222 .222 2 22222 .222 2322 .. 2322
222 22222 .22222 2322 25220222222
22.2 22222.22 222 2222222222 2322 22222222 22>
22222 .22 2 22222.22 2222 - 2322
222 22222 .22222 2322 262202822
22. 22222.22 222 22222 .22222 3.222 202222-02 2 22> 222 .22
.2 929.22% Eovmmfi m®2mzvm 20 83m 2025 222222. 2232222222 2229220222222
2.802 20 292.2me 22ch25an

 

1222222200 1 w 222.22

96

.moo . Va

2222.2222222

.222.v2

21.222.
.20 . v22
um.
.2962

we . @222 22 2222202222292 2225 2202202223222 2202222252 22$ 9: 22022225 222 3220222 22 .202 22222222222 20 :32 20.2222 22.2.20

.2 2229220222222 202 292223222 2o :52 .2028 9: 022m 22 22222.22

2202202282222 2202223212 v2.25 2222202222292 @222 20 mmswomn @323 20c 2.2.295 22033 222228 220222me

 

22222222222200 II 0 mqmdfi.

97

mo. m5 22 20222202222292 2on2

.222nzv 22252222U 2222 30222 22 2 222 .222qu 22252222V 2222 2:622 22 2 222

.>2®>22om222m2 622222262: @222 mm. . 22:22 om . 20 2222222022300 2202222292200 .3222

H

 

 

22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 2 22>
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 2222>
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 2 22>
22 22 22 22 22 22 22 22 22 22 22 22 22- 22 22 2 22>
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 20222
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 2222
22 22 22 22 22 22 22 22 22 22 22 22 22- 22 22 02222
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 202222
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 a 222
22 22 22 22 22 22 22 22 22 22 22 22 22 22- 22 05222
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 292222
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 a 222
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 022222
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 “U2
22 22 22 22 22 22 22- 22 22 22- 22 22 22 22 22 m:
MMMM
c. c. 70 TL m U2 w m. u. w m U2 2
1 O 1. O

 

 

HHZm§H<MmH mmDHqu HEB ZH mHZmQDHm mOm mmqmfimfix ZOHmmHHmO Q24

.DHEMQ¢O< :DZHQ2qmm JANDHOMAAMHZH 0203322 mZOHHEWEmOOmMHZH

m mam/22H

98

mo . .2222 22. 20222202222522 20m

{22232022223 69222222222 2:2 mm . vcm 2N . 20 2222222022300 2202222292200 \2962

222nzv22202222222222m2 2222‘222nzv22202222222m>022m: 2222

2

 

 

22- 22- 22a 22- 22- 22a 22- 22a 22- 22- 22- 22| 22- 22- 22- 22- 2
22a 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 2 22>
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 2 22>
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22! 22 2 22>
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 2 22>
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 20222
22- 22 22 22 22 22 22 22 22 22 22 22 22 22- 22 22 2222
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 0:222
22- 22 22 22 22 22 22 22 22 22 22 22 22 22- 22 22 202222
22" 22 22 22 22 22 22 22 22 22 22 22 22 22- 22 22 22222
22- 22 22 22 22 22 22 22 22 22 22 22 22 22- 22 22 022222
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 202222
22! 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 Am 222
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 022222
22 22 22 22 22 22 22 22- 22 22 22 22- 22- 22 22 22 22
22 22 22 22 22 22 22 22- 22 22 22 22- 22- 22- 22 22 02
22' 22 22 22 22 22 22 22 22 22 22 22- 22- 22- 22 22 o2
MMMM

A: cw 7” TL w I. o m u” m m mu m

1. 01. O

 

 

HEEUOmm MmZZHMm mEH 02328 mHZmDDHm mOm mmqm<Hm<> ZOEmHHmO Q22.

2OH2mQ<O< \OZH04mm JdDHOMAAmHZH 020224. mZOHHSMmmOOmmHZH
H mqmdfi

99

mo . @222 22 2022202222822 2202

92233209228 62222222292 22.222 22 . U222 22 . 20 2222222022300 220222228200 2222262

.22 2H7: 22282222 2222 262222 22 2 .222 ‘221,222 2802222 2222 25222 22 2 222

2

 

 

22- 22- 22- 22.. 22a 22 22." 22.. 22a 22- 22a 22a 22a 22 22 22.. 2
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22.3
22.. 22 22 22 22 22 22 22 22 22 22 22 22 22a 22.. 22 22.22
22... 22 22 22 22 22 22 22 22 22 22 22 22 22 22: 22 22:5
22.. 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22:5
22.. 22 22 22 22 22 22 22 22 22 22 22 22 22- 22a 22 2822
22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 222
22a 22 22 22 22 22 22 22 22 22 22 22 22 22- 22.. 22 as 22
22- 22 22 22 22 22 22 22 22 22 22 22 22 22.. 22- 22 28,222
22a 22 22 22 22 22 22 22 22 22 22 22 22 22.. 22- 22 2:22
225 22 22 22 22 22 22 22 22 22 22 22 22 22 22- 22 05222
22a 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 28222
22.". 22 22 22 22 22 22 22 22 22 22 22 22 22 22- 22 22,222
22» 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 92,222
22- 22 22 22- 22 22 22 22.. 22 22 22 22 22 22- 22 22 22
22 22 22 22- 22 22 22 22a 22 22 22 22 22 22a 22 22 02
22- 22 22 22 22 22 22 22 22 22 22 22 22 22 22 22 02
L 2 2 m 2 .2 2 2 2 2 2 2 2 2 2 2 o
9 8 z I m I m m n, w m U 2
1. O 12 O

 

 

12422002222 mmD>>OmO Mme UZHMEH. mHZmDDHm mOm mmqm<§<> ZOHmmHEO Q24
2OH§mQ<O< 202202.232 2A<DHOMAAMHZH 020222. mZOHHfimmmOOKMHZH

H mg mam.

APPENDIX B

SKINNER PROGRAM

100

 

{3

‘ OHM’S LAW AND POWERS or TEN

 

,N_,AVAL AIR TECHNICAL TRAINING COMMAND

L a

L'

 

O
Acknowledgements
This program was adapted by the Naval Air Technical Training center,
Memphis, Tennessee, from a larger one developed under contract No. N 61339-
f ' 1000 between the U.S. Naval Training Device Center and Teaching Machines,
Inc° In a letter dated 29 December, 1961 from the U.S. Naval Training
Device Center, unrestricted use of this material by any governmental agency
{ was confirmedo Appreciation is expressed to the U.S. Naval Training Device
} Center and to Teaching Machines, Inc. for assistance in making the original
: program available.
)
I
i
\
i
i
\
x

 

v

OHM'S LAW and POWERS OF TEN

1. If your answer to a question is a single Word, you will see a sincle

 

line, like this:
George washington was the first president of the United.______________..

2. If your answer is two words, you will see two lines:_________

 

. George Washington was the first president of the .
3. If your answer to a ouestion is in your own words, you will see a

dashed line: ___________________

 

Who was the first President?

 

What did he chop down?

b. When the answer is a number, symbol, or letter, it is marked like

 

this: ( )

5. A multiple choice Question has a series of items separated by slashes.
George Washington was the (lst/hui/Brd) president.

6. Some questions are test cuestions. The answer is not given, but you will

see the word "TEST” where the answer should be.

The steps in this program are fixed so that you should be right most
of the time. If you are doubtful about how to answer a question, read the
whole step over. It is important that you read and understand every word of
instruction before you answer. The answer to the question is on the next
page. For the answer to any question coming at the end of the book, turn
the page -— and then turn the book upside—down. The answer to your question
will be in the appropriate answer Space. Continue as usual. If you write
a wrong answer, study the correct answer carefully. Compare the incorrect
response with the correct one. Careful comparison of answers will help you
to move smoothly through the entire course. Always read the answer in the
answer frame. Do not neglect or rush by the answer frame, even though you
may feel certain that your answer is correct. COMPARE.

 

 

I.

' . . Georg Simon Ohm was the 'first man to study resistance, and the unit of resistance is named for him.

Units of R are called

 

35 . A news r

current and voltage (either order)

 

36.

\
Ohm's Law in symbols states I = H.

 

71. Answer (E)

(I)
(R)

 

 

72.
Here is an important word you should know: ”ampere" (AM-peer).
One coulomb passing a point in 1 second equals 1 ampere.

How many electrons flowing past a point in 1 second equal an ampere? { )

 

107 .. A nswer

1o2

 

108., ‘
(2 x 10‘) (3 x10“) = 6 x105,because2 x 3 = GandIO‘ X 10“ = 105.
We check the answer by changing the numbers back to the originals.

(2 x 10‘) (3 x 10‘) = (20) (30,000) = ( ) 600,000 = 6 x ( )

1.

 

09€

 

 

( ) = ,01 x 3.01
L01 = ,0: x ,01

'siuauodxa an; ppa ‘01 J0 sxamod ss passadee 9.12 min 5.10132} [fidmnux OL

'LOI

 

Lox x 92:1 000‘009 ‘SL
JGMSUV '901

 

 

 

( ) 30131532'
’4ng F22“ 2‘9 ( ) manna
gtq'uav“) 9.3"” ( ) Jule
loquIAE eumN
°IL

qwomoo ( 83'9 )

.IaMsuV "0L

 

pu’e ‘ ‘aou‘eistsax uaemiaq diqsuonmax aqi saAIB nun sauqo

“SE

 

'um aouazuas
sq: buot moq mom! Luptp aq pun uad sq: m; S'DM unul p10 em esnnoag

.I emsu v \f-qg

1. Answer

ohms

 

2.

Ohms are units of

 

36. Answer

 

37.

Georg Ohm found that current varies directly as and inversely as

 

72 o A nswe r

TEST

 

73.
If you say one coulonb per second is passing a point. it is the same as

saying ( )_ of current is flowing past this point.

 

108. Answer

 

600,000 105
, “109. . .
(1)10WX10=10 o 9 : 0(A)105X10-3:102
(2) or (10,000) (.01) = (100) (B) or (100,000) ( )2 (
This checks with equation ( 1 ). This must check with equation ( A ).

2°

°SE

 

'01 1° umod< )=( )- ooo‘zc x 008‘:
‘saequmu mufiuo an; auundmmu Aq uonuoudmmu mo nooqo Ann: 0M

'90I

L01 x 9n. = ‘01 (3': x at)

= (H c)0! x (m x m) = (.01 x an (m x 8'!)
= (,0! x 2'8) (,0! x 8'8) .xensuv '501

1 a uonoap "at x ( )

'01.

( "or x are )
' .xansuv '69

up. doxp am cum at: In- 4“!

 

V‘WE
( 1) aaudum wanna
( 3 ) salon aficuon .Io nus
( H ) “NO acumen”

s'ioams sum mum “MW "if

2 . Answer

(. Resistance

 

3.

The symbol used for resistance is ( ).

 

37. Answer

TEST

 

38.

1 ohm 1 ohm

In which circuit does the most current flow? ( )
Which has the highest emf? ( )

What is the value of R in A? ( ) ; in B? ( )

 

73. Answer

(. ( l ) ampere

 

7140

The number 6, 000, 000 can be written 6 x 10(

 

109 . A nswer
. 001 100

Note that .001 = 10'3 Note that 100 = 102

 

110._

(2 x 10‘) (4 x 10'”) = 8 x 102 = 800; becauseZ x 4 = Band 10‘ x 10‘2 = 102
Check, using the original numbers.

2 x 10‘ = 20,000; 4 x 10'2 = .04; 20,000 x ( ) = ( )

3»

oflg

 

 

( '){.10m [12 moqs)

'000 ‘zc x 008 ‘2 math Kidmnw pus 01 J0 smmod 01 asatn afiueuo

L01 >< zs‘sz = L01x(z'9 x 91>)

(:01 x ‘01) (3'9 x 9'17) = (€01 x 2‘9) (tor x 9'17)
'01 30 Siamod item 01 main eBueqo ‘isxu

‘ooz‘s x ooo‘sb steqwnu aseui Kidmnw 011mm am

.50-E

 

(I)

lei/is“ mg “1701

 

‘puooas .Ied suonaaIa ( ) 2 puooas .xad qwotnoo auo

'suoxioaio 3101 x QZ'Q = qtu01noo auo

'69

 

qtuopioo

.Iamsuw '09

 

'a1qsi aux aiaiduxoo pus A1163

 

 

I wanna
( ) 9110A ——
( ) — eaumsisea
m m sown
"it

(100.1109 5; Jaime) .mnmiii ‘OJOUI

.tamsuv mg

3 . A nswer

(R)

be
Current varies directly with the voltage and inversely with the resistance; this is called Ohm's Law.

Copy Ohm's Law.

 

38. Answer (B)

 

 

 

(B)
1 ohm
1 ohm
39.
‘ I c _ e
Ohm 8 Law written in words is. current — m
: Ohm's Law states in symbols: ( ) =
711. Answer
( 6 )
75.
One ampere equals ( ) passing a point each second.

 

110. A nswer
. 04 800

 

111 .

(2.4 x 10‘) (2.0 x10°) = 2.4 x 2.0 x 10‘” = 4.8 x 101°
(3.0x 10’) (4.0x 10“) =( )

h.

'CC

 

 

 

 

( ) = 001'
_’qot.'yx
(,0! x 9!) _
' Jansuy '{01
If
$70.31;,
43/ ' ° I = smmm mm x 92's
’99
3:01

Jemsuv 'Lg

 

W

nna
qécmmflfimm: '
9 “wine-nu ual so own a! um um «and x 11.. momma uaafi a

qflnonu (wanna) mo" muons am on: ‘01:») smut! minnow can ”out at“,

 

Ti

H)
JQMBUV '3;

h. Answer
Current varies directly with the voltage and inversely with

the resistance.

 

 

5.

The law relating to voltage, current and resistance is law.

 

39. Answer

 

E
I ‘ if
hO.
1

In which circuit is the voltage the greatest? ( ) ‘ ampere
Which has the greatest R? ( )

O Q
What is the current in A? ( ) in B? ( )

A

1
ampere -
2 ohms
O 0

B

 

75 o A nsweI‘

( 1 ) coulomb

 

76.
Very small numbers can also be changed to powers of 10.
Moving the decimal point 3 places, .\0(ll’ = 1 X 1/1,000 = 1 X 1/103_

.01 = 1 x 1/102 .00 = 1 x 1/10‘ )

001
a \ /

 

111 . It nswer
12 x 10'1 or 1.2 x 10° or 1.2

 

 

112.
(14x1m)(10xiw)= ( )xaoxw(
A. B

(3.0 X 103) (4.0 x 10") = ( )
C.

°Z€

 

 

( ) = (9-01 x 9'1) (901 x 9) (.01 x z)
:sxoioe} asaq; Kldmnw
.01 xo9 = (901 x V01 x E0t) (9 xvxc) = (90! x9) (*0! X») (:01 xc)

'paux’eel aAeq am spoqieui sq; Aq pandmnw aq Ram ‘01 10 sxamod s'e passaidxe ‘sxoio'e; [axe/lag

'EOI

 

8'6 = I x 8'6
= 001 x 8'6 = cOI x a.01 x (9'1 x 2'9)
Jansuv °30‘[

 

'qwomoo I = ( ) x 83 '9 s9 uamm aq ptnoo sum.
'qwomoo I = suomeia 000 ‘000 ‘000 ‘000 ‘000 ‘oez ‘9

(mm —00)[) .. 'qw01n00.. :moux plnoqs noK pJOM iu‘euodun us s; aJaH

 

0L9
i do 001 '9 S01 ‘2
901 '9 801 'z

201 '9 :01 '1 .zensuv '99

 

        
 
 
 

3110/1 L ' 5110A L ‘v'

 
       

( ) 1.511% 190w 91119811 11mm uoium

'ZE

seiadum 01 seaadure 001

 

eiedum 1
aJadunz 1

( El )
"a; Jensuv °‘[E

O

5. Answer
Ohm's

6.

t..-..-.u.-._-uL........-- . .... 4......

140. Answer (11')

(BE:

1 ampere
hl.

E stands for __ _

I stands for

R stands im- ‘ ._

ma?“
(s)

 

' 77..
625 = 6.25' x‘ 1'1"“

\0325 --‘ 6. :53 1". 1., Ii.-

112-. Answer
A. 2
'c. 12 1; :0“

W113. .

10‘ x 10“ = 10”

We know 013210” =- ..

"

-u-n—n-n. A"n:-| . . .

if the Voltage is not chaizged, 21.-iii resistance (load) is increased, the electron current flow

 

.i 1.113;; ere

 

 

 

 

4 ' — ‘d 3‘1 ><( )
1.2.5: :2.55x ( )
l "a

 

22.-.3 -,: .5301 = ( )

 

L931.
msuv ~15:

 

) = (5:01 x 9'!) (‘0! x 2'9)
stem; asatn “damn

I = 001

”3012

 

(:01 X 8'9)
.mmsuv °'[o'[

 

( )XS'I z: s"! ‘9
( )x z = ooo‘ooo‘z ‘9
( ) x saw a zsv‘i '9

( )xsz'x = ooo‘ozx '2
( )xZQ'f' = oza‘v 'z
601 x em = ooe‘zt '1

'99

 

0

JGMBUV 959

 

“12:33] V

medium 1

 

Mi “1 8M0” momma m
M! “1 9M0!) wanna mum

( ) (Juno woufim mu mm mm

) w wow mu mi amfln imam

 

9mm 99
sumo g1
soaodum g

.Ismsuv '0:

“€67.—

“ ‘ "hairless '—

6. Answer

decreases

7.
The single letter for resistance is R. 2.

The single letter for voltage or emf is ( );

 

for current, the letter is ( ).

E = emf or voltage
I = current
R = resistance

h2.

Write Ohm's Law in symbols. ( ) = H

 

In the diagram, I = I )amperes.

 

77. Answer
(10")

( 1/102 )

For ease of writing,‘ we say that 1/102 = 10'?” ,
and 1/10“ = 10'3

then 1/10“ = ( ).

 

113. Answer
10° or 1

11h.

(8x103)(3x10‘3)=ax3x103'3=24x10°=24x1—24
(6x103) (2x10‘3)=( 1

I

70

‘OE

( ) 0.19mi 11
( ) )1
_0I u!
c ___.—
( ) w
0! :10 REMOd SWVN TOHWAS .

°L€I

 

'e c_oI x0 = sdummiuip

.IBMSLIV °9£I

( ) = (9-01 x 0'1) (.01 x 3'!)

1.10102} 3531;; Axdmnw

'10! X 98'8 = (9+9-)OI X (2'1 x 9'3) = (.0! x 3'1) (9-01 x 8'z)
25.10192; asatn AIdmnW

'IOI

( a.01 )
Jemsuv 'QOI

x . = .
( )Ol 99 I 99 I

sand 0 iutod reuiiaap am paAouI am 2001 X OH = OH

 

 

   
 
   

 

 

'59
LSSL
.ISMSUV ”79
_

+ _ ( ) " 3
‘( ) = H
l 1 1 =1
sures unions swim
‘OE

( v) ( 9)

.IBMSUV '63

7. Answer
(E )

(I)

 

 

8.

Ohm's Law relates to , , and

 

 

E
1‘s

I = 1/2 ampere

 

i 112. Answer

 

78. Answer

D (10”)

One ampere (one amp. or one a.) equals one coulomb per

 

1111. Answer

rwflr‘m
115.).

 

115 9

When we multiply a string of numbers, we add alZ the exponents and multiply the numbers remaining.
(2x102) (4x 10“) (3x1o*=) = 2 x 4x 3x 10(“‘”> = 24x 10"

(8x103) (6X10'z)(2x13“\=( )( )( ).x10(3-2+4)=j )x(

30

.63

'3 ( ) = “01“!“ ‘ ( ) p234 s; em p

'9 90! x v = qomm ‘sdureflaw 0 pear s: 2w 7
'9€I
( 9 - )

.Iamsuv '5CT.

( )= 9-01x .01

‘01 yo SJBMOd asaq; Aldntnw

 

° OOI

 

201 x 91 = (2-01 x .01) (c x 9)
Jensuv '66

 

'( )OI X 69 'v = “[631, uain ‘saoeld {ewioap g BunguI Aq

A
c01 x L'ZI = 'OOLZI

 

 

"79
V
8
z .Iemsuv_;£_9_

 

'saJadum ( ) 01 83“le ( )

(110.1} saseadaep 1 uaq; ‘swqo 09 o; 01, won paS'eeJou; 9! H pm! 5110A 008 S! 3 H
H
' I

3
‘63
H

(a)
.Iemsuv '83

:9"

8. Answer

voltage, current, resistance (any order)

9.
Ohm's Law states:

The number of electrons that flow in a. circuit (the current) depends on the circuit resistance and

the

 

L13. Answer
(I) Ohm's

 

1 ampere
What is the emf in this diagram? ( ) / \V

 

79. Answer

second

 

 

80.
1246.78 = 12.4678 x 10+2
v

If we move the decimal point to the left, the power of 10 is +.

3,579. = 3.579 x( )

 

115. Answer

8 6 2 (any order)

96 x 105

 

116.
(2x104)(2x10°)(4><104)(8x10‘“)=(2x2><4x8)x10 4+6+( )-(
=< )><( )

9o

 

(adoosoxoiui Jaquiauiad) 71 = 0.1on = )01

(

(auoqdufiam .xaquxauiei) w = 233m = 90!

'SEI

 

 

283w

0.101111

ISMSUV ~qg1

 

( )=(z-01xs)(,01x9)

£10102} asst]; Atdmnw

.01 X 9 = (s-e)OI >< (8 X z) = (...-01 X901) (8 x z) = (9-01X z) (.01 x e)
'pandmnw are sums; .xeqio sq) ‘01 10 saamod sq; Suifldmmu Jam!

066

 

901 x (9 x 9) = (.01 x 201) (9 x 8)
.IGMSUV '96

 

' ,0I x H ' ui snnsaJ saoeld ( ) JUIOd lewloap 3‘“ BUFAOW
' gOI X I? 'I "l Silnsal 9303K) ( ) iutod {ewpap sq; BUIAOW
' 201 x {71 in sunseu saomd ( ) iuiod [ewioap aqi Sumow
' (OI X OH "I sunsai aoexd I iuiod [ewioep aqi fiumow
'oon renba sxequmu 999111 )0 nv

o€9

 

(Z)
.Iemsuv '39 '

( ) gswqo

‘— <__
SIX! 0
sauna g -1 “”0 I sdm 9 '1’ 190w 91" seq 11mm 110mm

082

ssal

JGMSUV °L3

 

as

9. Answer

 

 

voltage
~10. _.
MEASUREMENT SYMBOL _U_N_l'£
current ( )
pressure ( )
resistance ( )

 

hh. Answer
( 3 ) volts

 

AS.

Ohm's Law in symbols is I = %—.

Write Ohm's Law in words. (current) =

 

8O° Answer
( 10+3 )

 

 

81.

_ -2
1,246.13( _ 124,678. x 10

If we move the decimal point to the right, the power of 10 is —. 3.579 = 3, 579.

X (

 

116. Answer
4 8 128 x 10°

 

117.

1,000 x 31,000 x .000002 = 103 X (3.1 x 10“) (2 x 10'6) = 3.1x 2 x 101

2,000 x 40 X .00012 = ( ) = 6.2 X 101

=62

10¢

4 ‘V'W‘wfm. f—‘fi'gu 4

Ole

 

l t _ _—_
0000001—9+01— —w
000‘000‘1 _ _ ------ _

I —9-OI_ — fl

"(ET
n
at

“ JSMSUV '{gt

 

( ) = (.01 x 9) x (201 x a)
'01 10 sxsmod sq; Aldmnm pu'e suns; sssq} aimedss
.01 x (r x z) = ( c1,)01 x (v x z) = (:01 x I01) (7 x z) = (:01 x v) (.01 x 2)

'anms psumsl sm s2 ()1 Jo sxsmod sq} KIdDIDUJ ‘psnnedss ussq sAeq suns; sq) usqm

'86

 

ppe
Jensuv °L6

 

'( )OI X 9? '53 “3111““ 9Q pInos 009 (I7

‘SOI X 31 29!!“ 8X“ 1] all-1M 1120 BM SBOBId 9 JUIOd [BWIOBP am 9A0“! 3M H '000 ‘ZI

'39

 

000‘31

OOZ ‘I
.Isnsuv “[9

 

'(mou uonssls)

iusuns sq; _ sq; ‘(m soueiszssx sq) sxoui qu, 'sssesmsp iusuno sq; ‘sssusxoui
8011815153.! sr. neqi suesw qsgqm teammates; sq; s1; KISSJSAUI semen manna naq) ssi'eis meq s,1uqo
'uonse .10 ways u; susoddo suesux aSJGAU'I

’LZ

 

9110A 81
sumo 9

sstduuz g
asmsuv °93

 

 

10. Answer

current I ampere
pressure (emf) E volt

resistance R ohm

 

11.
In symbols, Ohm's Law states I = 11%.

If E = 100 volts and R = 20 ohms, I = 13% = 5 amperes.

If E = 175 volts, and R = 25 ohms, compute the value of I.

 

hS. Answer

emf voltage
current =

————.—— or —————.————
re51stance resmtance

 

 

A6.
Which circuit has the greatest resistance? ( )

Which has the greatest current? ( )

 

 

"0'2".

81. Answer
( 10'3 )

 

246. 810 = 2. 46810 X 10+2 , moving the decimal point ( ) places to the

246. 810 = 24, 681. 0 X 10‘2 , moving the decimal point ( ) places to the

 

ll 7 . Answer
(2x1w)(4xiow nleofi) = 2x4x12x10'1

= 96><10_l = 9.6

 

 

118.
In electronics we have names for the powers of 10.

What would be the exponent of 10 for 1 million? ( )

11.

'98

( ) 01 = (”0111101
( ) 01 = 111111101
( ) 01 019101

W01 = 3611101

(msiu) mu panes 11 Janet nests)

in news
>1 News
IN [Baden

11 = 0.101111
"1 = nuw
3i = Olpl
w = 289111

:ssuieu sq; siegAquqe am ‘suxm oi Jayses Lusq‘i swam 01,

“siusuodxs sq;

)= EOIXZI
( )= a01><zx
031:,01xz1
z1=001xz1

31101‘ M

sstdum g
* *

 
   

 

'ECI

1mm
.Ismsuv °gft

am ‘01 30 sxsmod Buyqdmnm usqm

(.01 x

usmlm sq p1nos

'L6

201) (9 x b)
.ISMSUV '95

ooo‘zI = 000‘1 x 21
0021 = 001 x at
021 = 01 x Z!

31 = I x Z!

a-[g

.IGMSU‘J '09

'snoA ( ) $1 nus sq;

‘suiqo ( ) st ssu’eisissa sq;

‘ssasdunz ( ) $1 iusuns sq)

Imam Sun “I

.98

(iv)
.Isnsuw '53

F‘)

(54S

11. Answer

E - 175 _
1: Ti" 'fi —7amperes

 

l2.

1 = 2.

When E was 100 volts and R was 20 ohms, I equaled 5 amps.

Ii E is increased from 100 to 200 volts, but R is unchanged at 20 ohms, then I = %

E has increased from 100 to 200, R is the same, and I has

= 10 amps.

from 5 to 10 amps.

 

hé. Answer

 

 

 

(A)
(B)
117.
Which is the correct formula for Ohm's Law?
_ R _ E _ R _ E
E — T R — I— I - -‘E‘ I — F
82. Answer
2 left
2 right
83.
1. .0625 = 6.25 x ( ) 4. 36.921 = 3.6921 x ( )
2. .000000125 = 12.5 X ( ) 5. .036921 = 3.6921 X ( )

3. 125,000,000 = 12.5 x ( )

 

118 Answer

 

 

12o

( 6 )

119.

NAME AMOUNT POWER OF 10 WRITTEN NUMBERS
mega 1 million 10“ 1, 000, 000

kilo 1 thousand 103 ( )

-3 1
milli 1 thousandth 10-0 T000—
, micro l millionth 10 ( )

.Sa

'sdumnnm amseaui plnom Jazauxure v

'sdumoaoxw amsnam plnom .xanmureontui v

“ZEI

OI I00000 '
Jamsuv '11];

('01 x 9) x (am x p) :asaq; arcades
'( :01 X 201) (v x 8)
:sun aim sums) .xamo 314; 1110.1} 01 10 smmod am excludes 33.11; am ‘sxaqwnu asain Aldmnw 01,

(90! x9) x (:01 x8)
.96

 

.LSEL

.Iansuv '56

 

'( ) = ooo‘OI 05 ‘SOJaH )sml 000‘01
'( ) = 00105 ‘SOJBH )qu 001
00‘ = I os-‘soxaz o seq I

009

 

(901 )
.IBMSUV '65

 

‘sdum ( ) o; pageaxoap s; 1 uaq; ‘swqo 93 01 pascamux s; a put: 3110.1 001 ye paBueqoun s; 3 n

’sduna g parenba 1 ‘swqo oz sum a we sum 001 sum 3 uaqm
. H

 

i=1
058
sum OI
8310A Z
Beladlb z'

JGMSUV We

6332‘

12. Answer
increased or doubled

 

13.

Current varies directly with voltage.
When E increases, 1 increases.

When E decreases, I

 

h7. Answer

3111-1

 

A circuit component designed to impede or resist the flow of current is called a resistor.

It is easy to remember that a component to resist current flow is called a

 

83. answer 1.10-2 4.10,

2. 10" 5. TEST
3. 107

 

811.
37.91246 x 10" = 379124.6 ' 26.9142 x 103 = ( )
46 x 10‘ = 4,600,000,000

 

119. Answer
1,000 Tfimfiffififi

 

120.

mega 1 million 10”

kilo 1 thousand 103

mm 1 thousandth 10' 3
micro l millionth ( )

13°

”la

'01 10 ”mod all: an
°dule ( ) .10 'dure ( ) st dmoxann v

'dlne “OX 10 'dlne 100' st dmmnu v
'IEI

 

punsnom c
.Iellsuv ‘OCC

( )= .0! x .-01

'01 )0 sxamod asaq; AIdmnw

 

 

056
( 3-01 )
JUMSUV ”T6
‘( ) = 000 ‘000 ‘I 09 ‘90192 9 9W 000 ‘000 ‘I
' c0! = 000 ‘I 09 ‘30193 8 3“! 000 ‘I '01 all} aAoqe

.zaqumu 12m and put 50.192 30 .laqumu 9m iunoa ‘01 30 Jamod e ow; or ;o aldmnul 'e afiusqo 01,

'65

 

( v )
.Iel-lsuv '95

 

'smqo ( ) = H
“51100 ( ) = 3

 
  

53.18de z‘

 

‘saxadum ( ) = I

 

 

O
'13
.LSELL
ssei
9.10“]

Jemsuv Ta

13. Answer

 

 

 

 

 

 

decreases
1h. - . .
E is emf. It does the pushing.
LOOK at Olun's Law. The larger E is, the (more/less) current flows.
E E f? ‘ I—> G I~>
— I /
l - fi- — R GXQIO 9 OXGXG G
E
E
148. Answer
resistor
119.
The symbol for a resistor is ——-’\/\/\/—-—.
It looks like the blade of a saw or a rough surface.
It shows that current does not easily through a resistor.
.. 814. Answer
«‘ ( 26914. 2 )
85.
37.91246 X 10" = .0037 91246 26.9142 x 10'3 =
V ( )

46 X 10" = .00000046

 

120. Answer

(10°)

 

121.

To remember micro, think of microscope.

A microscope is used to look at very (small/large) things.

 

1 .10 ( )0; su'eaul 01m

'OEI

lfliul onx efiaul
.IGMSU lj ‘03]:

( ) = 9-01 x 601
'01 J0 Slamod asaqi Kldmnw
z_OI =(z+ ,_)OI = zOI X ,_0!

Itlé

 

ooo‘ooz‘l

ooo‘ooz‘t
.2 311311 ‘J ° {6

 

( )OI = 000 or

.01 = 000‘!
201 = 001
101 = 01

'01 J0 SJSMOd Bin esn isni u'eo am ‘000 ‘1 .10 ‘OOI ‘01 Butsn I0 peaisul

'35

 

(000‘21)

.Iellsuy °LS

 

'papeau st 3 (ssoI/axoul) ‘3 8.10m par. 1 eunzs aql lulm
‘papeau st 3 (sseI/emul) ‘1 slow pus 3 sums an} 1mm
‘SMOUI (sseI/eioul) ‘3 9.10m pus a sums all) 1mm

°€8

 

ssol
emu:

9.10“]
.1 ensu V ' 33

1h. Answer
more

‘715.
R is resistance. It tries to hold back or resist the current flow.
'3 R gets larger, (more/less) current flows.

E I—>
I s 4%? £45695
If:
" ‘ R . E ~ R E

119. Answer

 

flow, pass, or move (any answer is correct)

50.

The symbol for ohms is 0 .

The :2 looks like a horseshoe.
A resistance of 50 ohms is written 50 ( ).

 

WBE. Answer
“* ( .0269142 )

86.

We call the'number above the 10 the "exponent" (eke-PO-nent).
The number 2 is the in 102 .

 

121. Answer
small

 

.122...

A microscope is used to look at very small things.

Very small numbers (10") have the name

 

15.

’82

'oaonn s1 r01

9! c-01

' Si c01
‘—___-_____'8l .0!

'6ZI

 

(9)
(8)
(9') Jensuv °ggt

 

 

( .01)
Jensuy ’86

( ) = 000‘! x at
0021 = 001 x Z!
021 = or x 21

 

at = I x 31
01.5

Z! 21
Jemsuv '95

 

 

'smou I (sseI/euoul) ‘3 emu: put 3 auras out 1mm
'11qu 31 mm; o; pepeeu s; a (seal/9.10m) ‘3 9.10111 put: 1 sures eq: lmm
‘qfinonu 11 qsnd 01 papaeu st 3 (seal /e.101u) ‘1 eaoux pus u sures all) 1mm

°ZZ

3J0“!

Jamsuv"ta

15. Answer
less

 

16.
The more the resistance, the (more/less) the current.

The more the emf, the (more/ less) the current.

 

50. Answer
( o )

 

51.

A resistor (draw synbol) is a circuit carponent with cl certain mount of

in it.

 

86. Answer
exponent

 

87.
In 103 the exponent is 3.
In 2‘ the exponent is 4.

In 129 the exponent is ( ).

 

122. Answer

micro

 

123.
To remember mega, think of megaphone

A megaphone is used to make a very (small/large) sound.

16.

)0: suealn 383m
)01 suseuu 01m
)0! SW3“! "IIul
'BZI
0.13mi
(r01 )
(8) Jemsuv 0L3];

 

( ) = [01 X 201

'01 to siemod asaqi Aldmnw

L01 =(‘ + ,)OI = ‘01 X ,0!

'pappc an szuauodxe eln ‘paudmmu an: 01 I0 sue/nod uaqm

 

 

 

'86
08b
.ISMSU‘] '16
000 '( ) ~ooo‘zr 00081
00 ‘( ) "oo‘zt 'aums all: 0021
0'21 G—Ism am 7921 x001 ll’e PIROM Kali) ‘Jaqwnu note “I z 3‘0 021
'31 >100] 1(an ueql 'ZI aplsaq .1er IUIOG [ewloep eln allow em n at
'SJequmu om ism ewes out out; snequlnu aseqi av
‘95
3
JGMSUV '55

 

°pepeeu I? 3 (sear/910m) 'esnemlrt 1 em 01,

I 5591
ti 3 'pepeeu er 3 8J0!!! 'AoU :uexmo sion eapul 0.1.
e I 9

"[8

 

 

9J0“!

.raMsuv '03

,.

. y . 16. Answer
’ less

‘ 1' more

 

\ 17.

Current varies directly with

 

Current varies inversely (indirectly) with

 

' 51. Answer

l(——wvvv-—~)

resistance

 

52.
NAME SYMBOL
Voltage or emf E

 

Current 2. ( )
3 (

UNITS
4.
5.

Ohms

 

 

87. Answer
1‘ (9)

 

88. .

When we multiply powers of 10, it is .just like adding their exponents.

We can see that exponent ( 1 )+ exponent ( 2 ) = exponent ( ).

10 x 100 = 1,000 101 x 102

123. Answer
large

= 10‘“ = 103

 

1.21“

A megaphone is used to make a very large sound.

0'

Very large numbers (10°) have the name

 

l7.

”‘1, ._ .mF. ,—;-— ,3— q_,_...—‘. . “...—....

'03

"W“
0U)!

'LZI

 

(9')
.IaMSUv

'9ZI

 

201 x 8‘v'=(

)

( )01 x 8'! = ooo‘zl

'01 go sxamod se Siequmu aseqi ssade3

‘16

 

8 9 r

IBMSUV '06

 

“(U /b/3/u) st 331mm .10; toquxlis elm,

J01Sisal
uxqo

'55

9011838158.! Jamsuvr Orl/S

 

 

'pepeeu st 3 (seer/em) 'sesnemug a an OS

 

H 3 "18ml: :1 «and 0: 3
(- lGXGIO. 9 9.10m peeu en 'elms sq: I deal 0: 196.1191 nob a II
’08

consists“ iueuno

Jensuv '61

Q.

\n.
.

17. Answer
TEST

 

78.
We know from algebra that we can multiply both sides by the same quantity.

55%men,(C)A s-g-“Hor AC =8 or =AC

Solve this equation for E.
I . g a .. ( )

 

:3}. answer
' A ' 2. (I) 4. Volts

..1. Resistance 3. (R) 5. Amperes
. 53. .
Thisisa --vav—-— , ithasmany 120111.
This is a 1. , it has many 2. of 3.

 

 

 

'Put in words for the symbols.

 

 

 

 

88. Answer ‘
(a )'

89 _

10‘x10’=10" (4+3=7) lotxlo’=()
10° x 10’ = 10“ 10’ x 10° a ( )
1211. Answer fl

mega

125.", A

mega 1 million 10’

kilo 1 thousand ( )

1111111 1 thousandth ( )
micro , 1 millionth ( )

18.

‘61:

M g‘ .1 U
5’ . E u I '( )01 = 1101qu ‘inuonuui-euo suseui “0.10m.

'921

 

OI OI cOI
.ISMSU v ° 521

ll
( )01 = ( )0! x ( )0! i0 ‘ooo‘ooo‘ooo‘l = ooo‘oor x ooo‘or

‘06

 

9'0! 001
.1 8M su \1 ' 69

 

Si ' 'stoqiulis ssaui JO seweu sq; in mg
91 U

'90 s so; seq Jaisuoi v
91 H um 1

~75

 

aoueistsea 'g sumo 'z .10181SBJ '{

Jemsu v '[5

 

sewn = Jlua

:spxom ut elnuuo; out our,“

 

'HI = 3 ss uamam aq use MB’] s,ulqo

'6T.

 

HI = El
Jailsuv fgt

SERIES CIRGUlTS
2-3

 

 

NAVAL AIR TECHNICAL 1RAINING COMMAND

f"

' 9

 

Acknowledgements

This program was adapted by the Naval Air Technical Training Center,
Memphis, Tennessee, from a larger one developed under contract No. N 61339—
1000 between the U.S. Naval Training Device Center and Teaching Machines,
Inc. In a letter dated 29 December, 1961 from the U.S. Naval Training
Device Center, unrestricted use of this material by any governmental agency
was confirmed. Appreciation is expressed to the U.S. Naval Training Device
Center and to Teaching Machines, Inc. for assistance in making the original

prOgram available.

' line, like this:

SERIES CIRCUITS

1. If your answer to a question is a single word, you will see a single

 

George Washington was the first president of the United

 

2. If your answer is two words, you will see two lines:

 

 

George Washington was the first president of the
3. If your answer to a question is in your own words, you will see a
dashed line: _ _-
Who was the first President?- _
What did he chop down?

A. When the answer is a number, symbol, or letter, it is marked like

 

thisi ( )

5. A multiple choice question has a series of items separated by slashes.
George washington was the (lst/2nd/3rd) president.

6. Some questions are test questions. The answer is not given, but you will

see the word "TEST” where the answer should be.

The steps in this program are fixed so that you should be right most
of the time. If you are doubtful about how to answer a ouestion, read the
whole step over. It is important that you read and understand every word of
instruction before you answer. The answer to the question is on the next
page. For the answer to any question coming at the end of the book, turn
the page -- and then turn the book upside—down. The answer to your question
will be in the appropriate answer space. Continue as usual. If you write
a wrong answer, study the correct answer carefully. Compare the incorrect
response with the correct one. Careful comparison of answers will help you
to move smoothly through the entire course. Always read the answer in the
answer frame. Do not neglect or rush by the answer frame, even though you
may feel certain that your answer is correct. COKPARE.

 

 

 

E =IR.

This formula is called

30. Answer

 

31 - we know.

 
 
 
 

First, put in the resistances A 29 B

  

battery voltage.‘
1 2

 

60. Answer
100v . 5a

P = IE = (.5a)(100v) = 50w

 

61.
A toaster draws 2 amps from a 100v line.
How much power does the toaster use? ( )

Which power formula did you use? ( )

 

90. Answer

. 20w

 

91.

 

You can see that all the columns in this table obey AC BD EF

Ohm's Law. ( )
8v _ 8v I 421 1a

In column AC, 252 (4a) = 8v or 29 = E or 421 — —2—9.

What is IEF'? ( )

 

 

 

 

 

 

 

 

 

°O€

 

 

 

 

 

BI - I ‘81 - 21
38
13:4: I ‘§-=zlos‘uzI-d
Mb sad 2 d Jemsuv '61I
Z) 2309 (1
1508 g E- A001
{ _A A A I , ) .oa .
\ ¢ d s: ieqm umoqs mimic sq; U1
8 2502 V ,
‘06
(A02) '0
(Hz) 'a
(2391) ‘v ~123M9uv '68
< l 'd mu: 01 an = a ammo; am 990
( )usnmm ( ) 33911qu
'91!!! RDA-001 “B [110.11 dum 9 ° smmp qlnq n18“ V .
09
= 3913.. = _fl =
Sdm 93 Sum 09 a I

 

Jamsuv ’6g

'alqm sun Kdoo
ems; B
anew 03 s; swalqoad nnono GAIOS oi ABM auo

°O€

 

Jamsuv '68

1. Answer

Ohm ' 3 Law

2.

As you remember, E —

T the total voltage applied to the circuit.

, You might guess that RT = the resistance in a circuit.

31. Answer

(4) (6) (12)
32.

We can find RTotal by adding all the resxstors.

Now find RTotal .
2 Q 4 $2 6 S2

   

61. Answer

P = (2a)(100v) = 200w
TEST

62.

I—andP=IE,thenI-( ).

9l._Answer

 

 

 

 

 

 

 

(4a)
92.
IAC _ ( ) AC BD EF
( R 29 ( ) 4o
Rso ) I ( ) 6a 3a
EEF = ( ) E 4v 12v ( )

 

 

 

 

 

 

'68

.-9—‘

( ) mum: an "mm
on

 

 

 

 

 

 

 

 

 

 

 

 

 

 

must - d .
'6'“
m - (oath: - "mm - ‘3
" ' 73% " JWW °8tt
( ) - ‘3 'o
( ) - I '3
Ave: ( ) . ,1 .v
mu Mou- zmom no u:
'68
T 183.1.
JOABUV '33
‘
( ) '1 mm 02 f} - Hummer an con
man - a
mum - d
'65
'1
A00!
JBABUV '35
or
”8 was“! 0m p and a:
W 8mm 8 :01 W cm was mom cm mm
or
'63
A9I-(us)(u)-“"m-“"a '
.1.u
'z=%%=T‘I JOMBLPV .83

2. Answer

total

 

 

3.
The total resistance of a series circuit is the sum of all the resistors.
1 Q 2 $2 5 Q 10 9
; Total ; Total
T 3 a T .

 

 

32. Answer

(12)

 

The total I is equal to

AB BC CD
29 482 69

Find I.

   

 

62. Answer

 

 

 

 

(g)

63.
I = E ; Complete the power formula.

92. Answer (2a)

(29)

( 12v)
93.

_ 2 35!

ECD _ ( ) IIJ ( ) / 1/29
REF = ( ) EKL = ( ) 6v

 

3

6i; 3 oz & av
11 _ .
_ 1 ..Iaquiauxeu
_L 3v
- 3 ( ) afilaimq am 10 631nm em S! mm
oz 8 of; V “9‘ = ova
'8II

 

Sum 81; = (31):: ‘snem 91 = (9):; ‘snem as = (8)17

 

 

 

Bb=21 Bz=iiz‘=§=z
W 3 Jamsuw 'LII
a 3 a
oov ooe
o0! —:
——’\/\/\r——I ( ) whom sun ur enumsrsal mo: aln 9111mm
3 oog 9 ooz V
°88
sum 00»

JSMSUV 'Lg

puv‘t )=a

GJOJBJBIH ‘eun “DA-001 2 mo.” dare I smexp uon fiuuaplos V

(s)

JSMSUV 'Lg

UV —: 1‘91

oz
'88

(A1) (A9) (A1)

JGMSUV 'Ag

3. Answer

(159)

 

L“ What is the total resistance of this circuit? ( )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

29 29 29
+ —
4Q 29
33. Answer
(1)
3h. The current is the same through all the resistors. 2 9
AB BC CD Total I What are all the rs?
R 20 49 69, 129
VI()()() la 12v—"—
E 12v
63. Answer
P 2
E °r E
61.. ‘
P _
r — < )
93. Answer (3/49) (8a)
(3V) (9a)
( 5/49 ) ( 2v)
91..
When we make a table that includes P, we can see that AB CD EF
the columns also obey the power formula [E = P. R 2 Q 69 39
In column CD: P = (3a) (18v) = 54w, orl = %;1 = I 23 3a 1a
3a,orE=54—w=18v. WhatisP ?( ) E 4" 18" 3V
3a EF P 8w 54w ( )
A.

‘AZ

 

 

 

 

 

 

 

 

 

Cl 3 J 'ZI PUB 1 Pull 1511.! ALNIH
oz 258 A” ( ) = JVd
2317 "
( ) = “3d
(IV
( ) d
C) OZ H 62 V
'LII
W = (0I)Ez = “"3
A91 = (8)93 = 3V3
98 = ‘13:? 1: I flex” 1 pug isnw no; JQMSUV .911-
‘pasn are .xsmod JO ( )
'aun A003 2 won sduna Z SM‘BJP aAois v
M0001 = (A001 ) ( sdum 01)
pan an: .xamod yo snem 0001
'aun A001 12 (no.1; sdure 01 smup aAois v
5L8
sdum z
Jemsuv '98
( ) _
d - 1
’25
HI = 3
Jamsuv '95
(13
( ) = 3
< > = 393
< > = “3
°LZ
I _ = 33. =
E — t1 HI 3 3 I

.I aMsuV ° 93

A. Answer

(120)

 

5.

If the voltage source is an 8 volt battery What is the current here? ( )
the current in this circuit will be;

252 252

I _ E _ 8v or 8v
I RT — 2+2+2+2 m

 

I=1amp

 

 

3A. Answer

 

 

 

  

 

 

( 1a) ( 1a) i la)
3 5 ‘ Now each voltage is equal to I times its resistance.
2 9
AB BC CD Total Find E E E A
, , - B

R 20 4e so 129 AB BC CD

I 1 a l a l a 1 a 12v ; 4 9

E ( ) ( ) ( ) 12v

 

 

 

 

 

 

 

 

6A. Answer

 

65.
If .' . . 200 watts
a ZOO-watt lamp 15 connected across a 100—volt line, 2 amps W111 flow. 0 volts 2 2 amps

What is P? ( )
What is I? ( )
What is E? ( )

 

9L. Answer

 

(3W)
95.
IA13 ‘ ( )
ECD ( )
PEF ‘ ( )

 

3 62 J oz
'31 paau noA n alqei e anew
”v T Ai’Z ( ) = 3V3
3V
( ) = 3
0 oz 3 ol 8 or V

‘llmoux an: 1 pm: a asn'eoaq ‘u 21 = d
sue/Ala = (8)3(8)
Jsmsuv 111:

'umexp em manna ;o ( )

'aun A001 '9. won sue/A 003 some; uou Buuamos v

 

.1 z A001 I = _3
I mos a
'umelp sl iuauno 10 dure 3/1 'eun A001 2 won snem 09 933121 uou Buliepms v
098
. g I
(Is) ( d) ( d)

.IGMSLIV '98

( )LH put; 1 IO sullen u: 3 saAifi qomm MB’] 2;,qu .101 RIF-(11101 3m SI 1.8%

a!“

JGMSUV °§g

 

( ) ( l ( )
'suuo; saw; no axiom

gmeq snuqo Si 1mm

 

(A9)

Jamsuv ' 93

5. Answer

RTotal = 2+2+2+4+2 = 12 9
24V

1:2: =2amps
R I29

 

 

 

 

6.

What is the total resistance? ( ) 129 49

2 S2 4 Q
35. Answer
(2v) (4v) (6v)
36.
Draw a table. Fill in all the blanks. A 2 $2 B
AB
R 20V ; 6 Q
2 $2

 

 

65, Answer p = 200 watts
I = {lamps
E = 100 volts
66.

The current in a resistor is 10 amps and the voltage across the circuit is 20 volts.

What is the power in the resistor? ( )

 

 

Which power formula did you use? ( )
95. Answer (5a)
(4V)
( 14w)
96.
1.1»:AB ( ) 5.PGH ( )
3.REF ( ) 7.RKL ( )
4.136H ( ) 8.PKL —( )

 

US

<—duie g

( ) épasn sq pmoqs emuuo; .Iamod qomm

( ) 010131391 sun u1 ss01 .Iamod at“ s; new

'SII

 

(CID (38

s)

3)

JBMSUV '711

 

 

 

 

 

 

 

( ) = d

( ) = I

'sA’em aalqiuamxm aq use ‘31 = d ‘BInuuo; .xamod 3111,
‘58

(oszl)
Jemsuv '78
( ) = pu'e 31 = d

’SS

21

d
JSMSUV °W§

( ) _ SIG

“Se

m = (Be)(ov)= (1)(°fla)= 3‘13

as = Qfll = UAJLiiLiZJLll-= —E§-= I

A09 A09 1.3

Jonsuv °vz

 

(5".

.6. Answer
Test

 

 

What is the RT? ( )
What is the current? ( )

36. Answer

 

37 .
In the circuit shown, the value R2 is not known.
But we know that

. ET
R2=RT-R1andRT=—:r—,sonz=< )

E =100v

 

 

 

66 . Answer
200w Test

 

 

 

67.
In an electrical circuit, some power is converted to heat.
A light bulb gets hot because some of the is converted to heat.
96. Answer V 1' (10V) 5' (m)
2. (2a) 6. (3a)
3. ( 2S1 ) '7. ( 39 )
4. (16v) 8. (27w)
97.

In the circuit shown:

E AB = the voltage between A and B

R AB = the resistance between ( ) and (‘ )
P AB = the between ( ) and ( )

 

7.

"78

“3+3a3.+( )+( )+8V:}=‘“3

'saBmIOA Innpwmur

 

am :0 ums am smnba nnono e u] 331mm [210; BILL

”TU:

 

.IaMSUV ° £1]:

’Bnu ( 7 ) sasn u

'aun “mi-001 '8 won sdum 9 '8I smup a dumq
'moos = ('99)(A00I)PWd = 13'

'su'am 009 seen u
°aun “mi-001 2 won sdum g smsxp v dung8

 

 

 

 

 

 

499m
.xensuv ~g3
'( ) = 1 ‘exoyezetn
‘31 = d 1me zuaxamp ow: ammo; sun efiuuqo use 9M
"15
(Japxo Jaime ) ( 1 ) ( (I )
.xamsuv °£§
i) (I 3
U 8 U z 8
:- A08 ( ) O 3
U 17 23 I
( ) = I
O a V .
’73
‘3 + an + ‘3 = “Lu
Jansuv 'CZ

 

 

 

 

 

 

 

 

 

 

 

. Answer
7 R =2+2+4+9+4+2+2=250
Total
E ' _ 25v _
I = .1? _ 250 — 1 amp
T .,
.8.
What is the value of RT (total resistance) in this
circuit? ( )
E ' 100
37. Answer = __T = __V = 1000
RT IT 121 -
RT=1OOQ=R1+R2=609+R2
R2 = 1000 — 609 = 400
38. . A 29 B
EAB = 4v
Fill in the table. 20v ; 60
252
D
C
67. Answer
Q .
power
68.
In the circuit shown, what is the power in the l = 221 .__>
resistor? ( ) E = 5v —.;
97. Answer
( A ) ( B )
power ( A ) ( B )
30v

 

98.
When solving E series circuit problem, make a table.

Fill in the table with the known values first.

Fill in only the parentheses.

 

'3 ,= aoumsisex pus] = manna 13101.1

“oisisal sq; u;
I ‘ sso1 .xamod at“ pug o; asn n05 plnom emuuo; rpm
' €II

Mzc = (z)(91) = i121
21291 = (HM) =31-
1

RV = JGMSUV ‘ZII

( ) amen S.un{o 10: BIanOJ aln 5! mum

(1)(3) 10 (3H1)

JSMSUV '38

'( ) pus ( ) ueam1eq saoB ufirs [enbe emu,

.. ‘EIIdu prom sin 1?. Bumool Aq paJaquiamaJ aq uea Jamod .10} emuuo; eql

MOI; wanna 10 53.19de guns .ro 3110A

JGMSUV 'gg

'meq sumo asn )ou 0(1 :ezoN
— ( ) 'umoqs unouo

It! . sq) u; ‘Lu JO amen sq; JO} uonenba us any“
. '63

AI: = (ozflz = 8V3 ’(QVHHI) = 8V3

B _zszI=J1=
z_/WZ 3

ozr=v+v+z+z=iu JGMSUV'ZZ

8. Answer

.\ _ T'_45V_
< RT-x—“s’I-m

 
 
 

l
\
A 9 . Use the current and voltage shown on the meters

to find the value of R2. ( )

   
 

3 8 . An swer

39.
i I = 2 amp A 20 B
‘ Remember: EEC = ET - EAB - ECD and
BBC 12v ; R
RBC = _I‘ D 29
Fill in the table completely. C

 

68. Answer

‘ P=IE=(2a)(5v)=10w

 

69 .
E the power in a resistor is converted to heat.
A resistor has 100 watts of power.

How much power is converted to heat? ( )

98. Answer

 

99.

From the information in the table, find current and voltage in the rest of the table.
Remember that E in each column = I X R and that the current is the same in each resistor.

Fill in the parentheses.

 

oz
<1————ep
( ) 2.1019183: SN: 10 9801321 an s; mm
’ZII
69=ob+oz=oau+avu=svu
+ + a
A31=(U9)(BZ)=OVHI=OVE{ laZ=‘9%=j—9zt_l=§=l

Jemsuv 'ITI

 

088
.2 = m 2 39:3 =1
““001 Bad

Jemsuv ' T8

sauesaJdaJ 1 put siuasarded 3 ‘Jequxauiax no/i sv

~31 = (J3M0)d s; emwxo; .xamod aim,

Jemsuv 'Ig

 

09 = (OI)(£)=(Z+17+Z+Z)(98)= s

JSMSUV ‘ {a

9. Answer

E
2 10v_
“2-=1— "Ta-1°“

 

10.

 

69. Answer
Test

 

70.
A lOO-watt light bulb is connected to a 100—volt line.

Put the values into the power equation and find_I.

( )

I = -———3 = ( )

tilltu

 

100.

When all of the currents and voltages are known, power for each column can be found from the power
formula,

P = (1)(E).

‘ Note: I?. has been added to the table. Finish the table.

10.

 

3 CI '1! paeu no!( 1! 9mm B anew f1 pun ism 1mm
oz = av
15? _:_A9I = $3 ( ) 3
c z
0vHI = W3
8 v
“III

 

maq .IO Jamod

 

 

(d)
Jar/SUV 'O[‘[
3 Eu (1
( ) ammo am u: ( I ) M015 manna an 5! mm
Z
8 ‘“ 'MOOI
st 33 u; .19/nod sq: pus ‘Aog = 083 ‘umoqs imam am til
8 I V
H
ӣ8
( Bz )
( or >
( Aoz ) VILMSU‘V' ’03

 

1 s: 3 pm: (1 J0 sumo} ui ‘1 10} uon‘enba Jamod .xo etnumo; .xemod aql

'( )=

u
ri
Lr\

(I)

Jamsuxf '0g

5qu ( ) - SI

:umoqs unsure em [11

 

qsag
JBMSUV '05

10. Answer

 

 

E
y (i)
1 11.
A 49 B
E AB (the voltage drop between A and B)‘ is 8v.
= —--->
E =(2a)(1$2)=2v I 2a
CD 29
E 49 D 152 C
EDE'( ) EBC=( ) EAE'()

 

LO. Answer

Al.

The path around which a current may flow is called a

70 . Answer

 

71.

The current through a resistor is 2 amps.
The voltage across the resistor is 50 volts.

Put the values into the power formula and find P.

P=IE=< >( >=<>

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

100. Answer
AB BC CD Total AB BC CD Total
R 69 59 49 159 E 12v 10v 8V 30v
I 221 2a 2a 2a P 24w 20w 16w 60w
101.,
I = ( )
29
PT : ( ) 8v T
If you have trouble solving the problem, make a table. 2 9

ll.

‘ssol = ssol a z]
'( ) = H 21 asnvoaq .. SSOI a 31.. em sun 1120 am

are sin 01 1301 put reaq 01 pafiu'eqo s; Joisisax 'e u; pasn Jamod at”,

 

 

 

 

 

 

 

 

'OII
(A91)
(mac)
.Ismsuv '601
O 139 CI
AOI ( ) = I
o8 A9 —: ( ) - “Va
A? CV
21 ( ) = 3
oz V
'08
Sad + svd
Q )
.IaNSU‘j ‘61,
_ l , _EL _ _EL . _
( ) _ d SBAIB 11131.1 am .10 $.39q13umaoueopmz Ell _ d Bunny/up 31 — d
:snn 93m ‘3 AQ saprs qioq fluipmip liq 1 .10; uonenba .xamod sq; amos ueo em
'05
HI = d
Jamsuv ‘67
( ) = 8
mar] s.uiqo
'OZ
2 1 _ I av V
or ‘ A91 avg ' H H

JaMsuv '61

we

in

11. Answer EDE = IRDE = 2a(49) = 8v

 

 

 

EBC = [REC = 2a(29) = 4v
EAE or ET = EAB + EBC + ECD + EDE = 22v
12.
EEC - ( )
hl. Answer
circuit

 

L2.

Power is the time rate of doing work; that is, when we do work for a period of time we use power

We use if we push a car for 5 minutes.

71. Answer

(2a) ( 50v) = 100w
72.

Power Formula Ohm's Law

E=() E=()

101. Answer

£_ 8 _8v_2a
R_2+20_4Q.—

P =IE = (2a)(8v) = 16w

 

102.
29
'EAB=(2a)(2SZ)=4v A B
__>

PAB = IEAB = 2a(4v) = 8watts 23.

P _ ( ) 16v T 49

BC _ 29

Make a table on scratch paper, if you need it. D C

12.

{I

 

 

 

 

 

 

 

 

(I
132
UV
4“ 133
V _ GV
o z ( ) — d
‘60I
m :
V8 svd
(159)2139 = (z+z+z)z(ez) = “821: “a
Jemsuv 'QQI
O
( ) = ‘Ld MIMI!) Sill) U1. J9M0d 1910) am SI 1mm
38d
'SJemod Icnpyupur
8 8Vd V sq; yo ums sq) 0; {cube s: unorro e u; Jamod [2101 BILL
“éL
C11)a + 38,d + SW8
'i
JUMSU'f '82, Q
( ) 'uornznba Jamod am 91mm
-uonenba .xemod am panes sq osm u‘eo nxnuuo; .xemod at”,
o 67
squ
.i QMSU‘rj '37
( ) flu Jo anx‘eA eq: 51 mum
mm = 8V3 unsure sun u1
‘éI
( Ava )
( A91 ) I"
( A8 )

JOMSUV '91

12. Answer

EBC = IRBC = (4a)(29) = 8v

 

 

 

 

 

 

 

 

l3 .
E _ ( ) A 49 29
AE
2a ---—-’>
E 4 Q 1 Q
Hint: Find RT and use Ohm's Law.
h2. Answer
power
L3 ' 1 :2
Figure A is a series circuit. ’\/\/\/
All the parts of the are in a line. 6v —L__ 2 Q
T 3 n
A
72. Answer
(13) (m)
73 .
In this circuit, the power lost in R1 is 50 watts, and
I = 2 amps. E L;
What is the value of E? ( ) I —> I

 

 

102. Answer

 

EBC = (23.)(49) = 8V
PBC = (1)(EBC) = (2a)(8v) = 16w
103.
EBC = IRBC = 43(2) = 8V
EAB =( )V
pAB =( )w

 

Make a table if you get stuck.

13o

321

M38.

aemsuv °LOI

 

033 + 38.3 + avg avg
'82.
(3)
(ti)
(1) .IeMsuv “M,

 

u; pernseaw s1! 3 pus seiadtue u; peJnseaui s1 1

$31 = d s; 'eInuuo; remod aim,

 

 

 

 

 

”8‘7
sue/n
aemsuv '57
< )= ova
( )= 39‘s
m =1 < >= “3
“8T
(21)
(v)
(8)' Jemsuv °LI

fl"

l

13 . Answer

 

 

 

 

 

 

 

   
 
 

22v
like
The symbol for the voltage drop between A and B is E AB‘ A 2 9 B
The voltage drop between B and C would be E( ) ( ). 4 Q
C
1.3. Answer
circuit
Ah.
is the time rate of doing work.
73. Answer
P _ 50w _
E — I— _—29._ — 25VOltS
715..
If we put another resistor in series with R1, would IT
IT be smaller or larger?
6v ;
R1
103 . Answer EAB = IRAB 4( 4) = 16
PAB = EABI
PAB = (16)(4)w = 64
10..
How much current is flowing in this series circuit? ( ) A B
The power in a resistor is equal to the current through the
. . . R1 = 5052
re51stor squared times the resxstance. _
ET = 300v ; R2 = 1509
What is the power in R3? ( ) —

Di.

 

 

 

 

 

 

 

8
'elnuuo; Jam0d( ) = d
23 Z sq; asn pInom n05 ‘usA1fi are u pur. 1 aours
‘— 2” av
V ( ) = d
'LOI
5119M91= (ov)(eb)= 2121 = d
3b =21
. '93 = I .IeMSUV °9OI
'( ) $1 afieuozi 10; quuuis an,
'( ) s1 eoueisisaa .10; [oqus 31”,
'( ) s1 merino .10; loquuis sq;
’LL
afieuon merino
.Iamsuv '9/_,
ur pamseeui s1 q1nq 111311 e u; pasn Jamod at”,
'snem u1 pamseaui s1 JaMOd
OAT?
( z) '2
g . Jo z/I II Jamsuv ‘97
A( ) _ (IO3
O 59 (1
oz A( ) - 383
‘-————Bz
_ 8V
8 or v "( ) 3

 

a pue v siuiod ueemeq doxp [enuaiod .10 331mm sq;

.IQMSUV ' 9‘1:

,1A. Answer

(B)(C)

 

15.

What is the symbol for the voltage drop between C and D? ( )

 

LL. Answer

Power

 

[$50

The symbol for power is P.

We use ( ) to lift a box.

 

7h. Answer
Test

 

75.

Power Formula Ohm ' 5 Law

 

 

 

10h. Answer

"U
u

I R = 100 watts

 

1050

15.

 

 

 

 

 

 

 

Sum 09 = (mm = d
( ) = d or oz
I 33:1 I 1 sdumsz=(s)g =21
'ulaIQOJd sun 00 sdumg = I
'9OI
(d) (a)
.IQMSUV 'SOI
“1019199.: sq; ssora'e 3‘”
pus .xoistsax sq; anonn sq; yo ionpord sq; o; ['3an s1 Joisrser e 111 JBMOd at”,
0911
ii. _3
(a) < d) a
.IGMSUV ~gg,
H o9 __ A( ) = 8V3 3
or 3.5 A9
4— I N ) = I I
68

r 1.1mm son ”I

 

(d)

.ISMSLIV ‘ gt]

 

saieotpui 3V3 quums 9111,

G33

.xemsuv ° gt

 

Acknowledgements

This program was adapted by the Naval Air Technical Training Center,
Memphis, Tennessee, from a larger one developed under contract No. N 61339-
1000 between the U.S. Naval Training Device Center and Teaching Machines,
Inc. In a letter dated 29 December, 1961 from the U.S. Naval Training
Device Center, unrestricted use of this material by any governmental agency
was confirmed. Appreciation is expressed to the U.S. Naval Training Device
Center and to Teaching Machines, Inc. for assistance in making the original

program available.

PARALLEL CIRCUITS

1. If your answer to a question is a single word, you will see a single

line, like this:

 

George Washington was the first president of the United.______________.
2. If your answer is two words, you will see two lines:

George Washington was the first president of the .
3. If your answer to a question is in your own words, you will see a

dashed line:

 

Who was the first President?-

What did he chop down?

 

it. When the answer is a number, symbol, or letter, it is marked like
this: ( )

5. A multiple choice question has a series of items separated by slashes.
George Washington was the (lst/2nd/3rd) president.

6. Some questions are test questions. The answer is not given, but you

will see the word ”TEST” where the answer should be.

The steps in this program are fixed so that you should be right most of
the time. If you are doubtful about how to answer a question, read the whole
step over. It is important that yOu read and understand every word of
instruction before you answer. The answer to the question is on the next
page. For the answer to any question coming at the end of the book, turn
the page -- and then turn the book upside—down. The answer to your question
Will be in the apprOpriate answer Space. Continue as usual. If you write
a wrong answer, study the correct answer carefully. Compare the incorrect
response with the correct one. Careful comparison of answers will help you
to move smoothly through the entire course. Always read the answer in the
answer frame. Do not neglect or rush by the answer frame, even though you
may feel certain that your answer is correct. COMPARE.

\ A
\

fix ‘
G

 

 

30. Answer

 

31.

Which circuit is different? ( )

 

 

 

 

 

 

 

(R2)
()1.
_ R1 R2
RT ‘ ( )( >( )
90. Answer
(R11)
91.
Solve
RL2=T—4—(4)+(4) :3;— = 29 €29 ‘ 4n '
_ (2) (2) _ 4 _ . Then 129
RT‘ 2+ ‘ E _m 1.2
29 29 0-
RT = < )

’O€

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

mo00‘01 M000‘9M000‘9AA000‘1 d
A0001 A009 A002 A001 3
1801 t:01 1201 B01 1
6001 009 608 2301 H J u
ems '
1210i (1:) 38 8V V 6“
’93 1mm 1811mm u1( )
2'1 z 1

31mm uam pm?

8 01 M09 H PW

u Bumalq

1511111011109110 910th 611110 ‘1‘8 atn pun um am

'06

 

 

UZ

JBMSUV '59

 

()+Iu_.L

 

61318 an pue V3
1mm x

undue mmpeu

 

 

 

 

- u
an In
009
_9_I=__8__-:_§_=
“‘19 (8)(8) la

 

xemsuv '53

 

 

uaqiafioi paioauuoo am 1015189.: qoea JO spua qioq asmaoeq ewes em are a pure v 331110.119

‘Of

(98)

JOMSUV °6Z

(’0.

1. Answer

(39)

 

2.

 

 

 

31. Answer

 

 

 

(D)
32.
AA B
BAD =( ) _L
E _( ) 6v; R1 R2
BC ' . I
'D c

 

61 . Answc r

 

R1 + “2
62.
P \\ N \“ f, [magi
IT = ( ) , ' j
r( / .1

 

91. Answar

(3Q)

 

92.

To solve any parallel circuit, take any two resistors and put

an equivalent resistance in their place.

 

Redraw this circuit with the R1 2 of the two 80 resistors.

2.

 

   

 

 

 

 

’62
1309 8u
3 CI
808 ‘2— A0001
zu
1.1113113 sun .10} 8161131 91” alaidwoi) V
' 8
‘61]:
i m ,. a _w ( M8 )
( Biz)
( 153) .Iamsuv ’91] _
O
".xoisgsal 238 am in?!“ Ian'enad 23 9 69 238
ill 8 “(H “NM ”“3110 3‘” MBJPBH 1‘8 an 88

 

6 9+8 _Z'IH

 

 

 

 

 

 

 

 

 

7.3?- : "_- : 8 ‘
8'1 ( 9 ) ( ) °SJOJSISGJ )0 UOIIBUIQWOO 5"? “MOP “‘3an “73° 9’“
068
(SH)
.Iemsuv “98'
We..i_.-_._ ._.-___i,__.___O Z I
Z . T t .( ) = ‘LH ‘aAoqe mnuuoy 31113111311 ‘68 = H PUB 238 = H H
21 *H z 1 .
(Ix .__§_3__§ = 1‘3 human Iausmd a 1:1
065
311.
L3
dengue °9§

 

'( ) = II ffiuenba

 

 

 

.I.
wanna am mds uses 81 pm: 81 ‘11 pm: €139 = In
’swed seam mu; smds

 

 

 

 

manna am ‘saqoumq IBHBJBd aaiqi an exam :1

068

 

 

 

 

...—.s-»w.~_._ -7 *A‘f. ..-,

 

 

(LI)

.zemsu ‘5’ ° 83

' 2. Answer

 

(39)

3.
RT=( )0
ET =( )V

 

 

33.

Remember that the voltage across each branch of a parallel
circuit is equal to ET .,
( 6V) ( 6v)

T ...

_LI
i M

 

 

62. Answer
is

RT
63.

Inaparallel circuit, RT = ( ).

E
Do NOT use ~13.

 

92. Answer

 

 

 

93.

After you find R2 3, combine this with the other resistor in parallel.

R Draw the RC
3 is the same as 49 452 of these two
S2

resistors.

3.

 

 

 

 

088

 

2517

 

a sum 9 =
i.

I39

(13

3

93 I
H
EIV

 

 

(Z)(f')='1'd
‘I‘HZ(J‘I)=‘Ld
= (2)2 =11
JILL-
(10(17)’ l3
'QII

.1 emsuv 'LII

\/<Z 18$ ><SH ( “DIM
i J Ianemd 111 sq pmom Z In

 

upue 13 ;o seen! 111 11 3nd pus

 

H

{i

z 18 pm} two am

.°99
(252)

(oz)
Jamsuv ‘19

 

H Butpug 10 SKI?!“ om eAeq am ‘imono {ensued e (11

‘LI L

“L3

2

'—-—-= spire H+IH=lH

{1‘3 Butpun 30 Site». can exist; em ‘nnono sagas e 111

. Z

'3 0: SAAOIJ (

IPW

I

°8§

JSH‘BILIS

.Iamsuv ° [,5

) 11811108 a mod 13 dn U10! 31 pm? II

1 om; v 1u10d 112 smds “1‘1 wanna Sill,

SBBHJIOA “f;

swarms '3

088
manna 'g

aBmIOA ‘1

.xamsuv °L8

3. Answer

(12) (24)

 

h.
When two electrical devices are connected with both ends together, we say they are in parallel.

Batteries A and B are r Resistors C and!)

in parallel. .2: A _=_ B are in . C

I_I_

 

33. Answer

(6a)

 

311.

 

 

63 . Answer

*3
931
m
*3

 

6h.

. 4 7 IfR1=69,R2=3n (6)(3)=_19§=20

=29 +

 

 

93. Answer O——\

 

91.1.

If we have four or more resistors, the. way we solve them is the same as with only three resistors

Redraw this circuit with an R for the R and R resistors
3. 4 3 4 ' R R
1 “2 R3 4

20 40 8!} 89

 

~12 =3

 

— “'93 91981 am U1 1113

m" 3
OVA/MTG
.. E.

H u-c
«AAA/003

0

   
 

3.211-..

 
 

€911

 

 

- _ V'E‘
zH+Iu=r1a ( )‘HH
28 Is ( 5 = 2 Is
.208
.33 pm? In JO sum [amend \

an 0; [cube s1 sum asoqm J01SISBJ 913mg 13 s; a 1};
z-[a aansuv '99

'68 Win 91 UV " ' ‘
39 us _— A91 ammo an: 111 m, = 3L = ill = '1‘“
____ A91 J.
2019193.! Rue usq; .19]me Meme 3

 

91 ‘I‘u am ‘nnom 19119:?!de 3’ = '3 + 2z ' ‘11

 

 

 

02‘s
( 25 z ) ( vz )
(Ev ) (Bz)
Jensuv '95
t
'99! ‘9 all: 1m '9 sum em emu Ire suns-no cams
'th 'z are put . '1 sures am emu 118 ammo 1311mm ,
'La \
(81») ~ ‘

.Iemsuv '93

it. Answer

parallel

 

....m-.—..-— ..-—...

 

 

5.

These reSiStOI‘S These batteries

These resistors

are in parallel. are in series. are in

31;. Answer

 

 

 

 

 

 

 

 

 

 

 

 

 

E- = 49
35’.
A C
EAB ' Eco = ET Eco = ( ) E l I1 3“ R1 I2 32
T“: 29 4o
EAB=IIR1=33(ZQ) ET=( ) .l f
B D
‘ r311. Anm-zer
_ (6) ( 6) _ 36 _
RT ‘ _6_+—6— - I2 - 3”
:45,
R = the product of R1 and R2 = ( )
T the sum of R1 and R2 ( )
9h. Answer
R1 2 3.4
29 452 452
G
95.
Now draw the circuit with the R2 and R3 4 resistors combined. 0
' R1 R2 gas 4
AR 452 $49

 

C)

y"!

 

 

 

 

 

 

 

 

 

'93 e
<1 s
‘ 83d + Hvd lenba mucus ‘I‘d ‘xoaqa a sv
259 i 69 l
T A 'a e a a a duo
all Z In I 9 I‘ll In 1 I O
_ 1 v I I 'mm JBMOd at“ sAaqo uumloo qoea u; .xamod
C) V
' ‘I
d 9L
A9 A9 A9 3
38 132 81 I
1561, US 159 H S V .
[B101 .1811 u, SIT

 

 

 

 

 

 

CID 8V
- 31mm Sm:
‘ so euros 9L!) Z ’( ) "1 1nd 1311’2
SI nnono s_ H
' “11 —> as pue Is mo 9m was am

 

 

'Z .111 [oqusz sq; Aq paiou s1 33 pus Ia 5.1013159; JO ‘1‘}; at”,
'98
.L ez[=eg+eg+29=ll
U; = 33»: -L= la 91.. 92.. or. L ‘
A31 3 A31 A31 A31 ‘ ”

Jamsuv °§g

 

017 612 )
i —T— ( ) = lI ( > = H

 

oL99' .10 ug Jo o§ (99)

(or) (A9) Jemsuv .55

 

 

 

( ) = z1
1I _ i1 = zl
Hath “L1 = a1 + ‘1 I
993
L831. ‘

.IGMSUV °§a

rs

S.lhumer

parallel

 

6.

 

These resistors are .:

in

 

 

These resistors are '

in .

 

 

 

35. Answer
(6V)
(6V)
36.
ET = EAB = E
ET = ( )
IT .. ( )

CD

To find IT , you must first find 12 .

65. Answer

R1 R2

R1*'Rz

 

E _L '33. R1
9' T 4:; I f 29
= I 1.

 

122(‘9 s

 

' 66.

 

 

2.3.4
252

 

 

96.

The final RT would be ( ).

 

1 “as“;
29 an

 

6.

‘SZ ,

_ at
o: $ no $
23 'I
u
z! * II
o v

1'...- A9
I 'alq'e; a u ‘I' “L ‘L : ‘6 '
_ ‘13 I H WU I PUB 3 30 89hr“ mu “IO-1.21

.6,- Sift.--

 

Jamsuv ‘3th _.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m “31
UV ( ) = I
’58
EB+B§+B§+98=LI
e = ‘LI
"1 32.6. ... .92 + 32.2 + 3.2.: .. m
A91 A91 AQI A91 "' I .Iemsuy .119.
'(I H °anm 2 exam ‘1; pasu no.4 n
I}! 3 ”Z #38 —:3 ( )=‘LH ( )=ZI
LI -—""‘3T ()ng ()=3
0 ‘ V
'SS
Ut‘l‘HL“““‘“139‘-=ZI I “ A8=‘I‘3
§§‘§L=LH 1ez - v9 = 31 (2mm) = “L3
' a , II - ‘LI = Z1 “3 = J‘s .Iamsuv ~11;
81 I; l
I“ V t“ ' 7 < ) = ‘1
3?
ta 1'1‘
058
AZI A21 A21
‘39 92 98
69

U V Jamsuv "13

6. Answer

parallel series

 

7.

~‘ If '0 took out the switch and connected A and

Bmcqwfimmmmfl

 

  

 

 

 

 

 

 

 

 

 

 

 

before
36. Answer ET 3 TEST
IT=11+12=3a+—2a=6a
_ 3V ..
_1 12 - m - Zafiwm‘d v
37.
1 ( ) ‘ A C
1 - a EAB = < )v A4——-
I ‘1- +
2 ' ( ’3‘ Ben = ( )v 80v-_—_ 1052 zoo I
I
Lr = ( )a l 2
B D
66. Answer
6 (any of these
2(4)_8 “4 *1,- ..
m _ B“ - 3:2. .. 139 _ 1.5352 answers)
67.

We have two ways of finding HT in a parallel circult.

Write them both.

$.—

96. Answer

TEST

2,. n... u gygu...r: ;-~ +4;

97.
What is the final R.r of this circuit? (

--v——-_ Ar...‘ ....

 

69

 

30

 

one ~ ‘2—

 

'SzuaJJno qoumq )0 runs = ‘11 mm xeqmauxeu

‘2'." A9
I I ' 3} pue 11 pug ‘amm sq: u; mou semen am 01033"

”III

 

.IEM'LSU V '51:]:

 

 

 

 

 

 

 

 

 

 

 

-.=- ( ) ‘= ‘1
v c z I T
I I I
ll f’1+“I+zI+II=‘I‘I
...p I
”w
_ gs _ v ... 3.5.2. = 8
Ba " Ag] " I 88 MI I
,__ 32.2 _._ z .
8" A31 I .Iensuv '{g
(I 8 'Jeded qeyeaos no one; ‘3 exam
La ethe-
I 99 V 033 a 3V3 2: 1‘3 :Jaqwemau ( ) a 1‘3
I) V
‘WS
(252) (m)
(29) (up)
.Iensuv °{5
(1 . El ’ WWI mmqo 233er names nose am «on

 

 

(1210: a mo awed) ammo sun «to: sum 8 exam

.Wz

 

A!) Afr - \\
m

'1“? us an: sail-uuon -
an pm: (as = 121 + 83) ppe szuauno em mm anon Jmsuv 'EZ

 

7. Answer

parallel

 

8 O ‘

Circuit A is a series circuit. Circuit B is a

 

 

 

 

 

 

 

 

37. Answer
TEST
38.
252 Sw B
. . = 1
IfSw1 15 mA, EAD ( ) 0‘1
. . A L
IfSw1 ismB, EBD =( ) — 12v 29 ::4Q
Hint: first find I in each case. " D
67. Answer
E R R
T 1 2 . ‘
RT — T_ and RT R + R (either cider)
T 1
68.
We work this problem both ways to check the answer. ‘
E R R I
R=_T=() R=#=( )LET so 39
T I T R + R
T 1 1 12v
—-—>6a

 

 

97. Answer

2/39 or .6679

 

98.

When we reduce a parallel circuit, it doesn't make any difference which pair we reduce first.

R1.2 =( ) R1.3 =( > R

R2. 3 = i l 1 69

8.

 

 

 

°€Z

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(1 8 '3 isencws eq) ueq; Jeu'euxs shame
68 339‘ s; inq ‘seoumsisea em )0 was em .LON 5; ‘1‘}; pu'e
% ‘eums em 3; 3 ‘nnoaie [entered e u; 112mm
as In —- "9
31 11 'squm eq) ow; senI'eA umoua em 1nd
3 V sumac leueaed u; Jemod pun 01 3mm 13 esn use em
‘CL'I
0N 30A
Jensuv '31:
l ( ) = i’1
_ 8
259 *2» ‘68 *cz * “A“ ( )—zI
b a Z I ( ) = 1
I I I I 29 = M = I1
A I
.€8
m z “”383 (0V3)(‘1)=0V3 .
‘8 = —-—— = I _ l
I 383 A?- 3 Af7=(Uz)(Bz)=aV3 J43=CIV3=DHH
.Iamsu‘g .38
(I 8 axeded qoimos uo ems; 2 exam
Z , I _::-_ _ .I. _ 83
a, 8! _—__A8 ()-8 ()—
0 ea 123 ( ) ___ .I.I ( ) = IE
3 V
.€S
-332.__‘_L_L_-LH .3- + =3§5I+§2=8HII=~LI
ZSZ - AZI " ‘I‘q " 9 ’ 1’ Z A31 A31
I .zerasux; '35
CI 8
i 'aiqm sun usmu 'ppe Suzanna am
“I? 3 put: ewes em us an; seBBnOA eqi ‘nnoaie Ian'emd e 111
] '00; Simona 1311mm Bug/nos .10; exqei e esn use em
0 V

(112161 a mo eAee'I) . {8

 

{enbe .10 sums eqi

.ISMSLIV °33

8. Answer - .- L

1 ”parallel circuit

9- . ,
f f -‘ 'A

In a parallel circuit, there is more than one‘
path for current flow.

   

, Which diagram shows more than‘one path

 

for current to flow? ( )

 

38. Answer

 

 

( 6V)
( 8V)
39. r 7 . _ 77
If the currents are all equal through one branch
and'IT = 20a, 11 = ( ). 1111 113%‘114

 

ITWVVWLI‘I

68. Answer RT_162_X_ 20
Note that the answer

 

 

 

 

 

 

 

 

RT = (73—1% = 1T8 = 252 isznbothways.
69.
Ra -
RT = ( ), useRT R2
E 16v 2:; 89 _ en
T
IT = ( ), useI = W v
98. Answer 1" " =' (2)(§""');‘§ éiilé ,
1.2 2+.5 5 5 R (2)(6) 12 ga
R _(3)(6)_18=m . _2_6_+ _ T 2
2.3 ' 5 + 6 — 3'
99. —""FW(_8‘)"(8)W 64’" ____,~ .
R2.3 = ’ia—rr = = 4“ _L
R! R2 R3
(4) (4): 20 8V;
RT: 3+ 3 =-§= , I 441 8.42. 842.
T
I'1‘=R.r= )
9.

 

 

. O

LIH J0 z/I “Lu 91
* ‘ “NEE"
(co) (oz)

.Iemsuv rift

 

031

(A8) '(A8)

Jamsu v ° {9

 

258

—’VV\/-

fl.

‘saoqsgsea em 10 ums em you s; 3‘3 linens renamed e u; :Jeqwewea

—_ A31

8
<>=L

H

I
691 = zu+

___________ In: em seBeuoli em imam) tenured '8 III

'83
('89)

.Iemsuv '13

 

L.)

9. Answer

(a)

 

100 '

Circuit A is a parallel circuit.

 
 
 

Circuit 8 is a ‘ circuit. ._'-

 

 

. 39. Answer

(52.)

 

ho.

Circuit A is a . Circuit B is a

_l_ A .

A

 

 

69. Answer

 

 

(452)

(4a)
70.
Infig-ure 1: Infigurez: 311—» :4a——p

_(6)(3)_18_ 69 E 49 4o
-RT'—6_+TT"§‘ZQRT=( ’ET 3‘”
G v o _

ET = RTIT = 2(3) = 6v ET = ITRT = ( ) Figurel Figurez

 

99. A news r
4%.

 

5 R
' nr=< >

100.
We could have started with any pair of resistors.

 

 

 

12 U
'anva 11an 10 an SI "u .

69 'anlue new :0 m We; J's

U9 ( ) = “‘8 ( -> = ‘u

'enle/l .11qu J0 z/I 37)::- s; in em uaqz ‘lanbe iou are 91019133.: em n
' .9III
( UZ ) V

Jemsuv 'o'['[

 

 

 

 

 

 

( ) 3 La : > (. V ‘):,= “as ._
A9 - (uz)<vv) = 9Va'
cm.i , :Jva _-~ 1.3
'V'IQ '
- 3V , . . \
AQI- H fizafgialalu Z'OVK 38:39+88="LI (\J
V JGMSUV '09
“3 258 _TAQ In 239 —_:-_- A9.
-————->iz

 

'= 3 I r-————>s§ ' ’ “
: 143+; '39‘69+6V°qmmmiomm
.sioisisex em 10 um em iou st SWJ’UZ = % ___ {11.71— ___ J.“ ‘Ieuu'ed nuts
3

 

 

 

 

(in) .(us) (no)
.191in ‘05 ~
7.8 2* II! 1* -_L J1 z ‘
I I ' _ -( ) = ‘3’ . “mung a II!
-‘_ $.11]— . -

‘I3

(89) W e

.Iensuv ~03

 

10. Answer

l . . parallel

11.

I Ohms Law may be expressed three ways: ..

   

 

’40. Answer

; parallel circuit series circuit

Lil. i
.‘ The separate paths for current flow in a parallel T g 4
circuit are called “WAGES."
There are two branches in figure 1.
' How many branches are there in figure 2?

 

 

 

70. Answer

(‘29) (8v)

 

E
In both series and parallel circuits RT = I—T’ but in a series circuit, RT = R1 + R2, and
T

in a parallel circuit, RT = ( ).

 

See, they all give

 

100 Answr' r
. (2) (2) 4
B. =——=—=m R 1152.
A'R'I“ (6)(6/5)_3_6/5=1sz RT (3)(3/2)49/2 To
3+ 675 36/5 C- RT = W = —9/2 ‘ m
101 . u; :26» J wwwm-M

(6H3) _ 18 _
9 20

 

.11.

.02

cs= (91)-¥="‘u-
a.“

 

SJOISlflal renbe om 1o ¥ = .-
°0II
9.’ is 22.1,
.I’o‘LlSUV .60-E

'11 peeu MA 3; ems; 2 exam

 

 

 

=1 i1 ,_. . .
V1 :‘AQI ( )= We ( )= ‘11:
z -

L» I I] 29 = Z1 tz = Ii

VV—i '
008
. _ W ”-.-.-.” mm mm
sures ppe

.ISMSU'V‘ ’61.

   

6mm sun qswm ‘05

a g i ‘I‘a 1031.103 on eseqi ;o Kuv
3

.tensuv '6‘?

 

 

 

‘ ‘ J- zI+.[I:‘I‘I '()a‘LI'301P°:oeuuooemngyqaoqn
<L .. ‘nc . 7'1 '& nomsod n: at ‘ms :1
i: v 1’ — '98 = II N mm“ It! 9! Pas )1
Le” 81 \0__:l— '11 zI +13% 3 J‘I 'pp swarms sq: '{anomd n1
——~—~——---—--——O »
' o . » , .
,. a... A .03
( A9 )
‘ ( A9 )
Jensuv '61

11. Answer

(if; (1%) (m)

 

12.

A resistor and a lamp are connected in parallel across a 100v battery.

Draw the circuit.

 

b1. Answer

(3)

 

h2.
In a series circuit, the voltages around the circuit
gig and the current flow is always the same through
each resistor.

ET = EAB + ( l + ( )

 

 

71. Answer
1R1 R2
R1 + “2

 

72.

452

 

101. Answer
(19)
(8a)
102.

To solve any parallel circuit, take any two resistors
and put an equivalent resistance in their place.

Redraw this circuit with the R1 2 of the two 832

 

resistors.

°6I

 

4_.

68 U8 b= ( ) X8
=Ql=:.ii;&_= l

4" ”9 99 (8)(8)
l

H .

 

 

°on1m .uoxn JO Z/I s; 3 am ‘lunbo am pm: {amend u; an: 31013133.! cm; 11

’ 601:

$83.1.
JGMSUV '90:

_—

qoumq qoea qfinoxm squanno

 

sq) put sq; s; qoumq qoea ssome 981mm am ‘unono {snared 2 UI
1033133.! 11023 qfinoxq; aq; s;
iuauno sq) DUB ‘unono am punom safimmA sq; ‘nnoxgo sagas B U} °6L
$83.1.

.xemsuv °9L

 

 

 

 

 

 

 

 

.L
1"— ( ) = B_9 = i1. = ‘LH
A8 .1.
139 $ 233 '1- 3
:‘AB 89=BZ+BV=QA%+—UX§-=ZI+II=‘LI
Z I l
I I . ‘SJOlSlSBJ at” us J0 ums am {53 3; ‘Lu unoaio Ian'emd B u;
’6W
(68) (68)
.xemsuv '8’"! '
.1 CI 8
“:13 ( ) = ”"133 A9 = 8V3
I < )= 033 A9=:l3
El 3 V ,

93+ZH+IH=LH°€

J8£~18UV “BI

F—4

12..Answcr

 

 

13.

In a c' '
1rcuit, the total power ( PT ) is the sum of the power across each component.

Which formula says the same thing as the sentence? ( )

 

 

 

 

 

 

 

 

 

 

 

1 P = I E 2 =
T T'r PT ITRT 3.PT=P+P2+p
h2. Answer
EBC + ECD
h3-
In a parallel circuit, the currents through each A C E G
branch add and the voltage across each branch
is the same. _ ET ‘2‘
E = E = E = ( ) = ( ) T
T AB CD B D F H
72. Answer
8 4
(6(2) gorgor 1.3352
73.
IT = < )
20v; 109 109
Hint: first find RT , then find IT .
102. Answer
4S2 49
103.
(6) (6) _ 36 _ - -
R13 ‘T+T ‘fi ‘ 3“
RT = ( ) 10v—1 R1 69 R 69 3 69
IT = ( ) _ _
13.
.———-———~——~—»-—ie _ “WA

°eun A03 us ssome Ianemd u; an: 1332314 (33 us pus ‘uon 253 me ‘1819801 2” v

 

 

 

 

 

°80I
(88)
(61)
.IOMSUV'LO'E
< >=lu
A03: 3 99=‘LI
.82!
. =§ =i=fl=_B_L_QI_=
23va {7" or 08 (8H0!) La

.xamsu if ° 1 1.

 

— A91
I——>BZ ——->sz
°8’l
.L
2317 = E‘g = IL: 1‘3
El

Jemsuv ° m

 

Ss-zs-Is=‘1‘a'z

.I.
Shit”
3

geouaiues am 8?. Sum: ewes sq: sites emuuo; qomm

'seoumstsea renpuuput

sq; JO ums am 01 [cube 9; ( ‘La ) soueisisa: 112101 sq; ‘smistsaJ g to unono 881.138 2 u;

.81;
us .xo ‘3 put: ‘8 ‘V

JGMSUV °L'[

”Kw"

.5)

13. Answer

3.P =P+P+P

 

T 1 2 3
lb.
Write three forms of Ohm's Law.
1.( )
2.( )

3-( )

 

113. Answer
EEF EGH

hh.

 

 

73 . Answer

 

 

_(10)(10)_1oo_552 {_E_T_2_0v_4a
RT" 10_+1o ‘W‘ T‘RT‘sn‘
7b.
4a‘— -
ET=() ETT 69 39

 

 

 

 

11.10

 

 

’LI

'KJaneq A8 us ssome

Impaled u; us are 10131531 15;; 1e pine ‘09 e ‘uz v

 

 

 

 

 

 

 

 

 

 

 

. 'LOI
259 259 69 —— A31
Jamsuv “901
258 ( ) = 1.8
’11
( W)
I-)
JOE-XSUV '9;
Izz _ A8 ( ) Hump smuo aoumswal 1210; an 5; mm
.L
l—1 = “Ls mu pun aoumsmel ream an $1 $3 112m moux em
a
0 L71
1831
Jamsu V ‘91]
‘ ‘ o a v
( ) “amend u; siuawels moqs Simona qomm
'LI
«
ZI ) (Hd ) ”V

d
(F)

.1 among; ' 91

"___ .

1h.

A nswe r

‘1.E=Ia

E
2.I——§

E
3. R2? (any order)

 

15.

In a parallel circuit there is (one/mane. than one/no) path for current to flow.

 

 

 

 

 

 

 

 

 

 

 

ML. Answer ‘1‘ = 11 + 12 I _ _T _ 8v 4
E - 2 — 2 _ To a a 1 — 6
< I __ T _ 8V -2 T - a
_I‘Tz‘f‘m'a IT=2a+4a
us,
I IT splits up atAinto 11 and 12'
12 splits up at B into :13 and I4
2" I3 andl4joinupat’Caggproduce Ix.
‘ So, ‘Ix = 12., Ix' = ( "').
'711. Answer~
" , _(e)(a) 18_
R'r'fia. “9“29
7S,-
4.8v"—L 69 452
I =( ) _-
T T
8)(8) 64 ”s ' ‘ 4
1011. Answer =—8—8_( = = 4n _ T _ 8v _
‘ R2.3 + 1—6 IT - TR'T— — m — 63
_(R1)(Rz.3) (2)(4) _§_g_119
RT”R1+R2 + ‘6'3‘5 ... ...
'105. l '
. _ .
.(4)(4) -16=23‘2F- 4
R =- —-B— irst R
1.2 3+3 . ,0
. .
(2)(2) _. 4 — 152 h R .
=‘T’T "I - Ten 29 1-2
RT + o 29

 

‘ '91 ‘ - C

‘nnona am mam
.1513an A21 1: ssome [auemd u; paoeld an 9.1033193: :39 aanu. .

.9OI

(us)

..1 emsuv ° 501;

73 .10 In Jaime

mun .xau'euxs 852mg st ‘11] snnono leaned in :Jaqmauxau

 

 

 

'( )=-‘u ...
Igz‘
. W 1 .
RZ=Z'I+B'=ZI+II=J‘I uz.t=%=zl 28.=.Ag?b=l 0

Jamsuv '51,

cea‘avuq‘ >- - 7‘ v A- ..

r - ”=11

’9’!

 

PW)

ISMSUV ogq

 

 

...-"uau f

0

L951. 1’ .
Jeusuv '51:

APPENDIX C

CROWDER PROGRAM

207

w—'fl._u

OHM’S lAW‘AND POWERS OFTEN

 

NAVAl AIR TECHNICAL TRAINING COMMAND

A cknowled geme nts

This program was adapted by the Naval Air Technical Training Center,
Memphis, Tennessee, from a larger one developed under contract No. AF 33
(6161-6983 CPFF between the U.S. Air Force and Western Design and Electronics.
In a letter dated 27 July, 1961 from the Aeronautical Systems Division,

Air Force Systems Command, United States Air Force, use of these materials
by other governmental agencies was confirmed. Appreciation is expressed
’00 the Aeronautical Systems Division and to Western Design and Electronics

for assistance in making the original prOgram available.

 

V

4V7 v

Page 1

In this course you do not go from page 1 to page 2 to page 3 etc, but

after reading the material given, a question will be asked and your answer

will determine the page you go to next.

You advance at your normal rate of learning. You cannot advance until
you.are right.

We will start out with Ohm's Law and Powers of Ten and go on to simple
series circuits in the next lesson, and then to simple parallel circuits

in the third lesson.

Ohm's law can be summed up: Voltage (EMF) equals amperage times
resistance. In mathematical shorthand, E is the symbol used for voltage,

I stands for amperage, and R for resistance. The formula for Ohm's law
is expressed as follows:
E = I x R or E = IR
That means electromotive force (in volts) equals current (in

amperes) times resistance (in ohms).

Now, if I equals .5 and R equals 12 ohms, what does E ‘equal?

6 volts Pate 3 w
24 volts P359 5
60 VOltS Page 7

Page 2

YOUR ANSWER: The unit in which I is measured is the ohm.

No. And since you are apparently still uncertain about
what unit measures which quantity, or which letter is the

symbol for which quantity, here they are again.
Resistance (denoted by the letter R) is measured in ohms.
Voltage (denoted by the letter B) is measured in volts .
Current (denoted by the letter I) is measured in amperes.

Return to Page 3 for the correct answer.

 

YOUR ANSWER: 6 volts.
IR = .5 x 12 =

. Right. What is the unit in which I is

The Ohnl Page
The ampere .

Pa ge

The volt

Page

Fame 3

6 volts.

measured?

Page 2.

YOUR ANSWER: An ampere is equal to the electrons that will

flow through a resistance of one ohn.
No.

You haVe begun to say something. But you have not

finished.

Amperes denote coulombs—per—second. Your answer is
an incomplete statement because it contains no mention of the

time involved or the quantity of electrons.

You can't just splash around with words and hope they

come out right -- not in electronics, at any rate.

One of the alternative statements is much more precise.

Return to Page 6 and select that one for your answer.

 

Page 5

YOUR ANSWER: 24 volts.
Wrong.

This is the problem again: If I equals .5 amps and R equals 12
ohms, what does E equal?

Here is the formula once more: E = IR.

You seem to have divided I into R, instead of multiplying as the

formula indicates.
The calculation you should have made is:
E = IR = .5 amps x 12 ohms = ? volts.

Complete the calculation and then return to Page 1 for the

correct answer.

YOUR ANSWER: The ampere.
Right. Current is measured in amperes.

What is an ampere equal to?

The electrons that will flow through

a resistance of one ohm. Page 1.

Resistance times voltage. Page 5‘,

40‘

One coulomb of electrons per
second. Page ll.

 

Page 7

YOUR ANSWER: 60 volts.

No. You apparently got a little careless about the placing of the
decimal point. The decimal point is important, to say the least.

The problem: If I equals .5 amps and R equals 12 ohms, what
does E equal? .

The formula: E = IR
The calculation: E = IR = .5 amps x 12 ohms = ? volts.

Taking care to be a shade less reckless with the decimal point this

time, return to Page 1 for the accurate answer.

Page 8 6

YOUR ANSWER: An ampere is equal to resistance times voltage.
No.

This is a wild attempt at rendering Ohm's Law in different

form. In addition it is a hopelessly imprecise statement.

If expressing any scientific law, you have to be precise
about the quantities involved. "An" ampere means "one" ampere.
Your answer seems to say, " One ampere is equal to an indefinite
amount of resistance multiplied by an indefiniteamount of

voltage" .

To make matters worse, the form in which you have met
Ohm's Law so far is E = I/R, current (in amperes) equals voltage

divided by resistance. That's something we'll come to shortly.

To get back to the question: Do you recall what an ampere
is? It is another way of saying electrons are flowing at the

rate of one coulomb per second, isn't it?

Bear that in mind as you return to Page 6 to select

another answer .

 

 

YOUR ANSWER: B stands for resistance.
Oh, come now!

B is the symbol for voltage, or, if you like, electromotive
force.

As you have just seen, I is the symbol for current, and R
is the symbol for resistance. . ‘

Return to Page ll. , vowing never to forget that I stands
for current, E for voltage and R for resistance.

‘0

Pa {5e 9

YOUR ANSWER: .. 25a

If you got that answer right the first time, you did well
because you discovered one of the two variations of Ohm's Law.

We began with this: E = IR. The variation is written:

_E
“a

It is used when E and R are known and you need to find I.

Of course, both the original statement and the variation

are the same thing. If E = IR, then

_E
“a

It is perfectly legitimate to perform the same mathematical
operation to both sides of an equation. To change Ohm's Law for
purposes of discovering amperage, we merely divided both sides

of the equation by R.

If E and I are known but not R, how would we vary the

formula?
R = g1:— Page 12
R = IE Page 21
You can't find R from the
other two. factors. l‘a go. 23

 

Pa {{6 10 . 6 ‘

 

Page 11

YOUR ANSWER: The unit in which I is measured is the volt.

No. And since you are apparently still uncertain about
what unit measures which quantity, or which letter is the

symbol for which quantity, here they are again.

Resistance (denoted by the letter R) is measured in ohms .

 

Voltage (denoted by the letter B) is measured in volts.
Current (denoted by the letter I) is measured in amperes.

(\ Return to Page 3 for the correct answer.

E

YOUR ANSWER: R = —1—
Right. In this case, both sides of the equation are
divided by I.
So, Ohm's Law can be written in three different ways,

depending on which variable is unknown. The three ways are:

E=.IR
_E
R—r
_E
“a

Now, if you are given E = 6 and I = .3a, what is R?

.0552 Page 19
28.2 Page 2),
2052 Page 26

rs

Page 13

YOUR ANSWER: Voltage.

Good for you. E stands for voltage. This is not as hard to
remember as it may seem. Voltage is the unit in which we measure

electromotive force (EMF, as it is called).

Which of the following would be an accurate statement about
voltage?

Voltage is the electrical pressure needed to move

a coulomb through one ohm of resistance. Pa go 15

Voltage can be found by multiplying resistance
19
times current. Page

Voltage is the electromotive force of a resistance Page 22

of one ohm.

YOUR ANSWER: One coulomb of electrons per second.

Correct. One of the wrong answers , " The electrons that
will flow through a resistance of one ohm " is not a
complete statement. If it had been stated as: " The number of
electrons that will flow through a resistance of one ohm.
in one second when one volt of electrical pressure is being

applied" , it would have been accurate.

Now, what does E stand for?

Resistance Yen-7w 9

0‘
Voltage Page 13 |
Current Page 17

 

 

0

Pa n6: 1‘)

YOUR ANSWER: Voltage is the electrical pressure needed to

move a coulomb through one ohm of resistance.
Almost, but not quite.

Somewhere in that statement you need a reference to time.
You can say, for instance, that a volt is the electrical pressure
needed to move one coulomb through one ohm of resistance in.

one second .

You could say that voltage is electromotive force and that

it is measured in volts .

Unfortunately, neither of these complete answers is

available to you .

There is another way of arriving at voltage, and that is
through Ohm's Law. If you know the resistance and the current

present in a circuit you can calculate voltage.

And that should be enough of a hint for you to return to

Page 13 and select the correct answer.

Pa ge 16

YOUR ANSWER: 4a
Wrong.

Here is the problem again: If E = 24v and R = 969, what
is the value of I?

Ohm's Law as we have considered-it so far says: E = IR.
Obviously we need it in different form since I is the unknown

quantity.

To isolate I, divide both sides of the equation by R. This

gives you:
13 _ m
R .7
Thus I = E/R = 24 volts/96 ohms = ? amperes.

Return to Page 19: with the correct answer.

 

 

(\

Page 17

YOUR ANSWER: E stands for current.

No. No. And again, no. You've seen enough of these ,
symbols by now to be able to recognize them.

B is the symbol for voltage, or, if you like, electromotive

force .

As we saw just a moment ago, the symbol for current is I.
And R is the symbol for resistance. '

Return to Page 11. , vowing never to forget that I stands
for current, E for voltage and R for resistance.

Page 18

YOUR ANSWER: Voltage can be found by multiplying resistance times

current flow.

Ohm‘s law, E 2 IR. Right you are. The statement that voltage is
the electrical pressure needed to move a coulomb through one ohm of
resistance, was almost correct as a definition of one volt. But it
lacked a time factor. If it had been modified in either of two ways, it would

have been correct:

I. A M is the amount of electrical pressure needed

to move one coulombthrough one ohm in one second, or

2. A volt is the amount of electrical pressure needed to

move one ampere through one ohm.

The Greek letter omega, which looks like this S2 , is the abbreviation
for ohm. The shorthand for volt is simply v or V; that for amperes is

a. We could say:

That statement is true. It is Ohm's law written in different
symbols. However, it is not proper usage. Now, here is a

problem which illustrates the proper usage:

IfE = 24vandR = 969 , whatisI?

I = - 253 Page 10
I = 4a Page 16
I = 2304a Page 20

Q,

YOUR ANSWER: . 059

Wrong.

You’ve got things topsy-turvy.

(R The equation you should be working with says:
_ E
R "- r

The equation you seem to have used says:

_ I
R "a
Go back to the version you should have used and when you

have the answer, return to Page 12

Page 19

YOUR ANSWER: 2304a.
Wrong.

Here is the problem again: If E = 24v and R = 969, what

is the value of I?

Ohm's Law as we have considered it so far says: E = IR.
Obviously we need it in different form since I is the unknown

quantity.

To isolate I, divide both sides of the equation by R. This
gives you:
.113 _ IR
R ’ 77
Thus I = E/R = 24 volts/96 ohms = ? amperes.

Return to Page 18 with the correct answer.

Page 20 (

 

l

 

Page 21

YOUR ANSWER: R = lE‘

No. You're getting tangled up in transposing this equation.
We know that Ohm's Law can be expressed E = IR.

In a case where we do not know the value of R the
equation can be adjusted to isolate R by dividing both sides by I,
like this:

You see that the top and bottom I in the right-hand group
cancel each other, leaving the final equation:

E

R=T

Work it over once more for yourself and then return to

Page 10 for another answer.

' ' Page 22 Q 1

YOUR ANSWER: Voltage is the electromotive force of a

resistance of one ohm.

No. This is one of those answers that doesn't really
say anything. It would be as sensible to say that voltage

is the electromotive force of a stick of licorice.

A resistance of one ohm , got into the conversation
because it was possible to pass a current of one ampere

throw—h it by applying a voltage of one volt.

And, you may recall, that led us to Ohm's Law in the

form E = IR —- voltage equals current times resistance. .

You should now be well enough informed to return to

Page 13 and select the cOrrect answer.

 

Pa ge 23

YOUR ANSWER: You can't find R from the other two factors.
Your mathematics is coming unstuck.

In an equation involving three factors, if you know any

two you can find the third.

You have an example in Ohm's Law when it is expressed
E = IR. If you do not know E but have values for I and R, then
multiplying together the known values will give you the value of E.

That is what the equation tells us.

And if we divide both sides of the original equation by I, we
can obtain an equation which isolates R and expresses it in terms

of the other values, like this:

E_/IR . _
I —Z—leav1ngR—

If you have trouble in following this, substitute simple

IE
1

numbers for the factors in the original equation, thus: Instead
of E = IR, write 12 = 3 x 4. If you want to isolate the factor 4,
divide both sides of the equation by 3, like this:
12 _ 3 x 4 _ _ 12
737— — T leavmg f1 — 3

This method of substituting simple figures for factors

 

within a formula is useful as a check in many cases.

And with that, return to pa pp 10 and select the correct

answer.

YOUR ANSWER: 252
No.

Decimals can be a nuisance, but ignoring them won't make

them go 'way.

The equation you need is the version of Ohm's Law with

which we began the last page, R = E/I.

If we apply that to the present problem, in which E = 6v

and I = .3a, we get:
R = E/I = 6 volts/.3 amperes = ? ohms.

Complete the calculation, this time allowing for the

decimal point, and return with your answer to Page 12

Page 2b,

7) , Page 25

YOUR ANSWER: . 2 volt.
Wrong.

Obviously you needed prompting after all on the form of
Ohm's Law to be used.

You should have been back with the first version of the
formula, the one you should have committed to memory. Here
it is again. Memorize it.

E = IR

In this case we know that R = 10 ohms and I = 2 amperes.
E = IR 2 2 amperes x 10 ohms = ? volts.

Return to Page 26 with the correct answer.

YOUR ANSWER: 209
“v Correct. R = E/I = 6/.3 = 209,

Now without any prompting on the form of Ohm's Law

you're to use, what is the value of E in this circuit?

Renal).

 

 

 

2 volts Page 25
5 lets Page 28
20 VOltS Page 30

Page 26

(l

Page 27

YOUR ANSWER: 1. 2 volts.
Wrong.

This looks like a guess that was guided by the form of the
other answers. And if it wasn't then your arithmetic isn't

working too well today.

Let's go over the simple calculation: The resistance is
6 ohms, the current is . 02 amperes. The equation you need to

find the voltage is E = IR.
E = IR = .02 amperes x 6 ohms = ? volts.

When you have the answer, return to Page 30.

Page 28 ‘

YOUR ANSWER: 5 volts.
Wrong.

Obviously you needed prompting after all on the form of
Ohm's Law to be used.

You should have been back with the first version of the
formula, the one you should have committed to memory. Here

it is again. Memorize it.

E = IR
In this case we know that R = 10 ohms and I = 2 amperes. ..
E = IR = 2 amperes x 10 ohms = ? volts.

Return to Page 26 with the correct answer.

f“

Page 29

YOUR ANSWER: Not 1.2 volts

Correct.

This is the calculation:

E =IR .02 x 6 2 .12 volts.

This form of-Ohm's Law, E:=IR, is the most important and most often
used version. In addition, it has the advantage that it is the easiest
version to convert to the other forms. Merely dividing both sides by
the unwanted quantity will leave you a desired quantity at the right.

You should memorize the law in this form.

Now lets check your understanding of Ohm's Law in the next four
questions. Just choose one of the distractors and keep going. These
will be checked at the end of the last period. Just put down a, b, c,
or d for your answer.

In using Ohm's Law, if R is 20 ohms and E is 100 volts, what will
I be? I
0 a. .2 amps

b. .5 amps

c. 2 amps

d. 5 amps

After selecting an answer go on to page 31.

Pa pp 30

YOUR ANSWER: 20 volts.
Correct. E = IR = 2 x 10 = 20 volts.

Keep in mind what the letter E stands for: It is voltage,
the unit of measurement for electromotive force. Voltage is
really a measurement of the difference in electrical pressure
between two points. It is this difference in pressure that exerts
the force to make electrons move from one point (the point of

higher pressure) to another point (the point of lower pressure).

What is the EMF of this circuit?
R: éfl.

 

1. 2 volts Page 27
Not 1. 2 volts Page 2‘}

I don't knOW. Page 32

)

:3 ‘ ' Page 31

What is the resistance of an electric soldering iron if it operates
on 120 volts d-c and draws 3 amps.

a. 360 ohms

b. 40 ohms

c. 25 ohms

d. .025 ohms

After selecting an answer go on to page 33.

Page 32 ‘
YOUR ANSWER: I don't know.

There's nothing new in the text at this point. Is it the

diagram that bothers you? Here it is again:
R = all

 

And the only thing that is new about this diagram is the
method of showing the voltage source. This is the reason for
the reminder that preceded it: "Voltage is a measurement of the Q,
difference in electrical pressure between two points. It is this
difference in pressure that makes electrons move from one
'point (the point of higher pressure) to another point (the point

of lower pressure). "

In this diagram we are showing a voltage source without
indicating just what form that source takes —- it might be a
battery, it might be a generator, it could be that this is an
offshoot of some larger circuit. You are not told what it is and
it doesn't matter what it is, as far as the problem is concerned.
The important thing is that voltage is being imposed on this

circuit and that it is causing a current of . 02 amperes.

Once again the voltage can be found by using the E = IR
form of the Ohm's Law formula. When you have a value of E,

return to Page 30 and select the correct answer. ‘1

Page 33

Solve for the unknown in the diagram below.

I R=15,000 ohms

 

 

E=..?. {—
-:f_. I =.005 amps
(A)
a. 3,000,000 volts
b. 3,000 volts
c. 750 volts
d. 75 volts

After selecting an answer go on to page 3h.

Page 3h

If the resistance of the resistor in the circuit below decreases in

value, what will happen to the current in the circuit.

  
 

 

E 10 volts

a. it increases
b. it decreases

c. it stays the same

After selecting an answer go on to page 55 and continue with

Power's of Ten.

 

Page 35

YOUR ANSWER: 8 X 1030 would be 800 amperes.

No, this answer is wrong twice. The omega (0.) is the Greek
letter that is used for ohms , not amperes. You will find amperes

shown: A, a, or amp.

Also, you have chosen the wrong power of ten. Remember
that 103 is the same thing as 10 cubed, 10 x 10 x 10 or 1000. You
must move the decimal point three places to the right when you
multiply a number by 103 .

Now return to Mm 38 and choose the correct answer.

Page 36

YOUR ANSWER: 8 x 103rt is 8000 amperes.
No, the Greek letter omega (I1) is used for ohms, not

amperes. You will find amperes indicated: A, a, or amp.

However, you did choose the correct numerical value for
8 x 103. You move the decimal point three places to the right

to multiply by 103.

Now return to page 38 and choose the correct answer.

In

Page 37

YOUR ANSWER: 2. 5 x 104v means 2500 volts.

No. When you multiply by 104 you must move the decimal point four
places to the right, as you would to multiply by 10, 000.

The thing to remember is: The exponent tells you how many places

to move the decimal point.

If you feel you need a more detailed explanation of powers of ten,

please go to Page 55

If this was ju:.- . a slip and you think you understand, return to P2 ge 1.3

and select the correct answer.

How would you translate 8 x 103.52?

800 annperes E. Page 35

28000 amperes Page 36
800 Oth Page 39

8000 ohms Page 43

C/

Page 38

Page 39

YOUR ANSWER: 8 x 1030 is 800 ohms.

No, you've multiplied by the wrong power of 10. Remember
that 103 is 10 multiplied by itself three times (10 x 10 x 10 = 1000).
To multiply by 103 you must move the decimal point three places to
the right, so that you multiply by 1000.

Now return to Page 38 ' and choose the correct answer.

 

0
Page ’20

YOUR ANSWER: 2. 5 x 104v = 25,000 volts.

Right. Since the decimal point is before the 5, moving it four places
to the right yields 25,000.

If you came here by the direct route and now find this isn't completely
clear, spend a few minutes reviewing powers of ten by going immediately
to Pam 55

If it is plain, let's consider powers of ten with negative exponents.

To multiply powers of ten, you add the exponents. Or, to put it more
precisely than we have done until now, you add the exponents algebraically.
We introduce that word "algebraically" to take care of the instances in
which negative numbers are concerned.

To divide powers of ten, you change the sign of the number you are

dividing by and then proceed as in multiplication, adding the exponents

0.2

algebraically.
An example will make this plainer. Say you want to divide 100 by
10, 000. You know the answer is going to be 0. 01. Let's do it with powers

of ten:

[0

100 _10 _ 2—4_ —2_
10,000"—”10_‘10 -0.01

10

You see what's happened? In dividing, we arrived at an answer that

 

was less than 1. Our answer is 1 x 10-2, which is 0.01. To increase
0. 01 to 1, the decimal paint moves two places, but this time it moves
to the _r_i_g_ht_.

To arrive at a positive exponent, we moved the decimal point to the
left. This time, by moving the decimal point in the opposite direction, we
have an exponent with the opposite sign —- a negative sign.

Now you try it. How would you express 0. 000001 as a power of ten?
1 x 105 Page 1.12 1 x 10'6 Page 59 Q9

_ P ' It
1 x 10 5 age 7 I don't get it Page 71

.-

Page Al

YOUR ANSWER: 9 .8 x 10'4A means 0.0098 amperes.

Well, it looks as if you have four decimal places in your
answer, but you. have actually moved the decimal point only

three places to the left.

To multiply by 10‘4 you must move the point four places
to the left. Pay no attention to the presence or absence of
digits on either side of the deCimal point when you multiply by
exponents . Simply reproduce the figures in the same order they

were in, and move the decimal. point the propernumber of spaces.

Now return to page M and choose another answer.

 

 

\
O .
Page AZ.
YOUR ANSWER: 1 x 105
Wrong.
Just as 1 x 102 is equal to 1 x 10 x 10 (using 10 as a factor two times)
and 1 x 103 is equal to 1 x 10 x 10 x 10 (using 10 as a factor three times),
so 1 x 105 is equal to 1 x 10 x 10 x 10 x 10 x 10 (using 10 as a factor five
times), which equals 100, 000.
But we asked you to express 0. 000001 as a power of ten.
How would you write 0. 000001 as a fraction?
. 1 ‘ ‘
1,000,000 _s
t

Then express 1 as a power of ten: 100 (If you don't know why 100 = 1,
‘ please refer to the question on
Page 11.3__ before proceeding)

[And express 1,000,000 as apower of ten: 106

100
So you have —-—6- .
' 10

Remember the rule for dividing powers of ten: Y0u change the sign
.of the number you are dividing by and then proceed as in multiplication,
adding the exponents algebraically.

6 0-6

0x10' = 10 = ?

Think of it like this: 10°/106 = 10

‘ Now return to Page 1.0 for the correct answer.

Page AB

I

YOUR ANSWER: 8 X103!) 4.; 8000 ohms.

Right. When the exponent (the number above the line) is positive,
. we add to ~the left of the decimal point the number of places indicated
by the exponent. In the expression 8 x 103, the 8 is read as 8.0; that
is, the decimal point is at the right of the 8, and the 103 ("10 cubed",
as it is read) has the effect of shifting this decimal point 3 more places
to the right. And, 52 (omega) is the symbol used for ohms.

Now, try this:

2. 5 x 10% ...
2500 volts Page 37
25000 volts Page 00
250000 volts Page 1,5

I don't understand Page 55

Page M.

YOUR ANSWER: 0. 00007 amperes.

Good. The rule with negative exponents is add to the left of the
decimal the' number of places shown by the exponent. Since the decimal
point was to the right of the seven and the exponent was minus five, we

count off five places to the left to locate the decimal point of the product.

Now, try this one:

9.8x10'4a1s. ..

0. 0098 amperes Page 1.1

0. 00098 amperes Page 1.9

0. 000098 amperes Page 52

Page A5

YOUR ANSWER: 2.5 x 104v4means 250,000 volts.

Whoops, you went a little too far! True enough you have four zeros
in your answer, but multiplying by 104 means move the decimal point
four places to the right. You have actually multiplied by 105 to get this

answer.

Multiplying by 104 is the same thing as multiplying by 10,000. This
multiplier may have four zeros, but your answer may not have anyzeros
at all. For instance, you could miltiply 3. 897645 by 106 and you still
wouldn't have any zeros in your answer. It would just be 3, 897, 645. The
exponent, remember, tells you how many places to move the decimal

point.

If you feel you need a more detailed explanation of powers of ten,

please go to Page 55

If this was just a slip and you think you understand, please return to

Page 143 and select the correct answer.

Page L6

YOUR ANSWER: 0. 0025
Good.
102 -3
You multiplied 2. 5 x —-§ to get 2. 5 x 10 .
10

To complete the multiplication, you move the decimal point three

places to the left.
What is 7 x 10‘5a ?
0.00007 amperes Page 1.1.

0. 000007 amperes Page 1.8

Pa ge 1&7

YOUR ANSWER: 10‘5.

Not quite.

It is true that your answer will contain a negative exponent --— it must

if you are expressing: a quantity that is less than one.

We asked you to express 0. 000001 as a power of ten.
This will be ten to the minus something.

Here's a clue: The first place to the right of the decimal point is 110°
The second place to the right of the decimal point is 1100- . The third

place to the right of the decimal point is 1000’ etc.

. . . 1
SO 0. 000001, written as a fraction, is 1’000, 000 .

 

Translating this into powers of ten: 1 = 100. (If you don't understand

this, please refer to the
question on Page 113
before proceeding)

1,000, 000 = 106.

0

1 becomes —0—
1, 000, 000 106 '
O x 10'6 = 100-6. All you have to do now

 

SO the fraction

Divide them: 10°/106 = 10
is to add the} exponents algebraically.

Return to Page 1.0 and try it out.

Page A8

YOUR ANSWER: 7 x 10‘5a means 0.000007 amperes.

You're too enthusiastic. Be careful when you multiply by powers of
10, negative or positive, that you don't count zeros. True enough, you
have five zeros in front of your 7 above. But when you multiply by 10-5,

you are supposed to move the decimal point five places to the left.

To begin with, the decimal point is on the right of the number, 7.0.
The first place moved to the left just puts the decimal point on the other
side of the number, 0.7, and this is the same "thing as multiplying 7. O
by 10'1.

The exponent, remember, tells you how many places to move the
decimal point. The sign of the exponent tells you which way to move the

decimal point.

Now return to Page 1.6 and choose the correct answer.

Page 1:9

YOUR ANSWER: 9. 8 x 10-4a is 0. 00098 amperes.

Exactly. The exponent is -4. That means that the decimal point

must be moved four places to the left.

Let's restate the rules again before moving on:

 

I POSITIVE To express a large number as a smaller number
EXPONENTS

 

 

multiplied by a power of ten, move the decimal point to
the left and count the number of places moved. The
number of places counted is the proper positive power of

ten.

 

NEGATIVE To express a decimal as a whole number times a power
EXPONENTS

 

 

 

of ten, move the decimal point to the right and count the
number Of places moved. The number of places counted

is the proper negative power of ten.

These are important rules. Make sure you understand them before
going ,on. Once it is plain to you that the exponent is an instruction about
how many places and in which direction the decimal point is to be moved,

you should have no trouble.

Here's some practice in using powers of ten, this time involving a
more practical use. Complete the calculation:
(2.5 x 10352) x (2 x 10'2a) = 7

5 watts Page SL.
50 watts Page 56
50 volts Page 58

500 volts Page 60

4 WW7 fi‘k l

Page‘SO

YOUR ANSWER: 300 volts becomes 3 x 103v, and 2400 ohms becomes
24 x 103 ohms.

No, both numbers are off a little. 103 is 1000, so 3 x 103 would
be 3000. Likewise, 24 x 103 would be 24, 000.

Now return to Page 58 and choose numbers that express the

values given.

Page 51

YOUR ANSWER: 0. 00025
Wrong. You're getting zero—happy.

Here's how you should have worked the problem:
102 —3
2. 5 x — = 2. 5 x 10

105

10'3 tells you that you must move the decimal point three places E
the left.
Study the following table, and you'll see why:

0
1.0 1 = 1X10 = lxlOw-m = 1::100

1 1x10
1 1 x 100 (0-1) -1.
01,=1T)=—-—1=1x10 =1x10
1x10
0

1 _ 1x10 1x10(0--2) = 1x10—2

1 10
1 1x100 (0-3) -3
0.001 = ————= —————— = 1::10 = 1::10
1000 3
1x10

Return to Page 59 for the correct answer.

k .

1%) [15;

YOUR ANSWER: 9.8 x 10‘4A means 0.000098 amperes.

No, it looks as if you are counting the number of zeros instead
of decimal places. To multiply by 10.4 you should move the decimal
point four places to the left. ,The first place to the left will give
you 0.98 which is the same as multiplying by 10—1 . There are no
zeros at all.

Count the decimal places in your answer again, and you will
see that you have actually multiplied by 10-5

Now return to Page 1.1... and choose another answer.

\n
In

Page 53

YOUR ANSWER: 300 volts becomes 3 x 102v, and 2400 ohms
becomes 2.4 x 104 Ohms.

Well, you' re half right. You have the right expression
for the volts, but the number of Ohms is wrong.

When you multiply by ,1 04, you are really multiplying by
10,000. 2.4 x 104 would be 24,000. To convert a large number
to a pow. of 10, be sure to count your decimal places very
carefully The exponent tells you how many places to move the
decimal point, remember.

Now return to Page 58 and choose another answer.

Page 51,

YOUR ANSWER: (2.5 x 1030) x (2 x 10’2 amp) = 5 watts.

NO, this shouldn't come out to 5 of anything. When you
multiply 2 x 2 .5 you get 5, but you have to figure out your
exponents . As you remember, you must add them algebraically.
When you do this, you come out with 101 or just 10. To get

the correct answer you must now multiply by this .

And where did you get watts for your answer?

 

You should recognize that ohms times amperes comes
directly out of the Ohm' 5 Law relationship. Something = IR. '
What is it?

Return to Page 1,9 and choose another answer.

i O

Page 55
In electronics we work with numbers which go from one extreme to the
other Numbers in the millions are used in equations with numbers in the

millionths.

Numbers mount into the millions when you're speaking of resistance
and dealing with ohms.
short cut.

It can get very cumbersome if you don't use a

For example, say you want to multiply 250, 000 by 2, 000. In long-
hand multiplication, your problem looks like this:

2 50000
2000

000000
000000
000000
500000

500‘000‘000

But it isn't necessary to cope with all these zeros. You can get the
same result another way.

The first step is to take each number concerned and keep on dividing
it by 10 until there is a single whole number between 1 and 9 to the left
of the decimal point. 0

For instance, if 500 is our number, divide it by 10 (which reduces
it to 50) and again by 10 so that you are left with the number 5.

We'll take care of the vanishing zeros in the second step. Before
we go any further, what is the result of applying to the number 4000 this
first step of repeated division by 10?

4 Page 63

40 Page 67

 

Page 56

2

YOUR ANSWER: (2.5 x 1030.) x (2 x 10— amp) = 50 watts.

Your number 50 is O.K. , but how did watts get in here?
You should recognize ohms times amperes as something right
out of Ohm's Law.

Something = IR. What is it?

Return to Page 11.9 and choose another answer.

0

Pag. 57

Ohm' 3 Law can be stated in three forms:

E
R

E = IR I = R =1?-
Since we want to find amperage —- we know the other
two factors —- we choose the middle form of the law as shown

above.

Once again, the resistance is 2.4 x 1030 and E is

3 x 102v. How will the answer be expressed in powers of 10?

1.25x lo'la Page 61»
1.25x 101a Page 68
o 1
7.2x10 a page 70
5
7.2X10 a Page 73

Page 58 n

YOUR ANSWER: 50 volts .

Correct. Here are the steps:

2

(2.5 x 1030) x (2 x 10_ amp)

Since the order of multiplication doesn‘ t make any difference,

we rearrange the problem to:

(2.5 x 2) x (103 x 10‘2)

2:101)

(2.5 x 2 = 5) and (103 x 10'
S O
(5) x (101), which is 5 x 10, and that equals 50.
The answer, of course, comes out in volts since this was
actually a problem involving Ohm' 5 Law: Ohms times amperes
equals volts .

Now, let's try another one using Ohm's Law.

We are given 300v and a resistance of 2400 ohms, and have to
find the amperage. _Which of the following translates these figures
into powers of ten?

3 x 103v and 24 x 103 Page 50
2 4

3X10 vand2.4x10 Page 53
2 3

3X10 vand2.4x10 Page 61

YOUR ANSWER: 10'6.
Correct. 0

. ‘ . . 1 10
When you write 0. 000001 as a fraction, 1t becomesm or 10?.
You add exponents to multiply powers of ten. When you divide
powers of ten, you change the sign of the number you are dividing by, then
you go ahead as in multiplication —- adding exponents algebraically.
100/106 = 100 x 10‘6 = 10(0—6) = 10

In division as well as in multiplication of large numbers, we rely
on this short cut. Here is a problem ill long division:
200,000 _ 200,000 = 200 = 50
4, 000 _ 4, 000 4

Now 100k at the same problem expressed in powers of ten:

 

5
2 x 103 2 2_ x 10(5—3)
4x 10

= 0.5 x 102.
Multiplying by 102 means moving the decimal point two places to the
right. So 0. 5 x 10 becomes 50. This time let's make our larger number
the lelsor: 25’ 000
5,000,000

Here's the same problem expressed in powers of ten. When you

= 0.005

change the sign of 106, you get a negative exponent. Adding algebraically,

you end up with a negative exponent for your answer:
3
25x10_: 255X 10(3-6) 3

5x106

— 5x10—

Multiplying by 10"3 means moving the decimal point three places to the
left. So 5 x 10‘3 becomes 0.005.

(Note: For simplicity, we didn't reduce 25,000 any lower than 25 to
start with, since it was to be divided further within the problem.)
2
What does 2. 5 x i equal?
5
10
O. 0025 Page [#6
0.00025 Page 51
0.025 Page 62

Page 60

3 2

YOUR ANSWER: (2.5 x 10 o) x (2 x 10' amp) = 500 volts.

You must not have added your exponents correctly. Add 3 and -2
together algebraically and you get 1, so you must multiply your answer

by 101, which is just plain 10.

Now return to Page 1.9 and choose the correct answer.

in»

Page 61

YOUR ANSWER: 3 x 102 V and 2.5 x 103
Right. You just use the same process backwards. The exponents

indicate how many places the decimal point was moved.

In using Ohm's Law if E=.3 x 102 volts and R=:2.h x 103 ohms, what
will I be?

a. 1.25 x 101

b. 1.25 x 10'1

c. 8 x 100

d. 8 x 101

After selecting an answer go on to page 65.

Page 62

13

YOUR ANSWER: 0. 025
No. You're trying to give back more than you got!

If someone had a coi‘ffi‘fi‘that wasn't even worth one cent, why give them

two and a half cents for it?

When you put only one zero to the left of the decimal point before

writing the number, you are indicating %00§' This is not the right answer.

Look at the problem:
2

22.5le = 2.5x10(2-5) '= 2.5x10-

105

Ten to the minus third power tells you that to complete this multi-

3

plication you must move the decimal point three places to the left.

. Study the following table, and you‘ll see why:

 

 

0
1.0 = %— = ”100 = 1x10(0’0) =1x100
1x10
1 1x100 (04) —1
0.1=1—O=——————1=1x10 =1x10
1x10
1 1x100 (0-2) —2
0.01.-1W: 2:1X10 :IXIO
1x10
1 1x100 (0-3) -3
0.001 - =1x10 = 1x10

 

1000 1 x 103

Return to Page 59 for the correct answer.

Pa ge 63

YOUR ANSWER: 4

Of course. If you repeatedly divide 4000 by 10 until you are left with
a single whole number to the left of the decimal point, you come down to
4.

And now for the case of the vanishing zeros, in which, via an

explanation and a question, we‘ll lead up to step two.
4000 = 4x10x10x10

The figure 4 is a factor once; the figure 10 is a factor 3 times.
(It's possible to factor even further, of course, but this is the end of

the road for using powers of ten.)

Using the first step already described (repeated division by 10 until
you have a single whole number remaining to the left of the decimal point),

split the number 100 into factors.
10 x 10 Page 75

8
1x10x10 page 7

102 Page 9h

Page 61,

YOUR ANSWER: 1.25 x lo'la.
Right. Let's go through the problem.
2 3
I=E/R, E=3x10 v, R=2.4xl01’;

SO, the problem is expressed this way:

=3x102
2.4x10

1 3
Now, a positive exponent below the line 0f'a division problem
is handled as if it were a negative exponent above, SO, the problem

can also be expressed as:

2 -3

_lxlo x10 =3x10 _3 —1
1’ _2.4 °r' I 2.4 °" ’fl x 1°
1

Since 3 divided by 2.4 equals 1.25, the result is 1.25 x10_ a.

Now, express the result in common terms .

0.0125 amperes Page 66
0.125 amperes Page 71.

12 . 5 amperes Page 76

6

ii

Page 65

Using Power's of Ten solve for the unknown value in the circuit

below. I

E: 100 volts “E‘ R =

 

to

 

a. l x 106
b. l x 102
c. 1 x 10‘2
2d. l x 10—6

, After selecting an answer go on to page 57 and continue on.

You moved your decimal point too far. When you multiply by 10'1

you are really just moving the decimal point one place to the left. This

is the same as multiplying by 0. 1 .

Return to Page 61. and try again.

I Page 66
1’
YOUR ANSWER: 1. 25 x 10‘1 amp means 0.0125 amperes.

if
0

’1-

Pa, gt; 67

YOUR ANSWER: 40.

You're holding back!

We asked you to go all the way —— to a single whole number which

is nine or less. So it could hardly be 40.

Think of it this way: 4000 is the same as 1000 multiplied by what? 7

Return to page 55 and select the Obvious answer.

Page 68

YOUR ANSWER: 3 x 102v divided by 2. 4 x 103 ohms is 1. 25 x101 amp.

No, the exponent in your answer is incorrect.

Your division was fine, but then you reached the point where something

had to be done with the exponents.

Apparently you subtracted the 2 from the 3. This would have been

correct if you had been dividing 2. 4 x 103 by 3 x 102.

But, of course, you weren't. As the problem stands, you should have

subtracted the 3 from the 2.
If you do that, what will the resulting exponent be?

When you think you know, return to Page 57 and choose another

answer .

(a

(a

Page 69
YOUR ANSWER: I = 2.5 x 100a, R = 4 x1040, so E = 0. 625 x104v.

It looks as if you tried to divide instead of multiply. You shouldn't

have done it. Your formula is E 2 IR.

Besides, if you had really divided, you should have subtracted your

exponent 4.

Now return to Page 82 multiply the two numbers, and add the

exponents.

 

YOUR ANSWER: Given 2.4 x 103 ohms and 3 x 102 volts, the

current is 7.2 x 101 amperes.
You've got yourself all turned around.

First Of all, you multiplied the two numbers when you should
have divided. Remember, I = E/R, so you must divide the
resistance into the voltage.

And then, how did you get " 1" for an exponent? Did you
guess? 1f yOu multiplied tO get part Of your answer you would have
multiplied the rest. And when you multiply, you should add the
exponents . '

However, in this problem there is no question of multiplying.
The formula tO be used is, once more, I = E/R. You have to divide
voltage by resistance.

And remember that when you divide using powers of ten, you

divide the actual numbers but use subtraction on the exponents.

Now return to Page 21,7 and figure it out again.

Page Tl
YOUR ANSWER: I don't get it.

All right. You know that you add exponents of ten when you multiply:

2x102x4x104 = (2x4)x(102x104) =8x106.

When you divide, you change the sign of the number you are dividing

by. For example:
(Jr—fl: = 5% x104x10_2 =2x104—2 : 2 x102.
2 x 10
In algebra, when you add a positive number and a negative number and
the positive number is larger, you get a positive answer:
(+4) + (—2) : +2
This is what we did with the exponents in‘the problem above. But
when the negative number is larger, your answer will be negative:
(~4) + (+2) = —2

’ Now we'll change the problem:

2 .
“——m4—=% 10zx10_4=%x102—4=—12x10_2.
4 x 10
You added the exponents algebraically: (+2) + (-4) = 2 - 4 = -2.
0 —1 1 —2 1
10 equals 1. , 10 equals 0. 1 or 1—0 . 10 equals 0.01 or W .
0.001 is -—1~ This is 10‘3.

1000 '

Notice that to increase 0. 001 to 1, we would have to multiply by
10 x 10 x 10. We would move the decimal point 3 places, this time to the
right. We say this is a negative direction and it gives us an exponent of -3.

How many places to the right must we move the decimal point to do the
same thing with 0.000001? 0. 00049103!

SO two steps will gi ve you the answer: (1) Count the number of places
you must move the decimal point. (2) Express your answer with a minus

exponent. You will have ten to a minus power.
Return to Page W for the correct answer.

 

 

Page 72
Q:

4
v.

YOUR ANSWER: I: 2.5 x 100a, R = 4 x 10 52, so E = 10 x 10
Your answer is correct, but it can be simplified a little further.

The number 10 is "shorthand" for 10 x 10 x 10 x 10, four tens
multiplied together. When you multiply 104 by 10, as in your answer,
you are just adding another 10 to the string of multipliers shown above.

and choose the answer that expresses

Now return to Page 82

five tens multiplied together.

Page 73

YOUR ANSWER: Given 2. 4x 103 ohms and’fl‘x 102 volts, the current.

would be 7. 2 x '105 amperes.

Well, you multiplied the two numbers correctly; this would be

720, 000 amperes, and that would be a lot of current!
Actually you should have divided the resistance into the voltage.
The formula is I = g . This will give you a much smaller figure for

amperage.

Now return to Page 5'7 and Choose another answer.

Page 7h

YOUR ANSWER: 0. 125 amperes.

Right.
Here is another problem. You are asked to find voltage when you

are given:
I = 0.004a, and R = 450052
Use powers of ten in your calculation. It will be much simpler.

Given the above, the voltage is:

0.18 Page 77
1.8 Page 80

18 Page 82

Q.

Page 75

YOUR ANSWER: 10 x 10
You're not being as specific as we want you to be.

It may seem superfluous to insist on a more precise answer since
it is obvious that 10 x 10 = 100.

The method you're following is to divide the number by 10 as many
times as necessary to leave a single figure to the left of the decimal
point. Don't just ignore that single figure. It should show up in your

answer.

The reason is very plain if you express 200 in powers of ten. Here
you divide by 10 and then by 10 again. That leaves you with a single figure
to the left of the decimal point.

2 00

.M w

And the factors of 200 are 2 x 10 x 10.

Return to Page: 63 for a more complete answer.

Page 76 a 1

YOUR ANSWER: 1. 25 x 10.1 amp means 12. 5 amperes.

You moved the decimal place the wrong way. If you started with
1. 25 x 101, your answer would have been correct. When you have a
negative exponent, you have to move your decimal point to the left

and make your number "smaller."

Return to Page 6A and try again.

Page 77

YOUR ANSWER: I = 0. 004, R = 4500, so E = 0.18 volts.

Your decimal point is astray. Put 0. 004 and 4500 in powers of ten.
Then you will have I = 4.0 x 10 to some power, and R = 4. 5 x 10 to a

certain power. Multiply 4 x 4. 5 and add the exponents algebraically.

Take care of the details and then return to page 71, for an

answer that corresponds to your result.

\

YOUR ANSWER: 105v.

Page 78

Right you are. 10 x 104 is a correct answer, but, since 10 is 101,

we can express 10 x 104 as 105.
deal with.

Now, 105 is how many volts?
10, 000 volts

100, 000 volts

Clearly, 105 is the simpler answer to

Page 81

Page 86

a I

 

Page 79

YOUR ANSWER: 30 x 106.

Only half right.
You worked the multiplication correctly: 3. 75 x 4 x 2 = 30.

But evidently you multiplied the exponents, too. That's about the

only way you could have handled 106 x 100 x 101 to come up with 106.

You had better back track to page 116 and go over this again.

 

Page 80

YOUR ANSWER: I = O. 004 amp, R = 45009, so E = 1. 8 volts.

Your decimal point is astray. First express the two numbers as
powers of 10. Then you will have I = 4. 0 x 10 to a certain power,
and R = 4. 5 x 10 to a certain power. To multiply these two numbers
you will combine the 4. 0 and 4. 5 in ordinary multiplication, but you

must add the exponents algebraically.

Now return to Page 71, and choose another answer.

Pag‘ 81

YOUR ANSWER: 105 volts is 10,000 volts.

No, you seem to have forgotten just what 105 means. It is the
expression for five tens multiplied together -- 10 x 10 x 10 x 10 x 10.
Start multiplying tens in your head, and you will see that the answer
always has the same number of zeros as there were tens multiplied
together. No tens make 100, and there are two zeros in the answer.
In your answer above there are four zeros, so this is the product of

four tens or 104.

Now return to Page 78 and choose the answer that expresses

fiLe tens multiplied together.

Page 82

YOUR ANSWER: 18.0

Correct. The solution is as follows:

 

E=IR,

I = 0.004a, which is 4 x 10'3a,

3

and R = 4500 [1, which is 4.5 x 10 1)..

So, E =4'x 10‘3 x 4.5x103,

3 x 103.

ThatmeansE=l8x100. ’

or,E=4x4.5xlO_

. Remember that 100 (or any number to the zero power) equals

1 . That was proved in this manner:

100/100'=1.
Now, lOO‘is 102, so 102/102 = l.

2 2. 2 —2
But, 10 over 10 IS the same as 10 x10 , so

2

10 x 10'2 = l, or 100

:1.

Now, if 100 = 1 , the answer to our original problem is 18 x l
or 18.

 

Let's try one more problem. Given I = 2.5 x 100A and

R = 4 x‘104n , what does E equal?

0.625 x 104v Page 69
10 x 104v Page 72

5
10 v Page 78

"U
g)
(li-
83

YOUR ANSWER: 300 x 106.

Wrong.

You're not sticking to the rules. Don't try to mix in too much
ordinary arithmetic. There are several manipulations you might have

gone through to get this answer.

Look at the problem:

1

(3.75x 106) x(4x100) x(2 x10 ) = '2

Did you take 100 as being equal to one and 101 as being equal to
ten and figure it this way?
(3.75 x 106) x (4 x 1) x (2 x 10)

If you did something like this, you would have multiplied 3. 75 by 80 to

get 300 -- then you had 106 left over, so you put down 300 x 106.

This kind of monkeying around will get you no place.

It is true that 100 equals 1 and 101 equals 10, but when you’re
working a problem, always handle the powers of ten as exponents, not

as their equivalents.

Return to Page 1116 and follow the rules.

YOUR ANSWER: One thousand volts is l kv.

Correct,

but you didn't read far enough down the list. It' 5

always a good idea to glance down the whole list of answers

given. Sometimes it gives you an idea as to what the right

answer will be, and sometimes you' ll be surprised at what you

find.

As we said, there are many different abbreviations used in

electronics .

Often there is more than one way of saying something.

Now return to Page 91 and look over the answer list again.

Page 8i.

 

Page 85

YOUR ANSWER: 6ma would be 6 x 106 amperes.
No, the abbreviation "ma" means milliamperes or one-one thousandth
of an ampere. The answer you've chosen would be a monster, 6, 000, 000

amperes.

Return to Page 97 and choose an answer that means 6 milliamperes.

YOUR ANSWER: 100, 000 volts
Correct.

Powers of ten are a great help in expressing figures of many
digits . You may have noticed as you went through the last few
examples that powers of ten also provide a simple and convenient

way for calculating with many digit numbers . It is much easier

to keep track of the zeros this way than it is to compute by longhand.

There is an even more convenient way of eliminating zeros when
writing out figures, although it doesn't help much in calculating.
This. is a system of prefixes . Probably the most familiar of these
prefixes is filo, as used in kilowatt. It is in kilowatt—hours that
electrical bills are measured. We will learn more about watts later
on. For the moment, we are interested in the prefix, kilo. Kilo
means one thousand. So, kilowatt means 1000 watts and one

kilowatt—hour is 1000 watt—hours .

You should develop the habit of eliminating zeros in expressing
quantities by using this and several other prefixes, and translate

such numbers into powers of ten for convenient computation.
See if you can translate this figure: 5 kilovolts .

That means:

500 volts or 5 x 103v Page 89
5000 volts or 5 x 103v Page 91
5000 volts or 5 x 104v Page 93

Page 86

\l

Page. 8‘7

YOUR ANSWER: 1 x 10 x 10.

Exactly.
Just as 400 is 4 x 10 x 10, and 40 is 4 x 10, so 100 is 1 x 10 x 10.

And this brings us to the second step which is to show how many times

10 has been used as a factor (as a multiplier, if you like). There is a

"shorthand" way of doing this. For instance:

4000: 4x10x10x10 or4x103
400=4x10x10 or4x102
0r4x101

40 = 4 x 10
The raised 1, 2, 3, etc. , used to indicate the number of times 10 is
used as a factor (multiplier), are called exponents. It is usual in
mathematics to omit the exponent 1 since it is understood.
3 is called ten cubed, or ten to the third power.

10
102 is called ten squared or ten to the second power.

10 (which is really 101) has the numerical value of 10.
Numbers written using this notation, are said to be expressed as

powers of ten.
In-Writing 3,000,000 as a power of ten, which answer is incorrect?

3 x 106 Page 103
3' x 10 x 105 Page 115
300 x 104 Page 115

None of the above is incorrect. Page 117

YOUR ANSWER: 1000 volts is 1 x 100 kv.

Correct, but you didn't look closely enough at the answer
list. Often there is more than one way of saying something. Be
sure that you look at all of the possible answers before you choose
one.

Now return to Page 91 and study the answers again.

Page 88

Page 89

YOUR ANSWER: 5 kilovolts means 500 volts or 5 x 103 volts.

No, the prefix "Kilo" means 1000, so a kilovolt would be
1000 volts .

There's something else wrong with the answer you've
chosen. It isn't consistent in itself. The expression 5 x 103
doesn't mean 500. If you have 103, you have 10 x 10 x 10 or

1000.

Now return to Page 86 and choose an answer that defines

5 kilovolts .

Page 90

YOUR ANSWER: 6 x 10.3 amperes.

Right. The small m means milli, and milli is 1/1000 or, 0.001 .
Six ma is 0.006, which is 5 x10—3

The table below summarizes the information on prefixes and
abbreviations. You should have paper and pencil with you. Make a

copy of the table to use for reference until you get it down pat.

NUMBER PREFIX ABBREVIA TI ON POWER OF 10
1,000,000 Meg M 106
1,000 kilo k ‘ 103
0.001 milli m 10-3
0.000001 micro )1 10-6
0.000000000001 pi 00* p 10-12

as . . . . .
The alternate prefix for pico is micromicro and you may hear micromicro

used some times in place of pico, so don‘t let it throw you. The symbol

for micromicro is ml.
Is the following true or false?

Both mf and llf mean the same thing, one—millionth of a
farad.

. True Page 99

False Page 101

Page 91

YOUR ANSWER: 5000 volts or 5 x 103v

Right. Now, kilo can be further shortened to k.
An example of" using the symbol k is kv ( 1,000 volts ),

kw ( 1, 000 watts ), and etc.

You will find as you go along that abbreviations sometimes
vary slightly from book to book or from engineer to engineer. It
should not concern you —— and won't once you get used to seeing
a small letter where you might have expected a capital, or vice

versa .

Although this course will be consistent in usage where
possible, it will also try to give alternative forms , such as might

be found in data sheets from different manufacturers .

One thousand volts equals:

1 kv . Page 81,
1 x 100 kv Page 88
1 Kv Page 95

All of the above Page 97

Pa go 92 .

YOUR ANSWER: 6ma in powers of 10 would be 6 x 10_6.

Too small. Remember that 6ma means 6 milliamperes, and a

milliampere is one—one thousandth of an ampere.
The power of ten you have chosen, 10_6, would show millionths.

Return to Page 9'7 and choose an answer that corresponds to 6

milliampere.

YOUR ANSWER: 5 kilovolts means 5000 volts or 5 x 104v.

Not quite correct. The first part is O.K. A kilovolt
is equal to 1000 volts; the prefix " Kilo" means thousand.

But when it came to powers Of ten, your thinking began
to wobble. When you say 104, you mean 10 x 10 x 10 x 10,

four tens multiplied together. This is 10,000, so 5 x 104 would
4

be 50,000. Note that when you have a number multiplied by 10 ,

you'move the decimal point four places tO the right. You begin
. with 5.0, and after you move the point four places tO the right,
you have 50,000.

Now return tO Page 86 and choose the correct answer.

‘13
a;

U

Page 9h

YOUR ANSWER: 10‘2

You didn't follow directions.

We illustrated that 4000 is equal to 4 x ‘10 x 10 x 10, and asked you

to express 100 in the same way.

While it is true that 10 squared is equal to 100, that is not the way we

want you to write it -— yet.

Later on, it will be helpful for you to remember that even in the
figure 100, 10 x 10 is being multiplied by another number, a number
arrived at by dividing the large figure by 10 until you come downto a single

number to the left of the decimal point.

Please return to Page 63 and choose a more precise answer.

Page 97

YOUR ANSWER: All of the above.
That is correct. All, including 1 x 100kv mean 1000 volts.
There are three other quantity abbreviations besides kilo

All four come from the Greek language. Mega or Meg means million.
Me gohm can be written meg nor M Q.

There are also abbreviations for one-one thousandth and one—
one millionth. Milli is the prefix for 1/1000; Micro is the prefix
for 1/1 , 000,000.

 

You will notice that all three prefixes begin with m.

M means mega. (It may help to recall that mega stands

for million, and, as the abbreviation for the largest amount,
normally gets a capital letter.)

m means milli.

,4 means micro. (Micro is the smallest amount and the

symbol/u is the Greek equivalent of the letter m. It is
called mu in English and pronounced "mew" .)

Now, how would 6ma be written in powers of ten?

6 x 106 amperes Page 85
6 x 10_3 amperes Page 90
6 x 10_6 amperes Page 92

10_6 amperes Page 1'00

Page 98

YOUR ANSWER: 30 x 107.
Correct.

You multiplied the significant numbers, 3. 75 x 4 x 2, to get 30, then

you added the exponents, 166+0+1), to get 107.

Again we will check your understanding of this material.
What is your answer to this one.
I ( 1.25 x 102 ) ( 2 x 108 ) ( 1.5 x 100 )
a. 2.50 x 1010
b. 3.75 x 1010
c. 2.50 x 106
d. 3.75 x 105.

After selecting an answer go on to page 38 and continue

on with Power‘s of Ten.

YOUR ANSWER: True

-Page 99

Right you are. Both mf and uf mean the same thing, one—

millionth of a farad.

Now, here are three grOups of abbreviations with translations .

In one group, one translation is incorrect.

an incorrect translation.

lva = 0.012 volts
216,ua = 2.16 X 10“?L amperes

371m = 37,000 watts

800kfl==81<1050hms

luw = 0.000001 watt
120kv = 120,000 volts

lma = l x 10_3 amperes
17MQ = 1700 Ohms
llOOkwh = 1,100,000 watt hours

Pick the group that has

Page 102

P9,. g8 lO/+

Page 106

YOUR ANSWER: 6ma as a power of 10 would be 10"6 amperes.

No, keep in mind that 6ma means 6 milliamperes ,- and

a milliampere is one—one thousandth of an ampere.

You seem to be confusing the number of milliamperes with
the exponent. If you use -6 for an exponent, you are saying that
the decimal point must move six places to the left. Your answer
would then be 0.000001 ampere. This is one microampere or one—

one millionth of an ampere, and this isn't what you wanted to say.

Now return to Page 9'7 and choose an answer that means

the same as 6 milliamperes.

Page 100

Page,lOl

YOUR ANSWER: False, mf and pf do not mean the same thing.

No, this is incorrect. Ordinarily you Mdist‘inguish between
m for milli or one-thousandth and ii for micro or one-millionth.
However, with farads , the units of capacitance that you will study
later, we seldom need to use any unit larger than microfarads . So
we use uf to stand for microfarads . And for a unit still smaller,
‘we use the ' ' picofarad (one millionth of a microfarad) which is

abbreviated pf” or uuf.

Now return to Page. 90 and choose the-correct answer.

YOUR ANSWER: One of these is incorrect:
12mv = 0.012 volts
216ua = 2.16 x 10‘4 amperes
37 in»: = 37, 000 watts

No, these are all correct.

To convert millivolts to volts , you multiply by 1000, and
that is shown in the first example.

Take 2.16 and move the decimal point four places to the

left, and you will have 0.000216. This is 216 microamperes.

To convert kilowatts to watts you would multiply by 1000.
So 37 kw is 37, 000 watts .

Return to Page 99 and find a group with a false conversion

Page 102

Page 103

YOUR ANSWER: 3 x 106.

You must have misunderstood the question.

We asked you to indicate which answer was incorrect. This answer

 

is a most concise way of writing the number 3, 000, 000.

Return to Page 87 read the question again, then consider all
the answers more thoughtfully.

Page 10h

YOUR ANSWER: One of these is incorrect:

800 k: a: 8 x 105 ohms

luw = 0.000001 watt
120kv = 120,000 volts

No, these are correct.

The quantity 800k awould be 800, 000 ohms , and this is

shown correctly as a power often above.

A microwatt is one—one millionth of a watt,- and the decimal

fraction above is correct.

There are 1000 volts in a kilovolt, so the last conversion is '

all right , too.

Now return to Page 99 and choose a group that has a false

conversion .

Page 105

YOUR ANSWER: 1v

Yes, one ohm is the amount of resistance that will limit current

to one ampere when the electrical pressure is one volt. This
statement can be turned around to define all three of our units of

meas ure:

One volt is the amount of electrical pressure that will
push one ampere of current through one ohm of resistance.

One ampere is the amount or current that will flow through
one ohm of resistance when the electrical pressure is one

volt.

Each of these expresses one of the three forms of Ohm' 3 Law.
Which of the following statements is incorrect?
a. 80l< ohms is the same as 8 X lOLt ohms.
b. .000056 volts is the same as 5.6 X 10-5.

c. .0008 amps is the same as 80 micro amps.

d. 98,000,000 ohms is the same as '98 Meg ohms.

After selecting an answer go on to page 133.

Page 106

YOUR ANSWER: lma = 1 x 10”3 amperes
17MQ = 1700 ohms
llOOkwh = 1 ,100,000 watt hours

That's right, there is a mistake among those three translations .

Let's see if we can find it.

One milliampere equals 0.001 ampere, and 0.001 equals 10-3,

so that one is correct.

And, of course, llOOkwh is 1 ,100,000 watt hours. One megohm
is equal to 1,000,000 ohms, so 17M§1 is equal to 17,000,000 ohms.

When you are given problems with prefixes attached to the units ,
the values will have to be translated into the basic units before you
begin calculation. Thus , if you have a resistance given in M0 and
a current in ma' 5 , you will have to translate to ohms and amperes

before you can calculate voltage.
Here is a problem. Work it with powers of ten.

Given, R.=12 k (oftenlt is used alone fork {ior lOOO ohms)

3

and I 15 ma, what does B equal?

125 volts Page 108
180 volts Page 109
1250 volts Page 112
1.8 kv Page llh

Page 107

YOUR ANSWER: One ohm limits the current to one ampere when the

electrical pressure is lkv.

No, with one kilovolt, or 1000 volts, you would have 1000 amperes

of current through just one ‘ohm of resistance.

Remember that you must translate all of your electrical quantities

into the basic units before using Ohm's law.

Now return to Page 110 and choose the correct answer.

Page 108

YOUR ANSWER: R = 12k , I = 15ma, so E = 125 volts.

No, you must have divided. The proper form of Ohm's law is E = IR,

so you must multiply.

Now return to Page 106 and choose another answer.

Pa ,cze 109

YOUR ANSWER: 180 volts.

Correct. If you worked it out by powers of ten, the problem went
something like this: .
R = 125k = 1. 2 x104 ohms
I = 15ma = 0.015 amps = 1. 5 x 10"2 amps

Since 13:13, E: 1.5x10'2 x1.2 x104, or

E=1.5x1.2x10‘2xio4'
=1.8xio2
=190v

There's just one more thing we want to add about powers of ten. So
far, we have encouraged you to reduce a large number by moving the
decimal point to the left until there was a single figure to the left of the

decimal point. We followed a smililar method for negative exponents.

In practice, it is often more convenient to move the decimal point
fewer places than this. For example, if you are given a value of 256
milliamperes and you have to use it in a formula for Ohm's law where a
value in amperes is called for, you may find it less trouble to use
256 x 10"3 amperes rather than 2. 56 x 10'1 amperes. Or again, you
may have to figure a resistance value of :17k‘. Properly expressed in
powers of ten, this is 1. 7 x 104 ohms, but it's usually more convenient

to use 17 x 103 ohms.

You should have a copy of the table showing these and similar
abbreviations and their equivalent powers of ten. If you neglected to

copy the table, return to Page 90 and make a note of them now.

When all this is taken care of, please go ahead to Page 110

Page 110

Now, let us go on to resistance.

Do you remember the definition of an ohm? See if you can complete

the sentence below.

The ohm may be defined as the amount of resistance that will

limit current to one ampere when the electrical pressure is:

lmv Page 96

1V Page 105

lkv Page 107

Page 111

YOUR ANSWER: 100: 0.
You're still having trouble. So let's go over it slowly: .

This is a problem in division. When we divide one power of.a
number by another power of the same number, we subtract the exponent

of the second from the eXponent of the first.

And when you tried a case in which the exponents were equal, you

got
103/103 =10<3 ‘ 3)==100

Now look at it this way: When you divided a power of ten by itself,

you got 100. But you also know that any number divided by itself equals

one o

0

Thus you see that 103/103 is equal to 10 by our rule for working

with exponents -— and is equal to one by ordinary arithmetic.
That tells you how to define 100, doesn't it?

Return to page 120. The right answer is the logical one.

Page 11.2 0

YOUR ANSWER: R = 12k , I = 15ma, so E = 1250 volts.

No, it looks as if you divided the two numbers . The proper
form of Ohm's Law for this problem is E = IR, so you must multiply.

Also, before you leap at another answer, take a second look
at your powers of 10 . You must have assigned one or both of them
incorrectly. Foerk -‘ohms you-should have 1 .2 x 104, and for 15
milliamperes you should have 1 .5 x 10-2 (because 15ma is 0.015
amperes). Now multiply these two numbers , and return to Pa ge

106 to choose the correct answer.

Page 113

YOUR ANSWER: 3 x 10 x 105‘.
It just looks wrong.

But it isn't. It would not be incorrect to write 3,000,000 this way --
merely awkward. You are using powers of ten, but obviously, you can

condense the notation to a more concise form since

10x105 = 10x10°10'10'10'10 = 106

Please return to Page 87 and this time give more consideration

to each answer.

Pa ge lll. 1‘

YOURANSWER: R=12k , =15ma, soE=l.8kv.

You multiplied the two numbers together as you should have,
but your decimal point has wandered. Your answer gives E as
1800 volts .

4 and I = 1.5 x 10“2 (because

15 milliamperes would be 0. 015‘ amperes, and this is just two

In powers of ten, R = 1.2 x 10

decimal places removed from 1 .5). Now multiply these two

numbers and you get 1 .8 x 102 .

Return to Page 106 and choose the answer that corresponds "

to this power of ten.

Page 115

YOUR ANSWER: 300 x 104.

Why do you think that’s wrong?

You can write 3, 000, 000 this way, although it's awkward and

doesn't follow the rules we've given you.

The whole point of using powers of ten is to be as concise as possible.

You‘re defeating yourself when you stop short of the short cut.

300x104 .= 3-1o.1o-x10-1o-1o-'1o = 3x106

For our purposes, the shortest notation is 3 x 106. Any other

coin‘bination is unsatisfactory —— but it isn't necessarily incorrect.

Return to Page 87 for another look at the question -— then study

the possible answers thoroughly.

’1935116

YOUR ANSWER: 100 = 1.

Correct.

Dividing any number by itself gives you a result of 1, and this is as

true of a power of ten as of any other number.

And we'll toss in one other little piece of information which may
come in handy some day: Just as the base 10 to the zero power (100) is

equal to 1, so any number to the zero power is equal to 1.

Just to give you a little practice in working with 100, try this

problem:

(3. 75 x 106) x (4 x 10°) x (2 x 101) = ?

What is your answer:

30 x 106 Page 79
300 X 106 » Page 83
7

30 x 10 Page-.98

Page 117

YOUR ANSWER: None of the answers is incorrect.
Fine. Each of the choices offered

3x106

3 x 10 x 10
300 x 104
is a method of writing 3, 000, 000 using powers of ten. There are, of

5

course, many ways of writing 3,000, 000 in this notation. However, the
idea is to express it as conveniently as possible and there is no point

in dragging in unwanted zeros or extra tens.

In reducing 3, 000, 000 to powers of ten, we used two steps
applicable to any number greater than 10:

1. Using repeated division by 10, reduce the number until

there is a single figure to the left of the decimal point.
2. Indicate how many times 10 has been used as a factor.

Thus 3, 000,000 becomes 3 x 106, or, in words, three times ten
to the sixth power. "Ten to the sixth power" means that we multiply
10 by itself six times, which makes one million. In the expression
106, the number 10 is called the bag and the number six is called

the exponent.

10x10x10x10 = $0: exponen

o ower of ten

And for a little practice, what does 8 x 104.equa1?

8, 000 Rage 119
80,000 Page 121
800,000 Page 123

Page 118

YOUR ANSWER: 103/103 = 1000/1000 = 0.

Now you're letting yourself get carried away with this zero idea.
It's just a matter of dividing one number by itself. Any number
divided by itself equals one —- so don't look for tricks. Just go ahead and

work the Obvious, ordinary arithmetic.

Return to Page. .122 and try again.

"A

I
f‘
-C

Page 119

YOUR ANSWER: 8, 000
Wrong.

When you started multiplying 10 by 10, you stopped too soon. The
number 8, 000 can be factored as 8 x 10x10x10, which is 8 x 103, not

8 x 104.

If you simply made an error in multiplication, you know what to do.

But if you were counting digits, and figured that 104 is the same as a

number with four figures, then you've forgotten the two steps to be taken.

. 1. Using repeated division by 10, reduce the number until there
I. is a single figure to the left of the decimal point.

2. Indicate how many times 10 has been used as a factor.

SO, in 8 x 104, you multiply 8 by 104 which is 10 times 10 times 10
times 10.

This means that your answer will contain the figure 8 plus the four
zeros that stand for 104.

Now return to Fags- 117 for the correct answer.

Pa ge 120

YOUR ANSWER: 103/103 = 1000/1000 = 1.

Naturally.

Solving the problem by ordinary arithmetic, We get one for an

answer. But when we solved the problem using powers of ten, we got

this answer:

What conclusions

1o0

10

103/103 = 10(3 ‘ 3) = 10°

can you draw from this:
= 0 Page 111

= 1 Page 116

’4

...—..~_--.

 

Page 121

0
YOUR ANSWER: 80, 000.
Correct.
104 is 1 x 10 x 10 x 10 x 10 -- or 10,000. You multiplied 10, 000 by
eight.
Now, just to fix the picture in your mind before we go on, study
this table:
10, 000 = 1 x 10 x 10 x 10 x 10 = 104 (ten as a factor 4 times)
1, 000 = 1 x 10 x 10 x 10 = 103 (ten as a factor 3 times)
100 = 1 x 10 x 10 = 102 (ten as a factor 2 times)
10 = 1 x 10 = 101 (ten as a factor 1 time)
A. .‘ “Apply these principles to the problem with which we began:

250,000 x 2,000 = 500,000,000

First, reduce each number until you come down to a single number

to the left of the decimal point.

What two numbers do you get?

0. 25 and 2 Page 125
2. 5 and 2 Page 127
25 and 2 Page 131

Page i122

YOUR ANSWER: 103/103 = 10(3 " 3) = 10°.
Right.
But now that you have 100 -- What can you do with it?

Hold on a minute and let's figure this same problem by ordinary

arithmetic, without using powers, and see what you get.
Using ordinary arithmetic, which answer is right?
3 3 j .
10 / 10 =.:1000/1000 = 0 Page -1138

103/103 = 1000/1000 = 1 Page 120

Page 123

YOUR ANSWER: 800, 000.
A stab in the dark?

You can't go on this way for long. Sooner or later you'll need more
light. It might as well be now.

If this was just an error in multiplication, return to Page 117 and

select the right answer.

However, if the whole idea is still giving you trouble, you may have
been skimming. Go back to Page 55 and make a new beginning.

 

YOUR ANSWER: 103/103 = 1000/1000 = 1
It is true that 1000 divided by 1000 equals one.

But we didn't ask for an answer in ordinary numbers. We asked you to

give the result in exponent form -- as some power of ten.

In division of values expressed as numbers with exponents, you just

saw that we subtracted the second exponent from the first.

So, return to Burma-r and do likewise.

f.

{'5

YOUR ANSWER: 0. 25 and 2.
Wrong.

Did you forget that we want you. to come up with a single Whole
number to the leftyofthe decimal point? Your answer, 0. 25, is less
than one. 0 I A ‘7 ' I '

When you divided your large number (250, 000) by 10, you went too far.

Remember toth while you're still ahead -- one whole number ahead. In

this case, your decimal point will be followed by a fraction, so that

you have a whole number and a tenth.

Now return to 2‘»:.ge 1121 and select the right answer.

YOUR ANSWER: 10.

Of course.

102 is 100. 101 is ten. 100/10 equals 10.

Simple enough. But do you see what happened to the powers of ten?

You subtracted the second exponent from the first.
102 + 101 = 10(2 ' 1) = 101
Now, following the same method, let's see what happens when you
divide a power of a number by itself.
' We'll divide ten to the third power by ten to the third power.
103/103 = 2
How would you write the result as a power of ten?
103/103 = 10(3‘3) = 100 Page 122

103/103 = 1000/1000 = 1 Page 12:.

A;

Page 12?

YOUR ANSWER: 2. 5 and 2.

Correct.

If you had selected 25 as your answer, you wouldn't have been reducing
250,000 the _way we asked you to -- you'd just be lopping off zeros. This ‘
could get you into trouble. Besides, we wanted only one whole number to

the left of the decimal point.

To select 0. 25 is to go too far. You would have a value that is less

than a whole number.

Our calculations won't require accuracy beyond one or two parts in a
hundred, so you don't have to worry about more than three non-zero digits
in a large number. To put it in the proper language: You don't need to
go further than three significant figures. This will give you a whole number

 

plus at most a decimal fraction rounded out to the hundredth.

I So now you have the number 2 . 5 to multiply by a power of ten and the

number 2 to multiply by a power of ten. What powers of ten do you use?

Count to the left until you have moved the decimal point to the correct

pOSltlon: 23.9999, x 2.9.9,;
5‘ places 3 places

The number of places moved to the left tells how many times
you divided by 10. So, five places means 105 (ten to the fifth power); and
three places means 103 (ten to the third power).

You would write it like this: 2. 5 x 105 x 2 x 103

How would you state this problem: 375, 000, 000 x 400, 000 ?
3. 75 x 106 x 4 x 105 Page 129

37. 5 x 10'7 x 4 x 105 Page 130
MIME. CORRECT" '1. age 132

Page 128

YOUR ANSWER: 100.
How did you do that?

You made an error which indicates too much speed or too little

thought.

Perhaps you mistook 101 for’one and divided 100 (102) by one to get
the answer. Of course this is wrong. 101 is ten. What does 100 divided

by ten equal?

Return to Page 1.1.3 for the Obvious answer.

(J

Page 129

YOUR ANSWER: 3. 75 X 106 x 4 x 105.

TOO hasty.

Study this problem again:

375,000,000 x 400, 000

Reduce the first figure as far as you can to arrive at a single whole

number to the left of the decimal point. You get 3. 75.

But what happened when you started counting the number of places
you moved the decimal point? Count again. Were you counting only

zeros?

Return to Page 127 and consider all the answers carefully. You

must count (emery ,pllace you move the decimal point to the left —- not just

the zeros.

9553 130

YOUR ANSWER: 37. 5 x 107 x 4 x 105.

No. Take another look. This is simply a matter of counting.
375,000,000 x 400,000

In the first amount, you didn't move the decimal point far enough
to the left. In order to arrive at a single whole number to the left of the
decimal point, you must divide by 10 once more. And that means you

will have moved the decimal point how many places to the left?

Return to Page 127 and don't settle for a wrong answer.

’1-

O

Page 131

YOUR ANSWER: 25 and 2.
No.

For our calculations, we want to reduce each number down until a

single whole number remains to the left of the decimal point.

The figure 25 doesn't fall into this category, since it is comprised of

two whole numbers.

We want you to divide 250, 000 by 10 until you come to a single whole
number and, in this case, a decimal fraction. Don't let the fraction stop
you. Remember that you're still after one whole number to the left Of the

decimal point.

Return to Page 1521 and find it.

Page 132

YOUR ANSWER: Neither answer was correct.
Right.

The problem, 375,000, 000 x 400,000, multiplied as powers of ten,

was, in both cases, stated incorrectly.

Each time the error was in the first half of the problem —- involving
_the figure 375, 000, 000. The last half was stated accurately:

400,000 is written 4 x 105

The figure 375, 000, 000, written correctly in powers Of ten, would
be 3. 75 x 108 -— not 107 or 106. (The decimal point has to be moved eight

places to the left.)
So the first half of the problem is expressed: 3. 75 x 108. You are
going to multiply this by 4 x 105 (400, 000).

The answer should have been: 3. 75 x 108 x 4 x 105.

Once more, let's pick up our problem:
250,000 x 2,000 = 500,000,000

What is 500, 000, 000 expressed in powers of ten?

5 X 108 - Page 13h

5 x 1008 Page 135

El

Page 13 3

Solve:

( 8 x 10'3 ) ( 2 x 109 ) ( 2.5 x 106 ) ( 6 x 10'1 )

 

( 5 x 10h ) ( 1.6 x 102 ) ( 3 x 10'7 )

a. 1 x 10“
5

c. l X 106

b. l X 10

d. 1 x 1076

After selecting an answer go on to page 136.

Page 131.
YOUR ANSWER: 500, 000,000 = 5 x 108.

Correct.

Let's pick up the rest of the problem now. Expressing it in powers of
ten:

250,000 x 2,000 is 2.5 x 105 x 2 x 103

And we've just figured out that

500,000,000 is 5 x 108

We can put these two parts together and say that
250,000 x 2,000 = 500, 000,000
can be expressed in powers of ten as
2.5x105x2x103=5x108 . 9|

We're on the verge of something important now and it'll be plainer
if we turn that equation around a little.

3 8

2.5x2x105x10 =5x10

A careful look should show you the procedure for multiplying in
powers of ten: Multiply the significant figures, add the exponents.

In this case, 2. 5 x 2 gives you 5; 105 x 103 = 105 + 3 = 108

Now try this: Given 1 = 3 x 103 amps, and R = 4. 25 x 106 ohms, we
find that E = 12. 75 x 109 volts.

Using Ohm's law and expressing this in ordinary numbers, how would
the problem look?
3000a x 4, 250, 00082 = 12, 750, 000, 000v Page 137
3000a x 425, 00052 = 1, 275, 000, 000v Page 133
3000a x 42, 500, 0009 = 127, 500,000, 000v page 11.0 f".

Page if] 5

YOUR ANSWER; 500,000,000 = 5 x 1008.

What!
You've got enough goose eggs there to start a hatchery!

You reduced 500,000,000 to a single number to the left of the decimal
point. But to arrive at
' 5.00000000

how many places did you move the decimal point? Or, to put it another

way, how many times did you divide by 10?

The answer is 8, as you're probably aware. But remember, you're
dividing by 10. That 8 is going to be the exponent for the base 10. You're
working in powers of 10, not of 100.

Incidentally, have you paused to figure what number your answer
represents?
50,000,000,000,000,000

Return to Page 132 for a logical solution.

Page 136

How much current will flow through a 680 K ohm resistor that is connected

across a 3A volt battery 7

a. 5 milli amps.
. 50 micro amps.
c. 2 milli amps.

d. 200 micro amps.

After selectina an answer go on to page l39.

 

Page 137

YOUR ANSWER: 3000a x 4, 250, 0009 = 12, 750, 000, 000 volts.

Correct.

If I equals 3 x 103a, then I equals 3000a.

3x10x10x10=3x1000=3000

And if R equals 4. 25 x 106 ohms -- then R equals 4, 250, 000 ohms.

When you're in the millions, multiplying 10 by 10 by 10, etc. gets
to be a nuisance. It's easier to move the decimal point to the right as
many places as are indicated by the exponent of ten. In this case, the

value of R is multiplied by ten to the sixth power, so you move the decimal

point six places to the right. Thus, 4. 25 x 106 becomes 4, 250,000.

Now look at it from the other direction, changing 4, 250, 000 back
to 4. 25 x 106. The number of places you move the decimal point (the
number of times you divide by 10, in other words) to reduce 4, 250, 000
to the point where there is a single digit to the left of the decimal point

will give you the figure for the exponent. You arrive at the power of ten.

Expressed in powers of ten, your problem looks like this:

3 x 103 amps x 4. 25 x 106 ohms = 12.75 x 109 volts.

How would you work that problem using powers of ten?

Multiply the first two factors, add the last two. Page 1L1

Multiply 3 x 4. 25, add the exponents of ten. Page 11.3

Multiply 3 x 4. 25, multiply the exponents of ten. Page 11.1.

Pa gr; 138

YOUR ANSWER: 3000a x 42 5,0009 = 1, 275, 000,000 volts.
Wrong.
It is true that if I = 3 x 103a, then I = 3000a.

3x10x10x10,= 3x1000=3000

But something went wrong when you tried to translate 4. 25 x 106 into

ordinary numbers.

To multiply 4. 25 by 106, you must move the decimal point fl places
to the right. '

Return to Pa ge 131. and start counting.

l

at

at

Page 139

If the current in the diagram below decreased in value, which of the

following could have happened to cause it to decrease.

_L
T a

a. The voltage from the battery increased.

 

b. The resistor increased in value.

c. None of the above could cause it to increase .

After selecting an answer go on to page 1h2.

Page 1L0

YOUR ANSWER: 3000a x 42, 500, 00052 = 127, 500, 000,000 volts.
Wrong.

It is true that if I = 3 x 103, then I = 3000a.
3x10x10x10=3x1000=3000

But when you tried to translate 4. 25 x 106 into ordinary numbers, you

lost count someplace.

To multiply 4. 25 by 106, you must move the decimal point six places
to the right.

Re turn to Page 131. and this time count slowly.

Page 11.1

YOUR ANSWER: Multiply the first two factors, add the last two.
Impossible.

You may be thinking the right answer, but it's coming through wrong.

In this problem, 3 x 103 x 4. 25 x 106, the first two factors are 3 and

103. So far so good. Multiplying them isn't the right way to solve the

problem, but it is still a recognizable operation.

The second two factors are 4. 25 and 106. You certainly don't add

them. That's clearly a multiplication problem, too.

It is true that you multiply and add in order to solve the problem.
But your terminology is incorrect. What do you multiply, and what do

you add?

We'll answer part of that for you: To multiply powers of ten, you

add the exponents.

Return to Page 137 for the correctanswer.

Page 1/12

This is the end of the lesson. Please write the length of time it

took you to complete this lesson on your answer sheet.

 

Page 1h3

YOUR ANSWER: Multiply 3 x 4. 25 , add the exponents of ten.
Correct.

To multiply with powers of ten:
1. Reduce each number in your problem until you have a single
whole number remaining to the left of the decimal point.
(a) 2,000 x 25,000
(b) 24W) x 2.5931553
2. Determine how many places you moved the decimal point:
2.951591 x 2.5151511
3 places 4 places

This gives you the power of ten for each number: 103 x 104.

3. Multiply the significant numbers in the usual way:
(2.x 103) x (2.5 x 104)
2. x 2. 5 = 5
Add the powers of ten of each number multiplied:
103 x 104 =103 + 4 = 107

4. Your answer is expressed as two factors: 5 x 107.

The first factor is the product of your multiplication.
The second factor is 10 with an exponent which is the sum of

your addition. But note that the 10 remains 10.

You should feel easy about working with powers of ten now. So let's

pursue the subject a little further.

You know what 102 equals and what 101 equals. So what do you think
102 divided by 101 would equal?

Page 126
10

100 Page 128

Page 11.1. ._

YOUR ANSWER: Multiply 3 x 4. 25, multiply the exponents of ten.
No!

You do not multiply and multiply. You multiply and fl!

Using another example, here's what happens if we follow your method:

(2 x 102) x (2 x 104) = 4x108

Now let's check this by doing it the long way :

2

10 =10x10=100 2x100=200

104 =10 x 10 x 10 x 10 = 10, 000 i 2 x 10,000 = 20,000
20000 .
200 .1

00000
00000
40000

4000000

Your answer is 4, 000, 000. Now if you reduce this to a single whole
number to the left of the decimal point, you get 4. How many places must
you move the decimal point? Six, isn't it? So your answer may be

expressed in powers of ten:
4x106 ——_n_g4x108

Return to Page 137 for the accurate statement.

2‘0

 

.NA‘VAL AIR TECHNICAL TRAINING'COMMAND

I

t'\

Acknowledgements

This program was adapted by the Naval Air Technical Training Center,
Memphis, Tennessee, from a larger one developed under contract No. AF 33
(616)-6983 CPFF between the U.S. Air Force and Western Design and Electronics.

In a letter dated 27 July, 1961 from the Aeronautical Systems Division,

‘ ‘ Air Force Systems Command, United States Air Force, use of these materials'

by other governmental agencies was confirmed. Appreciation is expressed
to the Aeronautical Systems Division and to Western Design and Electronics

for assistance in making the original program available.

 

Page 1
INTRODUCTION TO CIRCUITS

An electrical circuit is made by connecting with conductors a source
of electrical potential and one or more electrical devices, light bulbs or
motors, for instance. The word circuit comes from the same root as the
word circle. Its use in electricity emphasizes the need for a continuous
path for electron flow.

Until now, we have drawn pictures of the electrical equipment in
our circuits. This is inconvenient. There are standard symbols that are
used to represent this equipment in circuit diagrams.

Here is a diagram of a simple circuit.

+ LAMP

BATTERY OF THREE CELLS;

 

F/XEp RES/STdR
Two electrical devices are connected to the battery. The connecting

lines represent conductors. Three standard electrical symbols are
employed. Here they are labeled, but most circuits are not. You will
need to remember what the symbols stand for. Don't worry, it's not
difficult. A complete chart is available in your Basic hlcctronics Text Book.
‘ . Current actually starts to flow in all parts of a complete circuit at

the same instant. Let us consider those electrons starting at the battery.
Which is the correct statement about the circuit diagram?

The direction of electron flow is from the
battery through the fixed resistor and page A
through the lamp.

The direction of electron flow is from

the battery through the lamp and through Page 6
the fixed resistor.

The direction of electron flow is not

P
indicated. age 9

YOUR ANSWER: The voltage across lamp A is higher.

It's difficult to see how you reached this conclusion if you looked

carefully at the two diagrams.

' Chemical sources of voltage are usually indicated as if they were
three-cell batteries, even though the voltage is greater than could be
generated by three cells. This is done for convenience, since it would
be impractical to draw the number of cells needed to generate 90v or
even 2 2 -l/2v. After all, the most common cells generate only 1.3V

to 2v.

It is usual to indicate on an electrical diagram the actual voltage

. being generated by the seat of EMF so that the three—cell diagram will
not mislead. Actual voltage may be written either right next to the
battery symbol or in a footnote or table. Normally, in this course you
will find the voltage right next to the symbol, as it is on the examples
on Tags ‘7.) Return not to that image and select the circuit with

the highest voltage.

P3423

-;_YOUR ANSWER: Yes, the two circuits are electrically identical.

1 2)

Correct. In the diagram on the left, the negative pole of the battery
is at the top. The
direction of current is
from the negative pole
to the motor, to the

 

lamp, and then to the
fixed resistor and back to the battery. In the diagram on the right, the
negative pole is at the bottom. The current flows through three resistances
of the same value as those in the diagram on the left and in the same

sequence. These diagrams are called electrically equivalent because

 

analysis by Ohm's law would show the same relationship in both between
voltage, current and resistance.

Remember when you learned about batteries, you were shown two
‘ ways in which batteries could be hooked up? One of those ways was
called series, one parallel. Here is an illustration of one of the two
ways. Which is it? '

- -. _

Series Page 8

Parallel ' Page 15

t Page L.

YOUR ANSWER: The direction of electron flow is from the battery through
the fixed resistor and through the lamp.
Correct. We can think of current as an instantaneous flow of
electrons moving from the negative side of the battery, around the circuit
and back to the positive side of the battery.

The symbol for the battery is composed of pairs of lines like this:
..i'_{ l:-
Each such pair represents a single cell. ‘The short line is the
negative pole, the long line the positive pole. Sometimes the negative

line is drawn thicker as above.

Electrical diagrams do not normally label the polarity of the cell,
so you will 1need to remember which side is which. It may be helpful to

think of the symbol for a cell as an arrowhead pointing in the direction

\
\
\
/
/

l
I
|
|/

of electron flow. That makes the point end negative.

Now look at these circuits:

 

In each case, is the direction of electron flow through the
fixed resistor and then through the lamp?

Yes first:

N0 3:. 3% [L0

I don't know 5:1;

Page j,
YOUR ANSWER: The voltage across Lamp B is higher.

9 Yes, the power supply in circuit B was labeled "60 volts"
and that. in A "50 volts". The three—cell battery symbol was
used in each diagram simply for convenience. If this device
weren't used, a 60-volt source composed of 2—volt cells would
require drawing in 30 cells.

So far we have used the special symbol ——@_ for a
lamp. Sometimes the only electrical importance of the lamp
' e ' c it offers and we’ll sometimes find that lam s
' m n e own si 1' b 7 resistor
symbols (W). Whole circuits will be drawn looking
as if nothing was being done but the pumping of electrons through
resistances.

For example, the standard symbol for a motor is
d

We can specifically include a motor, a lamp, and a fixe
resistor in a circuit, or we can show the electrical equivalent
with three resistors.

(D

v

 

 

l '9

 

5
Te

If we are primarily interested in the resistance offered by
each device, the second diagram is just as good as the first
and, in addition, it's a little easier to draw. What kind of a
device each resistance represents may or may not be indicated
but the resistance of each will be given in ohms (11).

Look at these two Icircuits. Do you think they are electrically
identical?

 

 

Yes Page 3

No Page 13

Page 6

YOUR ANSWER: The direction of electron flow is from the battery
through the lamp and through the fixed resistor.

Sorry, but you seem to have forgotten one of those really basic

points, or else you didn't look closely at the diagram.

Electrons flow away from the negative pole of a battery or cell.
They do so because like charges repel. Both an electron and a negative

pole are negatively charged.

If you remembered that but just failed to notice which side of the

battery was marked negative, look more closely when you go back.

d)

Now, return Page 1 . and select the correct answer.

Page 7

YOUR ANSWER: Yes, in each case the direction of electron flow is
.3 through the fixed resistor and then through the lamp.

Correct. The two circuits are identical. It's just that one has been

turned over.

   

There is no standard position for a battery. In most simple circuits,
we try to show the power supply on the left side of the diagram, but you will

have to check each time whether negative is on the top or the bottom.

In both circuits you saw three pairs of parallel lines in the battery
symbol. Occasionally this form may be used to show a battery that
actually has three cells in series. In those cases, the total battery voltage
is found by adding the voltage of each cell. Three dry cells would be 4. 5
volts, three lead cells 6 volts, and three Edison cells about 3. 9 volts.

In most circuits the three-cell symbol will be used to represent
larger voltages, however. The actual voltage will be indicated beside
the symbol.

Which of these two circuits has the higher voltage?

506:: A 9...; 6 it

The voltage across Lamp A is higher. Page 2

 

 

The voltage across Lamp B is higher. Page 5'
I can't tell without knowing what type

of cell is shown. Page 12
The voltage is the same in each circuit. Page 16

 

 

Page 8

YOUR ANSWER: Series .

That is correct. .

And all of the circuits used so far in this section have
been series circuits. The series connection of electrical
cells looked like this:

An equal number of electrons flows in each cell. And, as
we saw earlier, in a series circuit the total current flows 0

through each component. This is the distinguishing feature of
the series circuit. There is no alternative path for current. It'

can go nowhere but 'round and 'round the circuit.

Now, here is a diagram of another simple series circuit.
Notice that it has both a motor and a resistor. If we take
out the resistor and put nothing back in to bridge the gap,
will the motor run better?

 

Yes Page 11

No Page 18 -

it;

YOUR ANSWER: The direction of electron flow is not indicated.

Oh, but it is. True, there is no arrow to show which way
electrons are flowing, but you should know because the poles

of the battery are marked.

Electrons can only flow away from a like charge and toward
an unlike charge. Electrons have a negative charge. They will
flow away from the negative pole. You should certainly know
that by now.

Now, return to Page 1 and select the correct answer.

Page 9;

Page 10

YOUR ANSWER: N0 .

You couldn't have looked very closely at the two diagrams .
To tell the direction of electron flow in a battery, imagine that
the combination of lines used to indicate a single cell is really
an arrowhead pointing in the direction of electron flow, like

this:
\ \ELEcTRoNs FLOW
}\> \TH a 5 WAY

III
I
l/ / Tms ISTHE NEGATIVE POL-E

If you have that clearly in mind, you can look at the diagrams

again and figure out the direction of electron flow.

Now, return to Page 1+ and select the correct answer.

it)

YOUR ANSWER: Yes , it will run better.

It might seem that if there is less resistance in the line,
the motor will run better. That is not necessarily true, since
you might have a motor that would burn out if too much current

reached it.

But that is not really the point. The sneaky, key phrase in
the question was, "and put nothing back in to bridge the gap" .
If nothing was put back in to bridge the gap, the motor will not
run at all. What we have then is called an ope—n or open circuit.
Remember, you recently used an ohmmeter to test for opens .

Well, an open is what you've got here.

Now, return to Page 8 and select the correct answer.

Page 11

YOUR ANSWER: I can't tell without knowing what type of cell

is shown.
You missed the point.

Chemical sources of voltage are indicated as if they were
three—cell batteries , even though the voltage is greater than
could be generated by three cells . This is done for convenience
since it would be impractical to draw the number of cells needed
to generate 90v or even 22—1/2v. After all, the most common

cells generate between 1.3V and 2v.

It is usual to indicate on an electrical diagram the actual
voltage being generated by the seat of EMF so that the three—
cell diagram will not be misleading. Actual voltage may be
written right next to the battery symbol or in a footnote or table.
Normally, in this course you will find the voltage right next
to the symbol. Only if the voltage were not written in would
you need to know the type of cells used in order to compute the

voltage.

In these diagrams the voltage generated by the batteries in
both diagrams is clearly indicated right next to the battery

symbols . Return to Page 7 and select the right answer.

Page 12

 

YOUR ANSWER: No, the circuits are not electrically identical.
You must have lost something, somewhere along the line.

The last image was devoted to explaining that for some
purposes it is sufficient to show all the components of a circuit
as resistances. Any appliance or electrical device —— a toaster,
a light, a motor, a radio set, or what you will 4. offers a
resistance to electron flow, no matter what kirrl of work it does.
For purposes of discovering current flow or voltage requirements,
the only important thing about the device is how much resistance
it offers. So, sometimes we simply diagram a circuit as if it
were a series of resistances. That is simple enough and should

not have misled you.

The other thing that may have tripped you is that the
components, though they are shown to have the same resistance,
appear to be in different order. Perhaps you failed to check the
polarity of the battery. If you check, you will find that in one

illustration the negative pole is up, in the other down.

Now, go back and follow the electron flow of the two circuits
starting from the negative pole of the battery. See if the flow
isn't through the same resistances in the same order in both

circuits. The page was 5..

Page 13

 

Page 11.
YOUR ANSWER: I don't know. .

That's honest, at least.

Current starts flowing at once in all parts of a complete Circuit.
We'll consider a portion of the current starting at the battery.

First, let's get it clear how to tell the direction of current from

looking at the battery or cell symbol. The symbol for a single cell

[\ D/RECT/d/V oF
\ EZéYrRo/V Flow

as at the right, pointing in the i ‘ \ \ ““9
I
direction of current flow. If you / /§ flEGAT/VE Poéé

can be thought of as an arrowhead,

I
had trouble with the symbol for a l/

battery (which is just a collection of cells), begin at either end of the
battery symbol and take the first two lines as the cell that forms the
arrowhead, like this:

Pox/7N5 POLE Now, let's look at one

0F W’MLE of the schematic a !
4 ,1 EATI‘E/er

/ dia rams and fi ure out

/ l / l/___,) g g l
...—(...: lfi<Lfi L— the direction in which
\
electrons will 0
\4 ‘J g

thro hits 1: .
4..., EzEcha/v F;0W U9 componen s

The arrowhead points upward
in this diagram, so that's where

_.___.

electrons leave the battery. '-

_—..~
—

Electrons move toward the lamp. ' l
The direction of electron move— ' J

ment from the lamp is along the

 

conductor toward the next component, the fixed resistor. From the 1
fixed resistor, the direction of movement is toward the positive side
of the battery.

Now, return to Page it analyze both illustrations in this

manner, and select the correct answer. q

Page 15

YOUR ANSWER: Parallel.
Afraid not.

If you stop to consider what parallel means , you will

see that this illustration could not be a parallel circuit.

Parallel means running along beside each other without

ever touching, like railroad tracks .

Here is an illustration of a parallel arrangement of cells:

30%

 

 

 

Now, return to Page 3' and select the correct answer.

YOUR ANSWER: The voltage is the same in each circuit.

You missed the point. The fact that each battery is shown
to have three cells does not indicate that they have the same

output.

Chemical sources of voltage are indicated as if they
were three cell batteries, even though the voltage is greater
than could be generated by three cells. This is done for con—
venience since it would be impractical to draw the number of
cells needed to generate, for instance, 90v or even 22-1/2v.

After all, the most common cells generate between 1. 3v and 2v.

It is usual to indicate on an electrical diagram the actual
voltage being generated by the seat of EMF so that the three
cell diagram will not be misleading. Actual voltage may be
written right next to the battery symbol or in a footnote or
table. Normally, in this course you will find the voltage right
next to the symbol, as it is in the two examples on Page 7
Return now to that image and select the circuit with the highest

voltage .

Page 16

 

4.

YOUR ANSWER: Yes . The battery alone is grounded.

Here is the illustration once more.

 

Why would you—say that the battery is connected to ground
but that the resistor is not connected?

Return to Page 20 and select the correct answer.

Page 17

Page 18

YOUR ANSWER: No, it won't run better.

Correct. It won't run at all. If, as the question has it,
“we take out the resistor and put nothing back in to bridge
the gap" , we have made an open circuit, so no current can
flow.

All the series circuits studied so far have been closed
circuits , because they had wires running from one component
to the next. Sometimes a circuit is completed (closed) by

grounding it .

And here's a statement you can preserve: Ground doesn't
necessarily mean ground of the dirt type. The term ground is
used to mean a path other than a wire conductor for returning
electrons or current to the power source. For instance, in the
circuit for an automobile tail light, the car's steel frame might
be used as the return. One pole of the battery and one side of
the light bulb are attached to the frame. When this is done,
the frame is called the ground and the battery and bulb are said
to be grounded. Here is the symbol for ground and the circuit
of an auto tail light.

...—L...

 

 

HIIIIF

The current is through the wire to the lamp and then back to
the battery through the frame. It is necessary to run only one
wire the length of the car. This is a closed circuit. The chief
advantage of a circuit of this sort is that it saves wire and makes
for simplicity.

 

Are the circuits below electrically equivalent?

 

Q

Page 19

YOUR ANSWER: Part of the current is flowing through the ground.
That, frankly, is a rather improbable answer.
Remember these two principles:

1 . In a series circuit, the current is the same in all

parts of the circuit.

2 . It it necessary to have a conducting loop for current

to flow.

The ground forms one section of the conducting loop. There
is no point in this circuit where current can split up and go in
different directions. That means it is a series circuit. Since
it is a series circuit, the amount of current going through the
ground has to be exactly equal to the amount of current flowing

in any other part of the circuit.

If different amounts of current were flowing in different parts
of a series circuit, electrons would be piling up somewhere along

the circuit .

Now, return to Page 21 and select the correct answer.

 

Page 20

YOUR ANSWER: Yes , they are electrically equivalent.

Right. Both are complete or closed circuits , and both

have equal voltage and resistance.

When wires cross each other in a circuit, it is necessary
to show whether they are actually in electrical contact. When

there is such a connection, it is indicated by a dot, like this:

CoMchT‘ED WIRES CmsSEb-OVER WIRES
NO conuzc'fiou

Wires that cross without electrical contact do not have a
dot. The symbol for crossed-over wires with no connection,
means that the two wires are insulated from each other at that
point. In an actual circuit they may not even cross , though
they do so in the diagram because the drawing of a complicated

wire arrangement makes it unavoidable.

In the circuit below, is the battery alone grounded?

 

Yes Page 17 0

NC Page 21

. 1
'31)

0:;
II)

YOUR ANSWER: No. The battery is not the only component
' that is grounded.

Correct. The positive side of the battery was connected to
the resistor at the bottom of the diagram and both were connected
to ground.

In drawing circuits , simplicity is important. Look at the
three drawings below. Diagram A represents the basic circuit.
Diagram B shows an equivalent circuit which uses ground for
the return. Diagram C, which looks rather like the circuit on
the last image, is electrically identical to Diagram B.

In C we have dispensed with one drawing of the symbol for
earth. This may not seem much, but in more complicated circuits
any simplification helps.

 

Remember that in electrical circuits, "ground" does not often
mean a connection to earth itself. It usually means a connection
to a conductor to which one side of the power source is also
connected. This common conductor could be the chassis of a
radio set, a single wire or bar within an electrical apparatus, or
something similar, just as the frame is used as ground in the car's
electrical system.

Now look at this series circuit, and choose the correct
statement below:

 

Part of the current is flowing through ground. 1.:g:

All of the current is flowing through ground. "gs; ,

None of the current is flowing through ground.

YOUR ANSWER: More than 2 amperes are flowing through R1 .

You weren't reading very carefully. It may seem at first

that because the resistance at R3 is greater than at R1, there

should be more current flowing through Rl .
R. RL

  

8.1L

_—
_—
.—

R

3

  

If we had two circuits like
those below, one with a
resistance of 8 ohms and
one with a resistance of 10

ohms, it is true that, according

to Ohm's Law, the current would be greater in the circuit with

the smaller resistance .

 

i

”if

 

5v; I0

However, in a series circuit the amount of current flowing

through any component of the circuit is determined by the

total resistance of the circuit.

More current cannot possibly

flow through one part than another. Think back to the analogy

of the water pump and the continuous pipe. The rate of flow

through any portion of that pipe cannot be greater than the rate

of flow through any other portion.

Now , return to

Dr;

in

g

.3

V -—:--L

and select the correct answer.

Page 22

Page 23

YOUR ANSWER: No, they are not electrically equivalent.

Sorry, but they are. Look at the diagrams again. Both
have a power source of 6v. Both have a resistance of 10
ohms, too. So, the current should be equal in both circuits,

too, if the circuits are closed.

 

 

 

 

L

3% CV 10!}. :E"__
I 1

In the left-hand circ’uit, it is clear that we have

4V 10.0.

 

continuity, so we have current. To be precise, we have a

current of .6 amps, according to Ohm's Law.

The last image tried to point out that a common ground,
such as an auto frame, for instance, serves the same function
as a conductor, and can be used to complete circuits in such
a way as to save wiring. That is the case in the circuit on the
right. It, too, is a complete circuit, so there is current. The

current in the right—hand circuit is also .6 amps.

Now, return to Page 18 and select the right answer.

Page 2h

YOUR ANSWER: All of the current is flowing through ground.

Correct. In order to return to the battery after flowmg
through the three resistances , all of the current must go

through ground. This is

  

one characteristic of the
E series circuit: the total
:1: current goes through each
T - device in the circuit. You

may think of current as a .‘ ,
quantity of electrons. In a series circuit the quantity of
electrons flowing per second through any one part of the circuit
is the amount that must be flowing through all other parts of

the same circuit.

Electric current, you will recall, is measured in amperes,
so in a series circuit the amperage will be the same through

each component in the circuit.

Look at the circuit below and choose the correct statement.

Each resistance in the circuit is given in ohms ($2 ).
RI R2 R3

 

If 2 amperes are flowing through R then

3! o .5
More than 2 amperes are flowing through R1 . Page 22
Exactly 2 ampr-res are flowing through R1. Page 27

Less than, 2 amperes are flowing through R1. Page 29

Page 25

YOUR ANSWER: Exactly 2 amperes are flowing through the
battery.

Correct. In a series circuit you find the same amperage
at every point in the circuit. If 2 amperes were flowing through
one resistance, 2 amperes were also flowing through ground,
2 amperes through the battery, and 2 amperes through each wire
in the circuit.

 

Of course, we can't make 2 amperes flow through a circuit
just by wishing for it. We must have a battery strong enough to
move that quantity of current through the given resistance. In
this circuit, for instance, with resistances of 8 ohms, 5 ohms,
and 10 ohms, we couldn't make 2 amperes flow with a flashlight
battery or even with a 12-volt car battery. There is too much
resistance. We can apply Ohm's Law to determine how many
volts would be needed to produce 2 amperes of current in this
circuit.

First, you will need to know the total resistance in the circuit.
In a series circuit, the total resistance is found simply by adding
together all the resistances present. The grOund conductor and
the wires, themselves, have a slight resistance, but generally
you may consider them as having "no resistance. " (The battery
always has an internal resistance to its own current flow which
affects its efficiency and life span, but doesn't affect the appli—
cation of Ohm's Law to circuit analysis. We will consider
internal resistance of power sources later on.)

Look at these two circuits:

 

Which statement is correct?

The resistance in circuit A is less
than the resistance in circuit B. Page 28

The resistance in circuit B is less
than the resistance in circuit A. Page 33

Page 26

YOUR ANSWER: None of the current is flowing through

ground.

Well, let's look at the illustration again:

 

.———.
_—
_—
.—
.-

There is current, improbable as it m"y seem, and that current 18

'flewing through the ground.
' We have a source of electromotive force and a conductor that is

at least part of a circuit. The question seems to be, does the ground
form the balance of the circuit? Can't it be a conductor? If it is, then

the circuit is complete and electrons should flow.

oBe sure you understand that both ends of the circuit are grounded
to the same piece of material, and that that piece of material is

capable of conducting.
So, current can and does flow through ground.

Now, return to Page 21 and select the correct answer.

Page 27

YOUR ANSWER: Exactly 2 amperes are flowing through Rl .

Yes, the size of the resistors makes no difference in
this problem. If 2 amperes are flowing through one component,

then 2 amperes are flowing through every component in the

circuit. That is true of all series circuits .

R\ R; R3
MW

 

 

'|,L MI]

Remember that the ampere is a measure of quantity. If
we go back to our analogy of the ping-pong balls in the tube,
we could think of a series circuit as a circular tube already
filled with ping—pong balls. We could cut out a piece of the
tube so that we could push additional ping-pong balls into the
tube. (Then our hand would represent the battery or electromotive
force.) If we push three ping-pong balls into the tube, exactly
three will come out the other end, of course. And if we cut a
small window in the tube at any point, we would see exactly

three ping-pong balls move past our view point.

In the circuit above, how many amperes are flowing through
the battery?

Exactly 2 amperes are flowing through

the battery. Page 25

About 6 amperes are flowing through
the battery. Page 32

I don't know. Page 31,,

Page 28

YOUR ANSWER: The resistance in circuit A is less than the

resistance in circuit B.
You didn't add very carefully.

The total resistance of a series circuit is found simply by

adding together all the individual resistances in a circuit.

Here is an example. This is a series circuit. To find out
the total resistance of this circuit, we add the resistance of all
its cemponents. Starting
1082 at the top, we have 10 ohms,
+ 1,25% next down the side, we have

m 12 ohms, giving us 22. On

 

the bottom we have another

6 ohms, so the total resistance is 28 ohms.

Now return to Page 25 and calculate the total resistance

of both the circuits in the same way.

 

Page 29

YOUR ANSWER: Less than 2 amperes are flowing through R1.

No.

In a series circuit such as the one we are discussing,

.2, Q1 R3 the amount of current
flowing through any
9.0. 51L no.0.
component is determined

by the total resistance

 

-—J—- of the circuit. More

.lHIM

current cannot possibly
flow through one part than another. The current is the same,

whether you measure it in a series circuit.

Now return to Page 2L. and select the correct answer.

Page 30

YOUR ANSWER: To find electromotive force, use I = 73°

You're not watching very closely. This is the same

problem as before.

Electromotive force is E. And the unit of electromotive
force is voltage. If you want to solve for voltage, you choose
the form of Ohm's Law most convenient for finding E. That

form will begin, E equals. . . . ..

Now, return to Page 35 and select the correct answer.

Page 31

YOUR ANSWER: To solve for voltage, you would use the

form I = 3.

You could solve for E starting with this form, but you
would have to perform an extra mathematical operation to
isolate E on one side of the equation. You want to find out

what E equals.

The easiest way is to use the equation that starts out
E equals. . . . . This way, you will get the answer you want
directly. The form you have chosen would be most convenient
for finding amperage (I), since I is already isolated on one

side of the equation.

Now, go back to Page 33 and select the right answer.

YOUR ANSWER: About 6 amperes are flowing through the
battery.

That can’t be true.

The battery is just as much a part of the circuit as is the
resistance. The current is the same through each component

in a series circuit.

To use the ping—pong ball example again, the battery is
located in the circular tube where we cut the hole to add more
ping-pong balls. The battery collects ping—pong balls at one
end (the positive pole) just as fast as it pushes them away

from the other end (the negative pole).

The same is true in our electrical circuit. The battery

gets back as many electrons as it pushes out at the other side.

And at any point in the circuit it is true, too. As many electrons

as are forced into the circuit at the negative pole will pass any

point in the circuit.

So, the current in the battery is the same as the current

in any other part of the circuit.

Now, return to Page 27‘ and select the correct answer.

.age 32

Page 33

YOUR ANSWER: The resistance of circuit B is less than the
resistance of circuit A.
Right. In a series circuit, you find the total resistance by

adding together the resistance of each part of the Circuit.

Ohm's Law tells us the relationship of the electrical units
known as amperes, volts, and ohms.

The ampere (a or amp) is the unit for measuring the
quantity of electrons passing any one point in one .
second. This electron flow is also called current (I).

The volt (v) is the unit of electromotive force (emf or
E). It is the measure of the difference in electrical
potential, the force that makes electrons move.

The ohm (S2) is the measurement of resistance (R) to
current flow.
The abbreviations, a, v, and Q, are used in circuit drawings.

In Ohm's Law calculations we use:

= Current flow in amperes.
E = Electromotive force in volts.

R = Resistance in ohms.

The answers are expressed, of course, in volts, ohms, or

amperes, or with the appropriate abbreviation or symbol.

If you know the amperage and resistance in a circuit and

want to know the voltage, which form of Ohm's Law will you use ?

E
I — R Page 31
E = IR Page 35
R =% Page 38

 

“Page- 34.,

YOUR ANSWER: I don't know.

Well, isn't the battery 6. part of the circuit? The
amperage of all parts of a series circuit, including the battery,

is exactly the same.

If we look at the function of a cell again, perhaps the
matter will become clear. A cell, you will remember,
consists of two poles , usually of different materials , that are
placed in a chemical solution. The nature of the materials used
is such that a potential difference is produced between the two .'
poles . Different combinations of materials produce different
amounts of EMF. That is why dry cells , fresh ones , produce
1 .5v, lead acid cells 2v, and Edison cells 1 .3v. Each uses
a different combination of chemicals capable of maintaining a

definite voltage .

When electrons flow away from the negative pole of such a
cell and return to the positive pole, they tend to neutralize
the potential difference. But the battery wants to maintain its
natural voltage level, so within the battery just enough electrons
move to maintain the cell's normal voltage. The amount of
electrons necessary to maintain that voltage is exactly equal to

the amount that flowed through the circuit outside the battery.

Now, return to Page 27 and select the correct answer.

Page 35 "

YOUR ANSWER: To solve for voltage you would use the form
E = IR.

Correct.

To find voltage, multiply amperes times ohms (E = IR).
To find amperes, divide volts by ohms (I = g).

To find ohms, divide volts by amperes (R =12: ).

One volt is agreed to be the potential difference that will
force one ampere through one ohm of resistance. If we double
the voltage and keep the resistance the same, the current
(amperage) will double. If we double the resistance and keep
the voltage the same, the current will be just half as great.
Ohm's Law is derived from this relationship. The above formulas
are three ways of expressing this same law; any one may be

converted to the others by multiplication or division.

Earlier, we showed the circuit diagram below and said
that 2 amperes of current were flowing in the circuit. We said
that a 12 volt car battery would not be big enough to cause
2 amperes of current in this circuit. To find out what electro-
motive force would be required to obtain two amperes, which

formula would tée used ?

R1 R3

 
   
 
  

5.0.
I 2A CURRENT

Page 30

3:3
R
E = II"9 Page 39
E
R f ...-i Page 1.1

a Page 36

YOUR ANSWER: .8 amperes .
You divided the wrong things .
There were three resistors with a total of 16 ohms.
The EMF was given as 20 volts.

The formula was agreed to be I = %, since we want to

find amperage.

If we substitute the known values in this formula, we

have:

I = %%,sinceR=l6andE=20.

Apparently you divided 16 by 20 instead of 20 by 16.

Now, return to ' Page [+0 and select the correct answer.

Pa ge 37

YOUR ANSWER: R =I§ is the correct formula.
Yes, that' s the correct formula. Cover the R in the
triangle and younfiind 3E lovet‘ily'rvoltsdivided, bye amperes .-'>'i' :nmvann SEEK)?
it?
It may help to keep the triangle in mind if you remember
that the E is always on the top and that the three letters are

binary: “my
in alphabetical order.

».:','\'€“a-N/{::5l-é

Wm Mum.

.;J

, urabd”3fifilfl:flijfifiuai

 

..(.- .unilti‘l [3,?
Let's review the three forms of Ohm's Law. It's important

to learn them by heart. Memorize at least one of them, PW'F'QIBRIX 9 MIT

the equation for voltage E = IR. You can discover the other? forms l "l
- . ,4». Huge

mathematically from any one form.

, a
r ‘,.‘.,‘\ ,

' ,, ‘ ... . .. . " '3
2‘ l- : ' ‘ ‘w‘m’l may man; 53:2":

Here they are again wi‘thstheir meanings: fr. ,2: :i 23.-”2'13: ,m- 43m:- ;,~,_.,

E VOltS ,.);'.. U...
= — =-———-—-—-————- ..h‘r uJL‘IL‘INT.’
I R Amperes Resistance
ire-tn 1' “vatsa*Ampe‘résvxmeg-‘llgé‘a’iew 0W?
— E - _ Joli
R I Re31stance Amperes

Now use the correct formula to solve the problem in this circuit.

’31].

What is the resistance?

.005 ohms Page 50
50 ohms ,ga 53
200 ohms

YOUR waives: To solve toe: voltage, you would use the. form

_E
R-T.

You could solve for E starting with this form, but youf'.
“wouldhave to perform an extra mathematical operationlto' I
isolate E on one side of the equation. You want to findout what
E canals. I W

The easiest method is to‘ use the equation that starts out
E equals.'. ...... This way you get the answer you want directly.

The form youhave chosen Would be most convenient for finding

resistance (1?), since R is already isolated on one side of the
equation.

Now, go back to "6.1;; ‘33! ' and select the right answer.

. l
l

 

Page 39
YOUR ANSWER: To find the electromotive force, use E 2 IR.

Yes, this would be easiest. Actually, you could use any
of the three forms, but the arithmetic is easier if you choose
the formula arranged with the unknown value isolated on one

side of the equation.

It may help you remember the three forms of the equation

if you visualize its components in a triangle like this:

Cover up the unknown quantity. What remains shows the
operation you must perform. For the problem above, cover up

E, because voltage is what you want to determine. Below it you

see Itimes R, the arithmetic necessary to find E. Cover up I,
R! R1 R3

and you find E over R; or

cover up R and you find

 

— ' E over I.
I 2A CURKENT
T - Now see if you can
solve the problem of this
circuit. What is the EMF of the battery?

11. 5 volts Page 1.2
23 volts Page My
46 volts Page 1.7

YOUR ANSWER: To solve for amperes, use I = g .

Yes , divide the voltage by the total resistance, and the

result will express the number of amperes in the circuit.
See how this works out with our triangle:
Cover up the letter I (since you want to find the number of

amperes), and you see E over R, that is , divide E by R.

 

Work out this problem now:

What is the strength of the current flow in the circuit above?

.8 ampere Page 36
1 .25 amperes Page A3
320 amperes page A8

Page [.0

 

 

Page Ll

YOUR ANSWER: To find electromotive force, use R = 21:.
You're not watching very closely. This is the same problem

as before.

Electromotive force is E. And the unit of electromotive '
force is voltage. If you want to solve for voltage, you choose
the form of Ohm’s Law most convenient for finding E. That

form will begin, E equals. . . . .

Now, return to Page 35 and select the correct answer.

Page 12

YOUR ANSWER: It is an 11. 5v battery.

No. Let's take a look at the calculation to see where you went
wrong. Since we said that a 12—volt car battery was inadequate, this

battery must be inadequate, too.

To get to this page, you had to decide that the proper formula for
finding voltage is E = IR. That is correct. Here, again, is the circuit:

R. R1 R3

  
 

811.

-:_I_—— 2A

  

.7

You agreed that its total resistance is 23 ohms. This you
determined by adding the values of the three resistors. So we know
that the resistance (R) is 23 ohms and that the current (I) is 2 amperes.

This is how the formula should look:

E = 2x23

Now, return to Page 39 . and select the correct answer.

 

YOUR ANSWER:

Correct .

and you divided that into the voltage to get

Page h3

1 . 25 amperes .

The total resistance was 16 ohms (8 + 5 + 3),

20

16 1.25 amps.

factor in the circuit.

Now let's suppose that the resistance is the unknown

_i_

CSA

IOOV ...;—

T “

The diagram above tells us that 100 volts force just .5

amperes of current through the unknown resistance.

Since we

know two of the values , we can use Ohm's Law to find the third.

We must solve for "R",of course.

Which formula is correct?

_E

R—I‘

R=EI
__1
R‘E

O

Page 37
Page h6

Page 52

Page 141+

YOUR ANSWER: It is a 23v battery.

You slipped up somewhere. Let's take a look at the calculation

so you can find out where you went wrong.

To get to this page, you had to decide that the formula for finding

voltage is E = IR. That is correct. Here again is the circuit:

R. R1 R3

 
 

211.

i 2A

  

7

You agreed that its total resistance is 23 ohms. This you determined
by adding the values of the three resistors. So, we know that the resistance
(R) is 23 ohms and that the current (I) is 2 amperes. This is how the

formula should look: .

E = 2x23

Now, return to Page 39 and select the correct answer.

 

Page [45

YOUR ANSWER: To find amperage, use the formula E = IR.

It's hard to believe that you didn't choose this page
just to see if we could find yet another way of explaining

the same point!

Let's look at the triangle this time.

Here it is: A

Now, let's cover up the factor we are seeking, amperage.

Remember, I is the symbol used for amperage.

 

Remember, too, that we want a formula that begins with
the form I = . . ', That will save us the mathematical step

necessary to isolate I on one _side ofvthe equation.

Now, return to Page 147 and select the correct answer.

YOUR ANSWER: R = BI is the correct formula.

No, it's not. In fact, that is not even a variation of the
correct formula. It says resistance is equal to voltage _tim_es
amperage. That simply is not the case. Resistance is equal
to voltage divided by amperage.

The only form of Ohm's Law in which you multiply rather
than divide is that used in solving for E. The formula then is
E = IR, voltage is equal to amperage times resistance. This
is the basic form of Ohm's Law. The two variations can be

derived mathematically from the basic formula..

Since we can perform the same operation on both sides of
an equation, to obtain the formula for amperage, you isolate
the 'I by dividing through by R. If you want to check whether
a variation of Ohm's Law is correctly put, see if it can be
derived from E = IR.

Of course, you can also check with the triangle by covering

up the unknown. If you don't remember it, make a copy.

A
an

Now, go back to Raga 1,3 and select the correct answer.

Page ’46

 

Page 1;?

YOUR ANSWER: It is a 46-volt battery.

Correct. To find E, yOu multiply I times R (you can tell

that by covering up the E in the triangle below). The total
P.
3 resistance (R), found by

.... adding the values of the
“7;." 1A Cums»? three resistors, was 23
t - ohms. Multiply 23 times

%

5.1L

   
 

‘

   

 

  
  

2 , the amperage, and you have solved for E.

Now look at the circuit below:

 

To find the amperage in the circuit, which formula should

you use? ‘
E
I = E Page (40

E = IR Page M

R =‘f‘: Page [:9

Page £18

YOUR ANSWER: 320 amperes .

Wow: That's about enough to burn whatever you've get '
to a frazzlel Apparently you multiplied where you should have
divided. .

There were three resistors with a total of l6 ths.
The EMF was given as 20 volts. I .

The formula 'was agreed to be I - g- , since we want to

find amperage .

If we substitute the known values in that formula, we
have the following:
I = %, sinceR=l6andE=20.
You must have multiplied 16 by 20 instead of dividing 20
by 16. Now, return to Page to and select the correct answer.

 

, Page 19

YOUR ANSWER: To find amperage, use the formula R =1;

It's hard to believe that you didn't choose the p.25;
just to see if we could find another way of explaining the

same point '.

Let's look at the triangle this time.

Here it is: A

Now, let's cover up the factor we are seeking, amperage.

Remember, I is the symbol used for amperage.

 

What have we left?

Remember,too, that we want a formula that begins with
the form I = . . . That will save us the mathematical step

necessary to isolate I on one side of the equation.

Now, return to Page h? and select the correct answer.

Page 50

YOUR ANSWER: . 005 ohms .
You seem to have divided the wrong things .

Here is the formula we agreed was correct, R =I§ .

We also know certain values:
E = 100v, and
I = .5a.

You apparently divided . 5 by 100 instead of the other

way around .

Now, return to 33:, and select the correct answer.

 

Page 51

YOUR ANSWER: . ZOma .
No, .20 is how many amps it draws.

You evidently arrived at .20 by the correct procedure,
dividing the voltage (8v) by the resistance (40 S2) . That gives

. 20 amperes .

Remember, that milli- is a prefix indicating l/lOOOth. A
milliamp is .001 amps, just as a millivolt is .001v. Since there
are 1000 milliamperes in every ampere, you translate amperes
to milliamperes by multiplying amperes by 1000. So, 5 amps is
5000ma, and .5 amps is SOOma.

Now, return to Page 55 and select the correct answer.

I

YOUR ANSWER: R =1;- is the correct formula.

No, it's not. It's not even a variation of the correct
formula. It says resistance is equal to amperage divided
by voltage. That's not so. Resistance is equal to voltage
divided by amperage.

The basic form of Ohm's Law is E = IR, voltage is equal
to amperage times resistance. The two variations can be

derived mathematically from the basic formula.

Since we can perform the same operation on both sides
of an equation, to obtain the formula for amperage, you isolate
the I by dividing through by R. If you want to check whether
a variation of Ohm's Law is correctly put, see if it can be
derived‘from E = IR.

Of course, you can also check with the triangle by

covering up the unknown. If you don't remember it, make a

CODY . 3

Now, go back to page [,3 and select the correct answer.

Page 52

 

 

 

Page 53

YOUR ANSWER: 50 ohms.

You seem to have multiplied where you should have
divided.

Here is the formula we agreed was correct, R =I§ .

We also know certain values:

E = 100v, and
I= .Sa.

If we insert these values in the formula, we have:

R=lo— e

‘5

O

Apparently you multiplied .5 times 100 rather than dividing
100 by . 5.

Now, return to ‘ g 37 and select the correct answer.

YOUR ANSWER: . 021 ohms .

Sorry, but no. You forgot to translate milliamps into
amps. When you use Ohm's Law, you must be sure that
you have all of the values in comparable units . They can
all be milli— units if you choose, or all kilo- units, or, of
course, all basic, unhyphenated units , just plain amperes ,
volts and ohms. As a rule, it is probably simplest to

translate all figures into the basic units.

You can, if you want, use powers of ten. Since to
translate amps to milliamps, we multiplied by one thousand
(moved the decimal three places to the right), to translate
back from milliamps to amps , we divide by one thousand
(move the decimal three places to the left). So if there are

300ma flowing in the'circuit, there is .3 amps.

Now, with that much of a hint, return to Page 59

recalculate the problem, and select the correct answer.

Page Sh

'2
\

 

Page 55

YOUR ANSWER: 200 ohms .

Yes, you found this by dividing 100 by . 5; that is, you
divided volts by amperes. R = -I—. You could have guessed
that the resistance was fairly high, since all that electrical

pressure could push only one-half ampere.

If one ohm will let one

 

.5A ampere flow with one volt

loov 73-

“ 200 ohms to limit current

[_l

of pressure, it will require

flow to .5 ampere when

HIM

there is 100 volts applied

to the circuit .

In the circuit below, a lamp of 40 ohms resistance is
connected to an 8—volt battery. How much current does the

lamp draw? Express your answer in milliamperes (ma).

 

F—IIMF—

 

8v 40.0.
.20 ma. Page 51
ZOIna. Page 56
200 ma. Page 59

YOUR ANSWER: 20ma .

You have evidently forgotten that milli— is a prefix

meaning l/lOOOth. A milliamp is .001 amps, just as a

millivolt is . 001V .

You arrived at amperage by dividing the voltage (8v) by

the resistance (4052). That gave you .20 amperes. The

problem now is to translate amperes into milliamperes .

Since a milliampere is l/lOOOth of an ampere, there

are 1000 milliamperes in every ampere. To translate amperes

to milliamperes , therefore, you multiply by 1000. Thus, 5

amps is 5000ma, and .5 amps is 500ma.

'Now, return to page 55

answer.

and select the correct

Page 56 .

,‘x.

Page 57

YOUR ANSWER: 3 x 105 amperes.

You have apparently forgotten quite a bit about the use
of powers. First of all, since 3 millionths of an ampere is
a fraction -- less than one ampere, it will be expressed with
a negative exponent. Only whole numbers have positive

exponents .

Secondly, since this figure would be written out as
.000003, its exponent would be a six, not a five. You. determine
that by counting off the number of places starting from the right
of the first significant figure and moving left toward the decimal
point. When there is only one significant figure, as here, that
is the first (and last) significant figure. You count like this:

‘654321
.000003

That gives you the exponent.

 

If the figure had been .0000032, we would have counted off
like this: ..
l6 5 4 3 2 l
.0 O 0 0 O
6

In powers of ten, .0000032 would be expressed 3.2 x 10" .

 

 

2 IE: significant figurej

 

Now, return to Page 61 and select the correct answer.

Page 58

YOUR ANSWER: 3 x 10‘5.
Well, you' re partly right. You got the sign of the exponent
as minus, as it should be since we are dealing with a fraction.

However, 3 millionths of an ampere would be written out
as .000003 , so its exponent would be six, not a five. You
determine that by counting off the number of places starting
from the right of the first significant‘figure and moving to the
left toward the decimal point. When there is only one significant
figure, as here, that is the first (and last) significant figure.

654321,

.000003

You count like this:

That gives you the exponent.

If the figure had been .0000032 , we would have counted off

like this:
6 s 4 3 2 1'
. 0 0 0 0 0 3 2
,. 4‘3“ significant figural

~ In powers of ten, .0000032 would be expressed 3.2 x 10—6.

 

Now, return to Page '61 and select the correct answer.

‘, ...—J

Page 59

YOUR ANSWER: ZOOma .

Correct. Using the formula I = E , divide 8 by 40 and you

get .20 amperes. Since there are 1000 milliamperes in one
ampere, you multiply this answer by 1000, which you can do by
moving the decimal point three places to the right, giving 200

Now let's find the resistance of a light bulb filament

milliamperes as the correct answer.

 

will?

which requires 6.3 volts and draws 300ma.

 

300 ma.
_I_ “*—
:; 6. 3V R
.021 ohms Page 5).:
21 ohms Page 61

21 ,000 Ohms Page 6).:

 

Page 60

YOUR'ANSWER: 3 x 106
You're right in thinking that the exponent is a six, not

a five, but you've forgotten about signs. Your answer would

be translated back into regular numbers as 3,000,000. But

3 microamperes is .000003.

Since 3 millionths of an ampere is a fraction -- less than
one ampere —— it will be expressed with a negative exponent.

Only Whole numbers have positive exponents.

Now, return to Page 61 and select the correct answer.

 

 

Page 61

YOUR ANSWER: 21 ohms.

Yes , it is necessary to change the milliamperes into
amperes before calculating so that all values will be in
comparable units . To translate milliamperes to amperes ,
divide by 1000 (move the decimal point three places to the
left), changing 300 ma into .3 ampere. Dividing the voltage,
6.3, by .3 amp (R =IE)’ you get 21 ohms.

In the circuit below, a 4-megohm resistor is passing 3
microamperes of current. Remember, a megohm (meg) is a
million ohms , and a microampere is one one-millionth of an

ampere, 3 m'acro amperes
‘—-—_

 

Amegn‘

Ill—“IMH—

Let's find the voltage across the resistor. This will be the
same as the total voltage produced by the battery since there is

just one resistance in the circuit.

Use your knowledge of exponents in solving this problem.
Don't mess around with a lot of zeroes when you don't have to.
Let's have a brief check-up on powers of ten before tackling the
problem .

We will want to translate 3 microamperes (three millionths

of an ampere) into powers of ten.

Three microamperes equals . (Complete the sentence)

3 x 105 amperes Page 57
3 x 10_5 amperes Page 58
3 x 106 amperes Page 60

3 x 10—6 amperes Page 65

YOUR ANSWER: 4 x 106 ohms.

Yes, this is the correct way of expressing 4 million ohms
or 4 megohms. The figure ” 4" is six places from the decimal
point so we use 6 for an exponent.

Now, solve the problem below using the values for ohms
and amperes that you have just expressed in powers of 10.
I=3x1o‘6

R=4x1$

What is the voltage across the resistor?
3 Micro amperes

 

_C “— --
, 4:. Ame 1‘-
i-i 9*
12 volts Page 66
36 volts Page 67

12 x 10'

12 x1012 volts ' Page 68

Page 62 ' .‘

Page 63

YOUR ANSWER: 4 x 107.
Whoops! When working with positive exponents we
count a little differently than when working with negative

exponents .

Four megohms is 4,000,000 ohms . When we count Off
for powers of ten with a whole number, justas we did with

a negative power, we start from the right of the first significant
figure.

Here is how we counted off the microamps:

654321
.000003

‘1

starting point

Here is how to count off the megohms:

1 2 3 4 5 6'
4'4——__
In both cases , counting is started from the right of the
significant number that will be the multiplier when the figure

is expressed as a power of ten, and from there proceeds toward

the decimal point.

Now, return to Page 65 and select the correct answer.

YOUR ANSWER: 21, 000 Ohms.

Afraid you went all wrong. You could have done a couple
of different things to get this answer, so we won't try to track
down the error. Here is what you should have done. Figure

out for yourself just where you left the track.

When you use Ohm's Law, you must be sure that you have
all your values in comparable units. They can all be milli-
units if you choose, or all kilo- units, or, of course, all basic,
unhyphenated units, just plain amperes, volts and ohms. As
a rule, it is probably simplest to translate all figures into the

basic units.
You can, if you want, use powers of ten.

Since to translate amps to milliamps, we multiplied by

one thousand (moved the decimal three places to the right), to

translate back from milliamps to amps, we divide by one thousand

(move the decimal point three places to the left). So, if there

are 300ma flowing, there is .3 amps.

Now, with that much of a hint, return to Page 59

recalculate the problem, and select the correct answer.

Page &

a

j

Page 65
YOUR ANSWER: 3 x 10”6 amperes.

Correct. Since 3 millionths is .000003, you counted
six places from the right of the significant figure, and since
.000003 is a fraction, the sign of the exponent is negative.

Here is a comparison of some negative exponents:

3 =3 x 100
.3 = 3 x 10‘1
.03 = 3 x 10'2
.003 = 3 x 10‘3
.0003 = 3 x 10“4
.00003 = 3 x 10‘5
6

.000003 = 3 x 10'
Now let' 5 consider the other value given in our problem:
4 megohms . How will you express this as a power of 10?

4 x 106 ohms Page 62

4 x 107 ohms Page 63

YOUR ANSWER: 12 volts .
Correct. The formula is E = IR, so the problem was computed
in this manner:
E = (3 x 10'6) x (4 x106)

E = 12 x 100, and, since 100 equals 1, E = 12 volts.
How much current. is flowing in the diagram below?

 

E=6volts “3' R3360hms

 

(39

a. 3. amps
b. 03 amps
c. .03 amps

d. .003 amps

After selecting an answer go on to page 71 and continue.

Page 67

YOUR ANSWER: 12 x 10'“.
That' s awful few volts. Remember, you have 4,000,000

ohms of resistance to overcome, even to get 3/1 ,000,000ths

of an amp of current. You went wrong somewhere along the

line.

Since E = IR, you were correct in trying to multiply the
amperage (3 x 10‘s) by the resistance (4 x 10 ), but you've
forgotten how to multiply exponents . You multiply exponents
by adding them algebraically. Although in algebra 6 X —6
would be —36, 106 x10—6

and the positive powers cancel each other out.

gives us 100, since the negative

And 100, remember, is the same as 1.
You do, of course, multiply the 3 x 4 in the normal arithmetic
manner. So
(3 x10—6)x (4 x 106) = 12 x 100 or ?

Now, go back and select the correct answer on Page 62’

YOUR ANSWER: 12 x 1012 volts.

That‘ s an awful lot of volts . Admittedly, you've got to
overcome 4,000, 000 ohms resistance, but you're only getting
.000003 amps through.

Your mistake came in the process of multiplication by
powers of ten. Since E = IR, you were correct in trying to
multiply the amperage (3 x 10_6) by the resistance (4 X 106).
But when it came to multiplication of the exponents you.
acted as if you were dividing instead of multiplying. To
multiply exponents, you add algebraically. The product of
106 and 10‘6
other out.

is 100, since the 6 and the -6 cancel each

6

'If you had been dividing 10 by 10'6, you would have

performed algebraic subtraction which requires you to change

the sign of the divisor and then add algebraically. The result '

of that process would be 1012. You can see it here:

6
£6 =106x106 = 10
10 ,

12
However, when multiplying,- you get the result 100. Of
course, 100 is 1. So,

0

(3x10‘6)x (4X106)=12x10 or ?

Now, return to Page 62 and select the correct answer.

Page 69
A common device in an electrical circuit is the switch. The simple

on - off switch is called a single pole, single - throw switch. The fancy

name means that it can Only make or break contact with a single electrical

connection. The symbol for such a switch is shown here. Usually, such a

'switch will be illustrated in this open position in circuit diagrams.
————o/ o———

If you are asked how much current flows in a circuit that includes
this kind of switch drawn in the open position, you will answer as if the
switch were closed. It is drawn in the open position only so it can be
easily identified.

Another device used in many electric circuits is the FUSE,
whose symbol is -of\/o- . It is used to protect electrical
equipment from damage due to excessive current. Fuses are
rated in amperes , according to the amount of current they will
pass through the circuit. If amperage rises above the level the
fuse is designed to pass , the current will create enough heat to

melt the soft metal used as a conductor within the fuse. That, of

course, breaks or opens the circuit and completely stops current.
5A FUSE

40V SJL

a?

 

Fuses are not designed to resist current flow, so for practical

purposes, we can assume they have no resistance.

Will the lamp in the above circuit stay lit after the switch
is closed?

Yes Page 72

NO Page 7h

I don't know Page 77

YOUR ANSWER: Current will flow in Circuit A only.

You are correct. With 8 ohms resistance, only 10 amperes
will flow. Since the fuse will theoretically allow 15 amperes,
it will not blow out. Circuit B with only 2 ohms would draw

30 amperes, but the fuse rated at 20 amperes should blow.

JLFL IS’A

20A

   

Most fuses are not absolutely accurate, so it is usual to
select a fuse rated lower than the maximum Current that can

be tolerated, thus ensuring that the components will be safe.

Let's say that a component in Circuit B would be damaged
if 25 amperes flowed through it, and we want to change the
circuit so it will work. What could we do to it?

, Put in a larger fuse. Page 73
Put in another resistor. Page 76 , (
Put in a smaller battery. Page 79 ‘

Page 71

Which resistor has the most current flowing throught it?

R1 = 12 ohms

 

R2 = 8 Ohms
R3 = h ohms
as R1
b. R2
c. R3

d. All resistors have same amount.

’After selecting an answer go back to page 69 and continue.

YOUR ANSWER: Yes, the lamp will stay lit.

It may light for an instant —— the time it takes for the

fuse to discover just how many amps are going through the

circuit -- but it will go right out after that.

The unknown, of course, is the amperage in the problem

presented to you.
volts and a resistance of 5 ohms. Since I =§ , orfl

R 5

The diagram showed a power source of 40

, I equals

8 amps. The fuse in the circuit can carry only five amps before

blowing, so, when you close the switch you will blow the fuse.

Return to

Page 69

and select the right answer.

Page 72 C.

YOUR ANSWER: Put in a larger fuse.

I No, indeed, not unless you were absolutely certain that
the fuse already in the circuit was too small. If you put in
a larger fuse, you would save the fuse from blowing, but you
might damage the components of the circuit. And, you were

told in this instance that you do not want in excess of 25 amps.

You are not trying to save fuses; you are trying to save

circuits.

Now, return to Page 70 and select the correct answer.

Pa 86 7h

YOUR ANSWER: No. The lamp won't stay lit after the switch

is closed.

 

Correct. The only important resistance in the circuit
was the 5 ohms of the lamp. Using the formula I = % (dividing ,
40 volts by 5 ohms) showed that 8 amperes will flow in the
circuit. The fuse will allow only 5 amperes, so the excessive

current will cause the fuse to blow. ‘

It is always irritating to have a fuse blow, but don't try
to prevent this by putting in a larger fuse. In the circuit we've
been discussing, the lamp might be damaged by excessive current.

The fuse warns when something is wrong and prevents damage by

turning everything off.
I 5

(8, .611 .512.

    

20A

 

In the circuits above, the fuses will allow current to

continue to flow in (complete the sentence).
Circuit A only Page 70 .
Circuit B only Page 75
Both circuits - Page 78
Neither circuit Page 80

Page 75

YOUR ANSWER: Circuit B only.

You've got it wrong. Now, we're not going to work this

out for you. We'll just remind you how to do it.

The question that must be answered for both circuits is
‘simple: Will the combination of voltage and resistance produce
amperage greater than the capacity of the fuse ? The capacity

of each fuse is shown.

You have to compute the amperage in each circuit. To
do this, you first add up the values of each cir'cuit's resistances

to get the total resistance of each circuit.

When you have the total resistance, then you determine
the voltage of the power source. In both circuits, the voltage
is given next to the battery symbol. Now, you have the total
resistance (R) and the voltage (E). You can calculate the
amperage with Ohm's Law, the variation that says
Name = 7_1§ (Zita...

When you have found the amperage the circuit is arranged
to deliver, compare that with the rated value of the fuse. That
rated value tells the amount of current that will cause the fuse
to blow and open the circuit. Compare the amperage of the
circuit with the rated amperage of the fuse. If the circuit

amperage is higher, the fuse will blow.

Now, return to Page 71; and select the correct answer.

Page 76

YOUR ANSWER: Put in another resistor.

Yes, that would normally be the easiest, fastest, and
safest thing to do.

The situation we now have is this: One device in the circuit

will be damaged if 25 amperes should flow through it.

We want to keep the 20 amp fuse in the circuit for protection,
but to keep it from blowing, we must add another resistor to

reduce the amperage of the circuit.

3 What is the smallest possible resistance we could insert

at "X" to limit current-to 20 amperes ?

.61).. .511.

60"

 

1 Ohm Page 81
3 ohms Page 82

 

YOUR ANSWER: I don't know.
Well, here is how you find out.

The question, really, is whether the fuse will blow out
when the switch is thrown. To find out whether it will blow,
you must calculate the amperage that will flow when the

switch is thrown.

You do this by applying good old Ohm's Law. That law
says amperage equals voltage divided by resistance. If we
divide voltage (40 volts) by resistance (5 ohms), we discover
that 8 amps is the current that will flow through the circuit -—

if the fuse will let it.

Now, go back to Page 69 and select the correct answer.

Page 77

 

YOUR ANSWER: Both circuits.

Sorry, in one of them the fuse will be blown. We're not
going to work this out for you. We'll just remind you how to

find in which circuit the fuse will blow.

The question that must be answered for both circuits is
simple: Will the combination of voltage and resistance produce
amperage greater than the capacity of the fuse ? The capacity

of each fuse is shown.

You have to compute the amperage in each circuit. To
do this, you first add up the values of each circuit's resistances

to get the total resistance of each circuit.

When you have the total resistance, -then you determine
the voltage of the power source. In both circuits, the voltage
is given next to the battery symbol. Now, you have the total
resistance (R) and the voltage (E). You can calculate the
amperage with Ohm's Law, the variation that says
I (amperage) = gflTTasi—gme)‘

When you have found the amperage the circuit is arranged
to deliver, compare that with the rated value of the fuse. That
rated value tells the amount of Current that will cause the fuse
to blow and open the circuit. Compare the amperage of the
circuit with the rated amperage of the fuse. If the circuit

amperage is higher, the fuse will blow.

Now, return to Page 7h and select the correct answer.

Page 78

 

Page 79

YOUR ANSWER: Put in a smaller battery.

It would certainly reduce amperage if you put in a smaller
battery, but it's not the usual approach, and it probably wouldn't

be a practical solution.

It's unlikely that you could change the battery, either
because you wouldn't have another one handy or because the

battery is also needed to power another circuit at the same time.

It would be more practical to seek a safe way'of modifying

some other part of the circuit.

Now, return to Page 70 and select another answer.

Page 80

YOUR ANSWER: Neither circuit.

No, the amperage in only one circuit is sufficient to cause
the fuse to blow. We're not going to work the problem for you,
just review the steps so you can calculate for yourself which is

which.

The question that must be answered for both circuits is
simple: Will the combination of voltage and resistance produce
amperage greater than the capacity of the fuse ? The capacity

of each fuse is shown.

You have to compute the amperage in each circuit. To
do this, you first add up the values of each circuit's resistances

to get the total resistance of each circuit.

When you have the total resistance, then you determine
the voltage of the power source. In both circuits, the voltage
is given next to the battery symbol. Now, you have the total
resistance (R) and the voltage (E). You can calculate the
amperage with Ohm's Law, the variation that says
I<amperage> =7———;§ 82:33.. '

When you have found the amperage the circuit is arranged
to deliver, compare that with the rated value of the fuse. That
rated value tells the amount of current that will cause the fuse
to blow and open the circuit. Compare the amperage of the
circuit with the rated amperage of the fuse. - If the circuit

amperage is higher, the fuse will blow.

Now, return to page 7), and select the correct answer.

__~,

U I ”Page 81

YOUR ANSWER: A 1 ohm resistance must be added at X.

Yes. This would bring the total resistance of the Circuit

to 3 ohms, so current would be just 20 amperes.

When a fuse blows , you have an open circuit. The
circuit is interrupted, Just as when a switch is open, and there
is no more current until a new fuse is installed. The circuit
would be open, too, if a connecting wire came loose, or if
there was a breakdown inside an electrical device. These

malfunctions that cut off current are generally called open .

Would the lamp light when the switch is closed in the diagram

      
 

 

below?
r.) , R2 = 10 OhJTlS
____‘[ R =

E = 10 volts _ fuse is

20 ma.
_c/
R1 = 35 ohms
a. Yes
b. No

After selecting an answer go to page 85 and continue.

Page 82 .\

YOUR ANSWER: 3 ohms.

You're probably on the right track, but that's not the right

answer. Here's why.

You have a 60 volt output and you want to limit amperage to
20 amps. To determine the amount of resistance necessary to
obtain 20 amps with 60 volts output, you divide voltage by
amperage (R = £13). The result is 3 ohms. So that is the resistance

necessary to limit Current to 20 amps. So far, so good.

. However, there is already resistance in the circuit, in
fact there are three resistors. So, to make a total resistance .
of 3 ohms, you do not need to add 3 ohms. You must add just
enough to make the total resistance‘3 ohms.

Now return to Page 76 and select the correct answer.

 

Page 83

YOUR AN SWER: None.

No, not none. There will be a voltage drop, since the
potential returns to zero. Just because we now have four
resistors instead of one doesn't mean that the situation has
changed. The positive side of the battery will still be considered
to have zero potential, while the negative side has a potential

of 40 volts -—relative to the positive side, of course.
The voltage expends itself somewhere.

You should know, we hope, that the total voltage to be
disposed of in the exterior circuit is 40 volts. It is going to
be disposed of in passing through the "working" parts of the
circuit —- the resistors. The question is, can you figure out

how much of the voltage is disposed of by each resistor?

Now, return to Page 85 and try again.

Page 811

YOUR ANSWER: 5 volts.

Come now! Stop and think about that answer for a moment.
You have four resistors. You say each drops the voltage by
5 volts. That means the total voltage drop around the circuit
is 4 x 5 = 20 volts. But, the battery is putting out 40 volts.
That means that by the time the current gets around to the
positive pole of the battery, it still has 20 volts push behind it!
The fact is, as we found not so long ago, the positive pole is at
zero potential.

Granted you have lower resistance. But don't be misled
into thinking this would produce a smaller voltage drop. The
total voltage drop for the circuit is 40 volts. The voltage drop

across each resistor is still 10v.

What has changed is the amperage. More current is getting
through now. If you calculate the amperage in this circuit,
with a total of 32 ohms resistance, against 40 volts EMF, you
will find that the amperage is 1.25. With 40 ohms resistance,
the amperage was 1. '

Now, return to Page 87 and select the correct answer.

 

Page 85

Now, let us consider what happens when there is more than
one resistance in the circuit. Below, the current flows through
each resistor in order, 1 through 4. Each resistor in turn
offers an obstacle to current, and each resistor in turn lowers
the voltage. We say the voltage drops as current flows in a
resistor. What would you guess to be the voltage drop across

each resistor below ? R z

 

None Page 83
10 volts Page 86
40 volts Page 88

Page 86

YOUR ANSWER: There should be a 10—volt drop across each resistor.

R.\ R1

 

That's right, and a good guess. Each resistor helps to lower the
voltage back to zero. When each resistor has the same ohmage, as here,
the voltage drop is equal across each one. Since there are four resistors,

you divide the voltage by four.

Let's get a shade more scientific about figuring the voltage drops.
There are two things we apply here:

(1) The current is the same at all points in a series circuit.

(2) The voltage drop across a resistor can be found by
multiplying the current through the resistor by the

=I ‘R.

value of resistance, E1 1 1

We know the value of R1 — — it's given in the diagram. We know that
11 (the current through resistor R1) is the same as the current for the
circuit. We have to figure the current for the circuit, using Ohm's law
I = E/R and the values given for the total circuit. '

I = E/R = 40 volts/10 + 10 + 10 + 10 ohms = 40/40 = 1 ampere.

Is that 1 ampere of current flowing through each of the resistors, or

just the one nearest the negative pole of the battery?
Each resistor. Page 87

Just the one nearest the negative

pole of the battery. Patio 89

 

Page 87

YOUR ANSWER: The same amount of current is flowing through
each resistor.

You are correct. One ampere is flowing through each
resistor, for in a series circuit the same amount of current is
present at any point in the circuit. The amount the voltage drops
across each resistor is found by using Ohm's Law. Since all
resistors in this particular circuit have the same value, you
need work the problem only once. If the resistors in the

circuit are of different values, you must solve for each one.

Using the formula E = IR, we multiply 1 (ampere) times
10 (ohms) and we see that there is a 10-volt drop across each '

resistor, just as we guessed.

Look at the circuit below. What is the voltage drop across

each of these resistors ?
K .0. 2 .0.

 

5 volts Page St.
8 volts Page 90

10 volts page 91

Page 88

YOUR ANSWER: 40 volts.

Whoa, you're overdoing it'. If each of the four resistors
disposes of 40 volts, we've lost 160 -— four times as many as

we had in the first place.

Perhaps you didn't read the question closely. You were
not asked for the total voltage drop, you were asked what would

be the voltage drop across each resistor.

Now, return to Page 85 and select the correct answer.

 

Page 89

YOUR ANSWER: Just the one nearest the negative pole of the battery
Where have you been?

The last page tells you quite plainly: "Current is the same at all
points in a series circuit . . . . 11 (the current through resistor R1) is

the same as the current for the circuit. ”

Current, remember, is amperes per second. It is a rate of flow.
Clearly in a series circuit, which is just like a continuous, one-path
pumping system, there cannot be a greater current flow in one part than
in another. That is, more electrons cannot be'moving past one point

than are moving past another.

I Now, return to Page 86 and select the correct answer.

YOUR ANSWER: 8 volts.
No.

We'll say it again:
(1) The voltage drop around a series circuit is equal to

the total voltage rise within the power source.

(2) The current is the same all the way around a series

circuit.

Granted you have lower resistance. But don't be misled
into thinking this would produce a smaller voltage drop. The.
total voltage drop for the circuit is 40 volts. The voltage drop

across each resistor is still 10v.

What has changed is the amperage. More current is getting
through now. If you calculate the amperage in this circuit,
with a total of 32 ohms resistance, against 40 volts EMF, you
will find that the amperage is 1. 25. With 40 ohms resistance,
the amperage was 1.

Now, return to Page 87 and select the correct answer.

Page 90

 

3

YOUR ANSWER: 10 volts. (Pageel

Correct. There are still four resistors of equal size, so the
voltage drop will be equal across each of them. The big difference
produced by lowering the resistance is to increase current.

This time I = g = g—ZO- = 1.25 amps. Last time, with four
10 ohm resistors, the current was 1 amp.

The voltage drop across each resistor, using the new values
for current and resistance, is: E = IR = 1.25 x 8 =10 volts. The
voltage drop across each resistor is the same as before, you see.

What we have just demonstrated is the third important fact
about series circuits: The sum of the voltage drops across the

resistors is equal to the applied voltage. This concept can be
stated mathematically:

Et = El '4' E2 + E3 +....En.

(

Et is the total or applied voltage. El, £22; etc. I stand for
the voltage drops across each of the resistors (which will

normally be numberedcorrespondingly, R1 , R2, etc.).

In the diagram below the voltage drops are : ERl = 10 volts,
ER2 = 20 volts, ERh = 15 volts, and ERS = 25 volts. What is the
voltage drop across R3?
R1 R2

E = 100 volts —~ ‘ R3

a. 10 volts
b. 20 volts
0. 30 volts

d. hO volts

After selecting an answer go on to page 93 and continue.

Page 92

YOUR ANSWER: E = 20v, E = 20v, E = 20v, and E = 60v.

1 2 3 t

The total voltage drop is 60 volts. But the values of the individual
resistors are no longer the same. We are now considering a circuit in
which the value of resistance varies from resistor to resistor. In this
case, the voltage drop varies too. It must, because the current can't
change. We have shown many times that current is the same in all parts

of a series circuit.

As we have said, the total voltage drop still equals the applied
voltage. Varying the resistances, just changes the proportion of the
voltage that had to be expended across each resistor.

You can calculate voltage drops in two steps. First, find the
circuit current by Ohm's law, I = %. Since the voltage is 60, and the
total resistance 30 ohms, the current is 2 amperes.

Then, you apply the formula E = IR, to each resistor. In all cases,
the current is 2 amperes. To find the voltage drop across each resistor,
multiply the individual resistance by 2 amperes.

Now, return to page 91 and select the correct answer.

"'3 Page 93
The power in watts is volts times current, or:
P ( in watts )=E ( in volts ) X I' (in amperes)
In some situations, you will not always know either the voltage or

current. Instead you may know the resistance. By using Ohm's law, the
formula for power can be altered.

As an example, how much power is being expended in this

 

circuit?
1. 5 watts Page 9L; 7
2 watts Page 95
’3 ' 18 watts Page 98

Page 9h

YOUR ANSWER: 1 . 5 watts are being expended in this circuit.
Correct. Ohm's Law works here, of course. It's simple to
find the current in the circuit.
I = E/R = 3 volts/6 ohms = 1/2 ampere
Then using the given formula for power:
. P=EI=3x1/2=1.5watts
Note that to find current, you used voltage and resistance,

then to find power, you multiplied current by volts again.

It seems reasonable that power might be found without ever
finding current. Here' 3 the formula for current: I = E/R. Suppose
instead of P = BI, you replace I by E/R. That gives you

p =3 x E/R = BZ/R.

That gives you power in one calculation without bothering about
current.

The same kind of approach will enable you to find P when you
don't know voltage but do have values for current and resistance.

The formula for voltage is: E = IR.
Replacing E by IR in P = BI, you get

P=IRxI, orP=IzR.
Thus P = E1 = Ez/R = IZR.

Which one of these formulas
is best for finding the number
of watts expended in this circuit?

 

P =nz/R Page 95
P = BI ' Page 97
P = 12R Page 99

‘l

 

Tags 95,»

YOUR ANSWER: 2 watts of power are being expended in this
Circuit.

No, but you can soon find out why it's wrong.

 

3v 6.0.

—~M|ll—

The formula for power is:
P = 131

If P = 2 watts, as you claim then
P = E1 = 2

. Since E is given as 3 volts, we have
3 x 1 = 2
and I must be 2/3 amperes .

Check this against Ohm' 5 Law for the current in the circuit.

I = E/R = 3/6 = 1/2 amperes.

Clearly the 2/3 ampere must be wrong. Return to Page 93 ..
and figure it again.

YOUR ANSWER: P = Ez/R is the proper formula to use.

No, it will work, of course. but why figure out the voltage?

Look at the circuit again. 50.11-

 

30 .0—
What are the values given? There are three resistances , and

the current is specified. You aren't given voltage.
at the formulas for power:

P=EI, P=IZR, P=E2/R

40.0—

N ow look again

Can't you find one which will give power without knowing
. voltage? Return to page 911 and make the correct choice this

time .

Page 96

U

Page 97

YOUR ANSWER: P = BI is the correct formula to use.

No, it will work of course, but why figure out the voltage?
Look at the circuit again. 50 0

40.0..

 

30.0-

What are the values given? There are three resistances ,

and the current is specified. You aren't given voltage. Now
look again at the formulas for power:

P =EI, P=12R, P=Ez/R

Can' t you find one which will give power without knowing

voltage? Return to and make the correct choice this
ti Page 9h
me.

Page 98

YOUR ANSWER: 18 watts of power are being expended in this

circuit .

No, but you can soon find out why it' 5 wrong.

 

The formula for power is:
P = BI
If P = 18 watts , as you claim, then
” P = 1:1 = 18
Since E is 3 volts, we have

3XI=18

and I must be 6 amperes.

Check this against Ohm' 3 Law for the current in the circuit.
I = E/R = 3/6 = 1/2 ampere

Clearly the figure of 6 amperes must be wrong. Return to
Page 93 and figure it again.

Page 99

YOUR ANSWER: P = 12?.

‘467_rx—

 

3(J.11_

That's right. The resistance and current are given. You
could stop and figure out the voltage, too, but by using the ‘
correct power formula this isn't needed.

Since the three units , volts , ohms, and amperes , are all
related, you may find the number of watts by using any two of
them. ’

' For measuring power in larger units, we have the kilowatt
(kw) which equals 1000 watts . In low-powered circuits we
sometimes use the milliwatt (mw) which is one one—thousandth
of a watt.

Now, using the formula you have chosen , solve the problem
above and give your answer in kilowatts .

1.08 kW Page 101
10.8 kW Page 105

108 kw Page 111

Page 100

YOUR ANSWER: 4 ohms is the resistance of a 25-watt bulb for
100 volt operation.

No, but it could be worse, the answer does start with a
41 Since voltage and power are given, and resistance is
wanted, the proper equation is:
P = Ez/R. Solving for R, you get
R = EZP = (1002)/25 = 10000/25

Return to Page 111 and check this computation, then
choose the correct answer.

 

Pa ge 101

YOUR ANSWER: 1 .08 kw is being expended in the circuit.

No. Bet you wouldn't have made a mistake here if it

were dollars instead of power:
Let's go over this carefully.

We'll use the formula, P = 12R. The current, I, is given
as 30 amperes. The resistance is the sum of 50 ohms, 40 ohms
and 30 ohms, a total of 120 ohms.

Then P=I2R= (30)2 x120 =900x120
= 108000 watts

So, if a kilowatt is 1000 watts , then the number of kilowatts
in the answer is just 1/1000 of 108000.

Return to Page 99 and select the right answer.

YOUR ANSWER: Yes, the power rating is being exceeded.

 

ralllli-

12.-BA

.———>

 

It certainly is! Using the formula P = 12R, you should
multiply 22 x 5 to get 20 watts. The lamp will burn very brightly
but only briefly and then burn out, for its power rating is being
greatly exceeded.

. Let's suppose that you have to make this lamp operate

properly. What would be the easiest way to do this?

Increase the voltage. Page 106

Put a resistor in the circuit. Page 112

Page 102

1,}

Page 103

YOUR ANSWER: Yes, the power rating of the resistor is being
exceeded.

No, it isn't. At least this mistake won't burn anything out.

Look at the circuit again.

.[

5° ““5" ' V4 wnr

' RES'STOR
T1 = 3’“ a.

To check this, all that' s necessary is to compute the power
J . being expended in the circuit and compare it with 1/ 4 watt, the
rating of the resistor.

 

Since voltage and current are given, use
P = £1 = 60 kv x 3 na
60 kv = 60,000'volts = 5 x 104 volts
3 ha = 3/1,ooo,000 = 3 x 10"6 amperes
Therefore:

P = (6 x 104) x (3 x 10’6)

= 18 x 10"2 watts
= 0.18 watts

Clearly, 0.18 watts is less than 1/4, or 0.25 watts.
Return to Page 112’ for the right answer when you' re sure you
understand this .

‘4

Page loll
YOUR ANSWER: 400 ohms 6
That's correct. Using P = Ez/R, and solving for R,
R = Ez/P
= (1002)/zs
= 400 ohms
Whenever power is expended in overcoming resistance, heat
is generated. For a given resistor, the more power expended, the '
more heat generated. The limit a resistor, or lamp can withstand
before it is damaged is given by the maximum power it can tolerate.
Thus a 4-watt resistor of 25 ohms cannot be used in a circuit having
a source voltage of more than 10 volts, {or I ,
P = Ez/R
= (102)/25
= 4 watts

Look at the Z—watt lamp in this circuit,

9 “‘5‘“

Is the lamp's power rating being exceeded? That is , is

 

 

more than 2 watts of power being used in the lamp?

Yes Pa ge 102

NO Page 108

Page 105

YOUR ANSWER: 10.8 kw is being expended in the circuit.

No. Bet you wouldn't have made a mistake here if it
were dollars instead of powerl

Let's go over this carefully.

We'll use the formula, P = 12R. The current, I, is given as

30 amperes. The resistance is the sum of 50 ohms, 40 ohms and
30 ohms, a total of 120 ohms.
Then p =12R = (30)2 x 120 = 900 x 120
O =- 108000 watts
So, if a kilowatt is 1000 watts , then the number of kilowatts
in the answer is just 1/1000 of 108000.

Return to Page 99 and select the right answer.

‘4

Page .195;

YOUR ANSWER: Easiest way to make the lamp operate would
be to increase the voltage.

Wrong. The only bright thing about this answer would be
the lamp going poofl

Consider the formula for power in terms of voltage and
resistance,
p = Ez/R
You should readily see the effect of increasing the voltage.
Return to Page 102 and choose the right answer.

Page 107

YOUR ANSWER: The resistance of the resistor is 2 x 10”2 ohms.

No. There are two ways to go wrong in figuring this resistor:
to make an arithmetical error, and to come up with the wrong power
of 10. You've found one. Let's find out which.

I

 

 

 

60 Kv : ' Va WATT
RESISTOR
I, = 3,4; 61..
‘ Prom Ohm' 3 Law,
R=ad
As before, E = 6 x 104 volts, and

I = 3 x 10"6 amperes.

Thu“ R= (6x104)/(3x10")
-= 2 x <1o“)/ (10")
Here' 3 where you went wrong. Remember that for any power
of 10, when you move exponents from numerator to denominator,

or vice versa, change the exponent from positive to negative, as
the case may be.

Thus, 102 - 1/10’2.

Complete this computation and return to Page 111: , and select
the right answer.

*0

.' (go 108

YOUR ANSWER: No, the power rating is not being exceeded.

Wrong. You have just burned out a bulb. Look at the
circuit again.

 

_i_
E —:—

A

E) R=sn

 

‘3};

 

The current and the resistance are given. To find power,
choose the formula using only R and L This is, P = IaR.

Then it becomes a matter of straightforward substitution.

P = (22)x5
=4x5=20Watts

This is considerably more than the 2 watts the bulb is
rated for. It Would last at most a few seconds. Return to Page 101.

and choose the right answer.

g;

Page 109

YOUR ANSWER: 2500 ohms is the resistance of a 25~watt bulb
designed for 100 volt operation.

N 0. There were two ways to go wrong on this , by doing
the arithmetic wrong, and by misplacing the decimal. You

managed to go wrong on both. Congratulationsl

Since voltage and power are given, and resistance is

wanted, the preper equation is:

Ez/R, solving this for R,

i!

P

R = EZ/P

However it appears you multiplied voltage times power,

or R = 3?. Return to .7 page 111 and recheck your figures.

Pa ge 110
Now let's try one a little more difficult. Suppose we have three
lamps in series as shown below. Remember lamps are sometimes used to

Show resistance in a circuit.

C?

_—L_

nov .1:—

T

:0
"a
P

9.22%

 

G)

as 251:.

To determine the total power expended must you find the
amperage of the circuit?

Yes Page 116

N0 Pa ge 117

Page 111

YOUR ANSWER: 108 kw 501).

-‘—_:— 30 A. 40—0-
4_._____

30 .0.—
Correct. Using P = 12R you multiply 302 x 120 and get
108,000. Since there are 1000 watts in a kilowatt, you have
108 kilowatts. '
J You should memorize the three power formulas that we got
from Ohm's Law:

P = EI 12R

*u
u
wlm
*v
u

Write them down to help remember them. If you can remember
only one, be sure to remember that P = EI, power (in watts) equals

volts times amperes. You can derive the other forms from this .

Using the power formulas and Ohm's Law, see if you can solve

this problem:

 

 

What is the resistance of a 25—watt bulb designed to operate
on 100 volts?
4 ohms Page 100
"‘ 400 ohms Page 101;
2500 ohms Page 109

Page 112

YOUR ANSWER: You could make the lamp operate properly by
putting a resistor in the circuit.

__|_.__
E '22"

a

1

 

ZWCO R=5_D—

 

Yes , increasing resistance would decrease the amount of
current that could flow in the circuit. If you chose a resistor
large enough, you could limit the power expended by the lamp .
so that it would be operating within its prOper limits .

' Let's suppose that you have a circuit with 60 kilovolts across
a resistor with a power rating of one—fourth watt. If the current is
3 microamperes , is the power rating of the resistor being exceeded?

_._ .

 

 

 

60 W : V4- wxr'r RISISTOR
I 1:=:1/4 4’
Yes Page 103
No Page 1111

‘a,

%.

Page 113

What is the minimum wattage resistor we can use in the diagram below?

 
  

E - 10 volts—Z- R = 200 ohms
a. 2000 watts
b. 100 watts
c. .5 watts

do .0025 watts

After selecting an answer to to pase 110 and continue.

Page 1111.

YOUR ANSWER: No, the power rating of the resistor is not
being exceeded.

60 Kv V4 WATT

RESISTOR

 

Correct. You can express 60 kilovolts (60,000 volts) as

6 x 104 volts, and 3 microamperes (0.000003) as 3 x 10—6 amps.

P = s1 = 6 x 104 x 3 x 10‘ or 18 x 10‘2; adjusting the decimal

point shows that 0.18 watt is being expended in'the resistor. This
is under its rating of 0.25 watt.

6

Just for fun (and to make sure you're still with us), calculate
the resistance of the resistor above. Figure the number of ohms
in powers of ten.

The 0.25-watt resistor above is:

2 x 10'2 ohms Page 107

2 x 1010 ohms Page 115
3 x 10.2 ohms page 118

3 x 1010 ohms Page 120

Pa ge 115

YOUR ANSWER: Given a resistor with 60 kilovolts across it and a
current of 3 microamperes, you could express

the resistance as 2 x 1010 ohms.
Correct. Of course this is an unrealistically high resistance.

We hope you didn't waste any time with the formulas involving P.
This time you were asked merely to calculate the amount of resistance
present. That called for Ohm's law, R = E/I.

The power formula wouldn't help you find the actual resistance. It
would tell you the maximum current you could pass at any given voltage or
the maximum voltage you could apply with any given current without

applying more power than the resistor is rated to carry.

The calculation as you should have made it, then, was:

6 10

R: E/I= 6 x 104/3 x 10' = 2 x 10 ohms.

It would take a resistance as large as this to hold current at such
a low level with this high amount of voltage.

If the resistance of the resistor in the diagram below decreases
what happens to the power?
R = 8 ohms
_£:—_V:/'
E = 16 volts '*
a. Power increases
b. Power decreases

co Power stays the same

After selecting an answer go to page 113 and continue.

Page 116

YOUR ANSWER: Yes, it is necessary to find the amperage of the
circuit.

No. If you were flipping a coin, it wasn't your lucky one.

The first definition of power you saw was
P = EI
But it was shown that if only current and resistance were

known in a circuit, this formula could be modified by substituting
1R for E, getting,

0.,

P = IZR
. Or, if only voltage and resistance were known, you could V
substitute E/R for I , getting,

1? = Ez/R

Put away that wooden nickel and return to Page 110

Page 117

YOUR ANSWER: N 0, you don't have to find the amount of current
to determine the total power expended.

 

 

 

_L {’9
NOV 3:...- RI 1011' (a R2201)...
T o
WRS 2511. 2

Correct. There are three formulas for power: P = EI. P = -E-

R

P = 12R, so you can find watts with any two of the three electrical

units -- voltage, current and resistance.

Use the correct formula, and solve for

P with the units given

in the diagram. (Be sure to write out your numbers in fraction form

and simplify your fraction before multiplying it out.)

What is the total power expended in the circuit?

110 watts Page 121
220 watts Page 12h
6050 watts Page 127

Page 118

YOUR ANSWER: 3 x 10—2 is the value, in ohms of the resistor in

the circuit.

No. Both the value of the resistor and the power of ten is

wrong in your answer. Look at the circuit again.

 

60Kv V4 WATT
RESISTOR
From Ohm' 5 Law, "I
R = B/I
As before, 60 kv = 6 x 104 volts
3 pa = 3 X 10"6 amps
Thus, R =(6x104)/(3x10‘6)

This can be written
R = (6/3) x (104 x 105)

Complete this calculation, then return to Page 111, and
choose the right answer. Make sure you understand this before

going on, however.

Page 119

How much power is consumed by the circuit, if the power consumed

by R2 is 20 watts?

:3
ll

2h0 ohms

E = 190 volts R2 = 80 ohms

 

a. .5 watts
b. 10 watts
c. 60 watts
d. 90 watts

After selecting an answer go on to page 136 and continue

‘11
Oil

Page 120

10

YOUR ANSWER: 3 x 10 ohms.

You' re wrong, but in an odd way.

The purpose of this question was to check your knowledge of
manipulation with powers of ten. You handled the powers. of ten
correctly, but somehow managed to divide wrong.

From Ohm's Law,
R = E/I

As before, 60 kv = 5 x 104 volts g
3 me = 3 x 10-5 amps

Thus,

' R =(6xio4)/(3xio’5)

This can be written

R = (6/3) x 1010

One more small step, and you have the answer. But that‘s
where you blundered. " Check your figures and then return to
page 111,. for the right answer. 3

Page 121

YOUR ANSWER: The power expended in the circuit is 110 watts.
No. Somewhere along the line, your arithmetic has come unstuck.

Here are the figures given: E = 110 volts, R1 = 10 ohms,

R2 = 20 ohms, R3 = 25 ohms.
These were the formulas from which you had to pick the one for
the job:
2 2
P = EI P =T P = I R

The formula you needed was the one which expressed P in the
terms given -- voltage and resistance. In other words, the one you

needed was:

9:

”I ”is

Using that formula, your calculations should have gone this
way: 2

2 11mm:

_E _
P—F— ’55, ?WattS

That calculation shouldn't be hard to finish. Return to Page 117‘
and read through it again so that you get the idea clear in your mind,

then pick the correct answer.

YOUR ANSWER: E

l = 5 volts, E? =10 volts, )3

3 = 12 . 5 volts
Wrong.

And if you had remembered that the sum of the voltage

drops is equal to the total applied voltage, you would have
seen this for yours elf.

2=20 ohms,R

Here is the information you had: E = 110 volts, R
R 3

l =.10 ohms,
= 25 ohms, P = 220 watts.

The next thing needed was the current.

You had two
ways of getting that: I = E/R or, since P = El, I = P/E.

I=E/R = llO/SS = 2 amps. I= P/E = 220/110 =2 amps.

Now you' re in business . The voltage drop across R
be obtained by Ohm's Law:

1 can
El

= IRl = 2 amps x 10 ohms = 20 volts
The voltage drops across resistors R2 and R3 are found in
exactly the same way.

Figure them out and then return to
correct answer.

Page 121. with the

Page 122
1% . - -

Page 123

YOUR ANSWER: 'P1 = 20 watts, P2 = 40 watts, P3 = 50 watts

No, you forgot at least one thing.

Add up the total number of watts (or wattage) in these resistors .
They total 110 watts , don't they?

If this doesn‘t ring a bell, it' 5 clear that you've forgotten that
the total watts us ed in the circuit was 220. Since you can account
for only 110, and there are no other circuit elements, something is
wrong .

In the previous question, you were asked to calculate the current
in the circuit and then figure out the voltage drops in the individual
circuit elements . Since you answered this question correctly, you
know the circuit current, and the voltage drop across each resistor.

Now all that is necessary is to use:

P = El

Do this for each lamp.

Now return to Page 130 , and when you have the individual
wattages, add them together and see that the total checks.

YOUR ANSWER: 220 watts are used in the circuit.

Correct.

Now go ahead and calculate the value of I, so that you can
find the voltage drop across each lamp. When you have figured

outthe three voltage drops , compare your list with those below,

KER
.L U

.... 11
‘1: m
“kc/R3

The calculation you needed was:

2
2
_E __ 110x =
P—R_-— ——?57£9_ 220 watts

:0
O

HOV "‘ 1“ R2203.

 

 

 

25.0.

and choose the correct one.

E1 = 5 volts
£2 = 10 volts
E = 12.5 volts

Page 122
3
£1 = 110 volts

E = 110 volts
E = 110 volts

Pae126
2 g

3

E1 = 20 volts
£2 = 40 volts
E = 50 volts

Page 130 .
3

None of these Page 132

Page 121;

YOUR ANSWER: p = 40 watts, p =- 100 watts. Page 125

1 = 80 watts, P

2 3

(E)

HOV 20JL

R3 2511

 

 

r—{llllt—
CE)

Correct. You had to apply the formula P = EI for each
resistor in turn, using the voltage drop across each resistor
for E, and using 2 amps for I (the current is the same through
each resistor since this is a series circuit).

As you remember, we found the total power expended in
the circuit to be 220 watts. Think over what you have learned
in this problem, and see if you could make some observation
about power in a series circuit.

What is the new thing this problem told us about power
in a series circuit?

You can use the power formula in

combination with Ohm's Law to

determine the power expended in

a series circuit. Pa ge 128

Only two of the three electrical

units (volts, amperes, and ohms)

are needed to calculate power in

a series circuit. Page 131

The total power dissipated in a

series circuit is equal to the sum

of the amounts of power dissipated

by the individual components. Page 133

YOUR ANSWER: E1 = 110 volts, E2 = 110 volts, £3 = 110 volts.

No, no and no. The sum of the voltage drops should be
equal to the total __applied voltage (in this case, 110 volts). Here
you have voltage drops which total three times the applied voltage.
Something' 3 wrong. Let' 5 do it the right way.

Here is the information you had: E = 110 volts, R1 = 10 ohms,
R2 = 20 ohms, R3 = 25 ohms, P = 220 watts.

The next thing needed was the current. You had two ways of
getting that: I = E/R or, since P = BI, I = P/E.

I = E/R = 110/55 = 2 amps. I = P/E = 220/110 = 2 amps.

. Now you‘re in business. The voltage drop across R1 can be
obtained by Ohm' 3 Law:

131 = IR1 = 2 amps x 10 ohms = 20-volts

The voltage drops across resistors R2 and R3 are found in
exactly the same way.

‘\
Figure them out and then return to Page 12b, with the correct

answer.

Page 12.6.

5‘

£04 141

‘1

Page 127

YOUR ANSWER: The power expended in the circuit is 6050 watts .

No. You really came unstuck. It looks as if you obtained
your answer by multiplying the resistance by voltage. And
that means you have made up a formula of your own for power,
which seems to say: P = BR. No. 'It won't do.

We'll begin at the beginning again. Here are the figures you
had to work with: E = 110 volts, R = 10 ohms, R = 20 ohms,
R3 = 25 ohms .

1 2

These were the formulas from which you had to pick the

one best suited for the job:
2

E. 2

P=EI P-'-=R P=IR

The formula you needed was the one which expressed P
in the terms given -- voltage and resistance. In other words ,

the one you wanted was:

2

-.El.
P‘R

Using that formula, your calculations should have gone this

way: 2

2
P=I; = 110x114) = ? watts
’52.;

R

 

Finish the calculation. That shouldn't be hard. Then return
to Page 117 and read it through again so that you really
understand it before picking the correct answer.

Page 128

 

YOUR ANSWER: You can use the power formula in combination with
Ohm' 3 Law to determine the power expended in a
series circuit.

Yes , but it' s not the answer we hoped for.

It' 3 true, of course, but you've known this for some time -- i
you couldn't have gotten this far otherwise. i

What the question aimed at was , which of the choices presented
contains a statement not previously stated?

Return to Page 125‘ and read the choices again. i

Page 129

YOUR ANSWER: P1 = 220 watts, P2 = 220 watts, P3 = 220 watts
Three times, "NO" .

When you figured the power expended in the circuit, you
used the formula: P = E /R to get an answer of 220 watts.

Since then you have found that the circuit current is 2 amps.
This gives you another check on the earlier figure by using
P = BI = 2 amps x 110 volts = 220 watts
What else do you know about the 110 volts in the circuit? It‘s

the sum of the voltage drops in the individual lamps, isn‘t it? So

we could have written:
P=EI==(B1 +E2 +33)XI= (20 +40+ 50)x2 =220watts

But this is the same as:
P=EI=20x2+40x2+50x2=220watts

Complete the problem in terms of each lamp, and return to Page
130, making sure you' re quite certain of yourself before going on.

Page 130

YOUR ANSWER: E1 = 20 volts, 132 = 40 volts, E3 = 50 volts .

m
J: Gallon? R2

“0V5: ‘ Q 20.0..
T mare-n—
) W

Yes, in a series circuit,the sum of the voltage drops is equal

 

 

to the total applied voltage. If you remembered this you didn't
have any doubts when it came to checking your arithmetic , for this
was the only group of numbers that added up to 110.

Using the formula I “1% , you should have found that there is
a Z-ampere current in the circuit. Using the values you have at hand
calculate the power dissipated by each resistor. In other words , how
many watts are expended in each resistor?

P1 = 20 watts
P2 = 40 watts Page 123
P3 = 50 watts

P1 = 40 watts
P2 :80 watts Page 125
P3 = 100 watts

P1 = 220 watts
P2 = 220 watts Page 129
P3 = 220 watts

Page 131'

YOUR ANSWER: Only two of the three units (volts , amperes,

and ohms) are needed to calculate power in

a s eries circuit

u
UA~ '1. .':~

" ? AS .4 ... i-
Yes, but it s not in“,

_ ’ .. .- 1 »
... We»: 5:35 ’tf-S'; ..3 $029.3". £03 0
v a .y ' V -- . ‘ *= .. .r o .
A}. .-_.. '- /,1 r _ ‘ :': a n .. A , ll \ -~o . ”,3 -,o‘. . u ._.~',a a. _ I~.
at 8 me 3 NJ; 5.; a: 5.11. S 3 <1 1.; in u .: L274: ‘3? 53 ' ' ‘3 A,“ ‘ 5:

M... -.s to: r one time ......
J '7 ‘ ' . . T... 4 . Jim“ . r.
you couldn t ha is genie... fins t

q" ‘ ‘ :9" n P’tfi '~**-"/\ -2'1“""‘ «. r \' :....
What the anemic... awn—....

u

d

_ , .. . .41....” , _. ... .._... .‘ f4 "
contains a statement as: ere-«gloat . .

.
x ., n- .
, Ago: ~_4\ a~ u
...

v , .
. flag“ (5 ”N“, ‘ I h, i" na .n -.- - f; ‘,~~.- .9
Rat-mi. .62 Pay-.- ~,:-; . i . ~

— i. F- ”A 0‘ A ...: 1
_ r 7 4" .v "r. “'-.a(' R’fi 'V.
t...\/ a... ‘5‘ ’"A o J‘J‘A’“ 4.8«1 ‘...1~. -VVL, ‘d‘... ‘flnw
hJ

rPege 132

YOUR ANSWER: None of these.

No. Why can't you determine the voltage drops? In the question it
was suggested you calculate the current. Look at the circuit again:

 

m
___L_ UR: 10.12.
HOV—E— o
T ' 3325.0.
\‘U

The resistances are given, R1 = 10 ohms, R2 - 20 ohms and
R3 - 25 ohms. It should be a simple matter to add these together to find
the value of RT’ and using I -= E/RT to find the current.

201).

 

Since in a series circuit the same current flows through each element,
1 the voltage drop in each element can be obtained by using Ohm's law for
each resistor, or lamp. And since the sum of the individual voltage drops
adds up to the total circuit voltage, you can check your figures.

Read this over if it still isn't clear, then return to _l’age 121; and
work the problem again.

\

YOUR ANSWER: The total power. consumed in a series circuit is . Page 133'
‘ equal to the sum of the amounts of power consumed
by the individual components .

Correct. You have also learned these facts about a series
circuit: .
1 . The current in a series circuit is the same at all points

in the circuit.

‘ 2. The total applied voltage of a series circuit is equal to
the sum of the voltage drops across the individual ‘

components .

3. The total resistance of a series circuit is equal to the sum
of the individual resistances .

l f

These last three facts are true of series circuits only. A
different set of facts applies to parallel circuits. But note this:

 

 

" The statement about power given in the answer at the top of the
' fpageis true of both series and parallel circuits. You should remember
that.
A How much power is consumed by R2 if the power consumed by R1 is T
320 watts?
R1 = 80 ohms
E = 300 volts -;-'_[_—- A \ R2 = SO ohms
R3 = 20 ohms
a. 50 watts
b. 100 Watts
0. 200 watts
d. 320 watts
L.)

-After selecting an answer go to page 119 and continue.

YOUR ANSWER: E1 = 5 volts , IE2 = 10 volts, E3
Wrong.

How did you do it?
Add up your voltages:

Etotal = E1 +E2 +153
5 + 10 + 15 = 30 volts

Pa ge 13h

= 15 volts

Long before you heard of Kirchhoff 3 Laws, you knew
that the sum of the voltage drops (or IR drops) in a series

circuit had to equal the source voltage.
in this case is 60 volts.

And the source voltage

Count to ten slowly, take a deep breath, and return to

page 1L3 and choose what you knew was the right answer
all the time!

Page 135

YOUR ANSWER: The algebraic sum of 60, ~10, ~20, and -30
is zero.

Correct. This fits Kirchhoff 8 second law for voltage:
The algebraic sum of the EMS and voltage drOps around a closed

circuit is zero.

That circuit had only one battery, so there was only one EMF
to consider. However, Kirchhoff's law applies just the same where
there is more than one source of EMF. Take a look at this circuit:

12.0. . 311

_Ea/V”V“/\/L**‘VAVNJ\/\¢:Ij?
2.4 __ '

V 6" -: 24v
3J2. T

€3\/

Here are two EMFs toproduce current. Again there are 2

12v

41'“!

 

 

amperes of current in the circuit, and the voltage drops are given
with each resistor. Remember, now, the negative end of the
resistor is the end toward the negative pole of the battery. If

you start at point Z, which of the lists below shows the numbers to
be added algebraically to analyze the circuit?

+24 - 6 +24
- 6 —24 - 6
+12 +12 -12
-24 — 6 -24
- 6 -24 — 6

Page 137 Page lhl Page ihh

Kirchhoff's secoud law‘is for voltage: The algebraic sum of Page 136
the EMF's and the voltage drops around a closed circuit is zero.
Often you will hear voltage drops referred to as IR drops .

This means that in any closed circuit the applied EMF is
equal to the sum of the IR drops around the circuit. This, of course,
is a form of Ohm's Law, E = IR.

It is what you have already learned about the voltage drops in
a series circuit. Shortly we'll see how it applies to parallel
circuits. But first, let's see how this would work in a simple series
circuit.

gov E

 

What is the current in this circuit?

0.5 amp IPage 138
2 amps Page 113
4 amps Page U45

U y,

 

Page 137

YOUR ANSWER: Starting from Z, the analysis of this circuit
according to Kirchhoff' 3 Law calls for you to
add +24, —6, +12, —24, and —6.
\ZlifL. i3.£L.

_fwngb 6v Z

.... 3.0. :24V

Yes , the result is zero, of course. And you analyzed the‘

I!!!

12V

!

 

circuit in the direction of current flow, as indicated by the arrow.

Now look at the diagram below. Again we have two batteries ,
but now the smaller battery is turned around so that its EMF
opposes that of the larger battery. Accordingly, there is now just
1 ampere of current in the circuit. Starting at point A below,

analyze the circuit and choose the list that gives the numbers you

shouldadd. _- QJL —l—_ 311 +

~WW“

24vۤ- {E-IZV
Z I=1A 3‘“— +
+ _—
+24 + 3 +24
- 6 +12 - 6
_ 3 —L 3 — 3
+12 + 6 -12
- 3 -24 — 3

Page lbO Page 1&9 Page 1h?

Page 138

YOUR ANSWER: 0.5 amps is the current in this circuit.

5.0. IOD.

 

You are wrong, and you shouldn't be at this stage.
This is a series circuit, you can add up the resistances in the
circuit:

R=5+10+15 =30 ohms
The voltage is given as:
E = 60 volts
By Ohm's Law, I = E/R
I = 60/30 = ? amperes
Now that's the way you've been doing it for quite some
time. Nothing has been changed,or will be changed for such
simple computations . Of course, if you get careless and use

Ohm's Law as I = R/E, it won't matter how well you know the
material, the answer will still be wrong.

Return to Page 136 and turn your formula right side up.

.1;

Page 139

YOUR ANSWER: E =10 volts, E2 = 20 volts , E3 = 30 volts.

__ JifL~I+ _. I<9_rL. +_

10v 20v

l—‘

60v “‘3
"7" 15211.
+ 2A
4-—.——.._ + ....
MM
2: Eflfii/

 

 

 

 

Yes , using E = IR, you multiply each resistance by 2 (the number

of amperes) to find the voltage drop across each resistor.

The diagram lists the voltage drops with the resistors and also

shows the polarity of the voltage drop.

Note that the battery is labelled + and -, indicating that the
direction of current is from the battery to the Si—ohm resistor and then
on around the circuit. The polarity of each resistor is given, too.
Note that they are given just as you would apply a voltmeter; the
negative end of each resistor is the end toward the negative end of the

battery .

If it helps you to remember the polarity of the resistor, think of
the resistor as a count er—force to the EMF of the battery. This means

 

that the polarity of the resistor is opposite to that of the battery.

To analyze this circuit according to Kirchhoff' 3 second law,
let's'begin at point Z, near the positive pole of the battery. Going
in the direction of the current flow, we pass through the battery and
then through the resistors , adding the EMFs and voltage drops
algebraically. We call them positive if we run into a + first, and we

call them negative if we find a — first.

Starting from Z, we add these numbers algebraically:
+60, —10, -20, ~30
What is the result?
0 Page 135
120 Page 1116

Page 1).].0

YOUR ANSWER: The list of numbers to be added in analyzing
this circuit is: +24, —6, -3, +12, —3.

61L 4. _ 3n +

'—' ' )Z\/

+ I: M 31L +

A + -

No, you were asleep at the switch here, and got caught

.—

_—
....—
_—

24V

on a real blooper.

Add up your list:
+24 —6 -3 +12 -3 = +24

It is certainly no news to you that the sum of the IR drops
must equal the source voltage, and that since they are equal, m
their difference must be zero. How then can their difference
add up to +24?

It looks as if you forgot that we reversed a battery on
you. Return to Page 137 and check the battery polarities .

J)

Page lhl

YOUR ANSWER: The algebraic sum of the components in the
series circuit shown is: -6 —24 +12 -6 -24

Wrong on two counts. In the first place it is customary
to read in the direction of current flow. As for the other, look

at the circuit again.

1211 3.11—

W2
_— Z4V éV

VLV --“.-:-- -—::"—.:— 24v
““ 31L. .._....

V

Note particularly the batteries . If you trace the direction

 

of current from each, do they take the same direction? If so

won't they have the same sign?

This was only a small mistake, but the effect was just
as wrong as if you didn't have the slightest idea of what was
going on. Someday this can be embarrasing so start being more

careful now. Return to page 135 and choose the right answer.

Page 1142

’2:
YOUR ANSWER: The list of numbers to be added in analyzing
this circuit is: +3, +12, +3, +6, -24.
24V -} —E~ lZV
A + —
No, but it's not an error that will ever cost you money.
The convention in analyzing such a circuit is to proceed
in the direction of the current. The result here will be the
same, but later on it might cause confusion. Return to page 137
and recheck the direction of current.
_V
'3?

Page 1).:3

YOUR ANSWER: 2 amperes are flowing in the circuit.
5.!)- 10.0.

I R R;

 

 

 

 

Correct. 1 =% = §% = 2 amps.

You know that there must be a total voltage dr0p in the
circuit of 60 volts , and that this voltage drOp is shared among
the resistors in proportion to their resistance. These voltage
drops can be labelled E1 , E2 , EB' etc. , with the subscript
(the figure below the line as in E1) corresponding to the resistor

number .

Calculate the voltage dr0ps above and choose the

correct list below:

131 = 5 volts

132 = 10 volts Page 13u
E3 = 15 volts

E1 = 10 volts

E2 = 20 volts page 139
E3 = 30 volts

E1 = 15 volts

E2 = 30 volts Page 1MB

E3 = 60 volts

YOUR ANSWER: The algebraic sum of the components in the
- series circuit shown is: +24, —6, -12, —24,
-6 O

No, you were careless in reading the circuit. Look at it

again. \Qfl 3fl

N

-: 24V éV __
\2V :- 3: 24V

W
év
Note particularly the batteries . If you trace the direction

of current from each, do they take the same direction? If so
won't they have the same sign?

This was only a small mistake, but the effect was just as
wrong as if you didn’t have the slightest idea of what was going
on. Someday this can be embarrasing so start being more careful
now. Return to page 135 and choose the right answer.

Fag

ellil.

YOUR ANSWER: 4 amps is the current in this circuit.
51L 1 0J1.
60V -—:-;—
“T“ I 5 _D.
W

 

 

 

 

It would be hard to be more wrong.

Add up the resistances in the circuit:

R=5+10+15=300hms

The voltage is given as:
E = 60 volts
By Ohm‘s Law, I = E/R
. I = 60/30 = ? amperes
Now that' s the way you've been doing it for quite some time.
Nothing has been changed or will be changed for such simple
computations . Of course, if you get careless and forget to add

up all the resistances, it won't matter how Well you know the
material, the answer will still be wrong.

Return to Page 136 and check your work before going on.

Page th

YOUR ANSWER: (+60) + (-10) + (—20) + (—30) = (+120)
Not this year.

Or next year either. This was carelessness on your
part, for you surely know how to add negative values . If
it makes it easier for you to total all the negative values
and subtract the sum of these, the computation becomes:

(+60) — (10 + 20 + 30) = ?

Perhaps you should take a breather before returning

to Page 139 and choosing the right answer.

Page 1146

 

 

Pa ge 1117

YOUR ANSWER: +24, —6, -3, '-12, -3 are the numbers you should
add to analyze the circuit-1_starting at point A.
._... -+ -

-W

24v 3 . —.-.":——- Izv

~—~u '—‘r-
+ W +
A - ... __
Good , The -12 is for the battery. Since it is connected
so as to oppose the EMF of the 24-volt battery, its EMF must be
called negative. It subtracts voltage from the larger battery.

Note that the lZ-volt battery acts just like a resistor in this

 

 

circuit .
Good,
You have done very well by completing this lesson in this amount of

time. On your answer sheet write down the amount of time it took you to

complete the lesson.

Page 1&8

YOUR ANSWER: E1 = 15 volts, E2 = 30 volts, E3 = 60 volts
Wrong. How did you do it?

Add up your voltages:

Etotal = E1 + E2 + Es

= 15+30+60 = 105 volts

Long before you even heard of KirchhoffI 3 Laws , you knew
that the sum of the voltage drops (or IR drops) in a series circuit
had to equal the source voltage. The source voltage in this

case is 60 volts.

Count to ten slowly, take a deep breath, and return to
Page 11:3 I and choose what you knew was the right answer all
the timel -

PARALLEL CIRCUITS

 

NAVAL AIR TECHNICAL TRAINING COMMAND

v

”It

Acknowledgements

This program was adapted by the Naval Air Technical Training Center,
Memphis, Tennessee, from a larger one developed under contract No..AF 33
(616)-6983 CPFF between the U.S. Air Force and western.Design and Electronics.
In a letter dated 27 July, 1961 from the Aeronautical Systems Division,

Air Force Systems Command, United States Air Force, use of these materials'
by other governmental agencies was confirmed. Appreciation is expressed
to the Aeronautical Systems Division and to Western Design and Electronics

for assistance in making the original pregram available.

Page 1

In a series circuit, the sum of the voltage drOps across the
resistors is equal to the applied voltage. This concept can be stated
mathematically:

Et: E

4' + P 4. 00. 0

Et is the total or applied voltage. E1, E2, etc., stand for the
voltage drops across each of the resistors (which will normally be
numbered correspondingly, R1, R2, etc.).

Calculate the voltage drops across each resistor in this circuit

and then choose the list that gives all the drops correctly.

W

 

 

 

 

 

 

60V ‘2:" R1 311 R2
-1_— R3 15Il 1251
-‘----CV“v“/\/VA+—~
Page 2 Page A Page 7
El: 20v El: 6v El: 30v
E2: 20v E2: 22d! E2: 214v
E3: 20v , E3: 30v _ E3 = 6v
E : 60v E ::60v E ==60v

t . t t

Page 2

YOUR ANSVER: E1 = 20v, E2 = 20v, E3 = 20 v, and Et = 60v.

The total voltage drop is 60 volts. But the values of the individual
resistors are not the same. We are now considering a circuit in which the
value of resistance varies from resistor to resistor. In this case, the
voltage drop varies too. It must, because the current can't change. We
have seen many times that current is the same in all parts of a series
circuit.

As we have said, the total voltage drop still equals the applied
voltage. Varying the resistances just changes the proportion of the
voltage that had to be expended across each resistor.

You can calculate voltage drops in two steps. First, find the
circuit current by Ohm‘s Law, I = g . Since the voltage is 60, and the
total resistance 30 ohms, the current is 2 amperes.

Then, yOu apply the formula E = I R, to each resistor. In all cases,
the current is 2 amperes‘ To find the voltage drop across each resistor,
multiply the individual resistance by 2 amperes.

Now, return to page 1 and select the correct answer.

(1

Page; 3

YOUR ANSWER: Diagram A is not electrically identical to
the others .

No. Here are the diagrams again:

ELM} m; . l

 

 

 

 

 

 

 

 

«P 2J1 (C)
.1. 3n. .
m8
( ) LM—L—

The voltage drop across each component connected in
parallel is the total voltage of the source.

When we told you that , we explained that the negative ends
cf resistors connected as in (A) above could be considered
electrically identical to each other and to'the negative pole of
the battery, while the-positive ends of the resistors could be
considered as electrically identical to each other and to the
positive pole of. the battery.

Let's put it another way: We join one end of each resistor
in a circuit. If it is possible to trace one end of each resistor
back to a single point on the path from the negative pole of
the voltage source and to trace the other end of each back to
a common point on the path from the positive pole, the resistors
are in parallel.

Check that out against the diagrams above and then return
to Page 10 for another answer.

Pa ge ll

YOUR ANSWER: E1 = 6v, £2 = 24v, E3 = 30v, and E: = 60v.

Correct. The total resistance is 30 ohms, allowing 2
amperes to flow in the circuit. Multiplying 2 times the value
of each resistor gives voltage drops across the individual

resistors .

As you can see, the voltage drops steadily in a series
circuit. The voltage drop is proportional to the resistance
and total drop equals the applied voltage.’

Now look at this circuit:

60 v.

 

R. JEJL

What is the voltage drop across R1?

6 volts Pa ge 6
18 volts Page 9
36 volts Page 11

\(J

Page 5

YOUR ANSWER: Diagram B is not electrically identical to
the others .

No. Here are the diagrams again.

2.0.
(i “1% J— 2:;
T (c)
F——’\/V\r-—-

LZJL ,
The voltage drop across each component Loonnected in

 

 

 

 

 

 

 

parallel is the total voltage of the source.

When we told you that, we explained that the negative ends
of resistors connected as in (A) above could be considered
electrically identical to each other and to the negative pole of
the battery while the positive ends of the resistors could be
considered as electrically identical to each other and to the

positive pole of the battery.

Let' 5 put it another way: We join one end of each resistor
in a circuit. If it is possible to trace one end of each resistor
back to a single point on the path from the negative pole of
the voltage source and to trace the other end of each back to
a common point on the path from the positive pole, the resistors

are in parallel.

Check that out against the diagrams above and then return
to Page 10 for another answer.

Page 6

YOUR ANSWER: The voltage drop across R1 is 6 volts .

You've picked the wrong resistor, apparently. The
resistor you were asked about is Rl .
If you simply chose the wrong resistor, return now

to Page L1 and calculate the voltage drop for resistor
R1 . It is labelled R1 , so there should be no trouble finding
it.

YOUR ANSWER: E1 = 30v, E2 = 24v, E3 = 6v, and Et = 60v.

You're guessing, or else you got your math mixed up a good deal.

The total voltage drop still equals the applied voltage, which is 60 volts.

You can calculate voltage drops in two steps. First, find the
amperage of the complete circuit by Ohm's Law, I = g . Since the

voltage is 60, and the total resistance 30 ohms, the current is 2 amperes.

Then, you apply the formula E = IR, to each resistor. In all cases,
the current is 2 amperes. To find the voltage drop across each resistor,

multiply the individual resistance by 2 amperes.

Now, return to Page 1 and select the correct answer.

Page 8

 

YOUR ANSWER: 6 volts applied to R3.

60v.

 

Correct. With 3 amperes in the circuit, E1 = 36 volts and
E2 = 18 volts; there is a 54 volt drop across R1 and R2. The 6
volts remaining are applied to R3, and this is equal to the

voltage drop across the last resistor.

All of the circuits we have discussed in this section have
been series circuits. You should know these things about

series circuits by heart now:

1. The amount oi current in each device in the circuit
is the same.
2. The total resistance of the circuit is the sum of the ‘,

resistances of the individual devices .

3. The sum of the voltage drops across each device in

the circuit is equal to the total applied voltage.

Another type of circuit is the parallel circuit. The three
statements above do not apply to parallel circuits; we must

learn a new set of characteristics for them.

Here are the same three values of resistance shown above,

but now they are connected in parallel.

_—£—_ RI R1 R3
60v...:_ 1241 61‘» 2J1—
T

Look at R3, at the far right. Is it connected directly

 

to the power source?
Yes Page 10

NO Page 15 ‘ ‘

YOUR ANSWER: The voltage drOp across R is 18 volts .

1

You've picked the wrong resistor, apparently. The

resistor you were asked about is R1 .

If you simply chose the wrong resistor, return now
to Page 1+ and calculate the voltage drop for resistor
R1 . It is labelled R1, so there should be no trouble finding
it.

Page 9

YOUR ANSWER: Yes, R38 is connected directly to the power source.
Page 10 e

 

.... R) Q). 23
60" _":_. 11.12. {._a. 2.0.

Correct. Points B,'C and D are all electrically identical
with point A, the positive pole. Likewise, points F, G and H
are electrically identical with point E, the negative pole. The
lower ends of all three resistors are all connected to the
negative side of the power source; the upper ends of the

resistors are all connected to the positive side.

If you had a struggle to visualize the above as a parallel
circuit when you first lookedat it, it may have occurred to

you that there are other ways of showing a parallel circuit.

For instance, the following is a parallel circuit that is
electrically identical to the one above. However, this new

diagram may make it plainer why it is considered a parallel

circuit. J):
éov -—- 1241

Now look at these three diagrams . Which is not
electrically identical to the others?

311 __L
311 31L

 

(A) (B) (C)
A Page 3
B Page 5
C Page 13
They are all the same. Page 18 \\
They are all different. Page 21 %x

1 eat 1/

YOUR ANSWER: 36 volts.

Yes, R1 is at the bottom this time, but it's labelled so you should
have had no trouble finding it. Usually the first resistor in the direction
of current flow is designated "R1”. The negative side of the battery is
toward the bottom of this diagram.

K: Q-(L

 

 

In this circuit there is a total of 20 ohms resistance, so 3 amperes
flow. You can multiply this current times the rating of each resistor (in
ohms) to find the voltage drop across it. Do this for all the resistors in

this circuit, so you can answer the following question.
How much voltage is applied to R3?
6 volts Page 8
54 volts Page 1.2

60 VOltS Page 16

Page 12

YOUR ANSWER: 54 volts.

That is an improbable choice. Here is the diagram again. An

arrow has been inserted to show the direction of current flow.

Ra ZJL

‘___.

 

We know that the current through each resistor is the same as the
circuit current (3 amperes) since the current is the same at all points

in a series circuit. v)

We can use the E = IR form of Ohm's law to figure the voltage drop
across each resistor. Thus the drop across R1 is 3 x 12 = 36 volts,
across R2 it is 3 x 6 =18, and across R3 it is 3 x 2 = 6 volts.

Let's add these together:

E1+E2+E3 = 36+18+6=60volts.

In other words, in a series circuit the sum of the voltage drops is

equal to the total applied voltage.

NOW return to Page 11 and select the correct answer to the

question.

Fag. 13

YOUR ANSWER: Diagram C is not electrically identical to the others.

No. Here are the diagrams again:

2 .n.
M) _ do «6)

The voltage drop across each component connected in parallel is

the total voltage of the source.

When we told you that,we explained that the negative ends of
resistors connected as in (A) above could be considered electrically
identical to each other and to the negative pole of the battery, while the
positive ends of the resistors could be considered as electrically

identical to each other and to the positive pole of the battery.

Let’s put it another way: We join one end of each resistor in a
circuit. If it is possible to trace one end of each resistor back to a
single point on the path from the negative pole of the voltage source and
to trace the other end of each back to a common point on the path from

the positive pole, the resistors are in parallel.

Check that out against the diagrams above and then return to

"Page 10 for another answer.

Page In

YOUR ANSWER: Each lamp in the series string has 110 volts
applied to it, and there is a voltage drop of
11 volts across each lamp in the parallel string.

No.

Consider the first part of your answer: Each lamp in the
series string has 110 volts applied to it. The total voltage
drop in the series circuit must be 110 volts . But won't there
be a voltage drop at the first lamp? If there is, how can the
second lamp have 110 volts applied to it?

As for the second part, it appears you've forgotten the
first thing you learned about parallel circuits, which was:

In a parallel circuit, the same voltage is applied

to each resistor, or in this case lamp.

This is the case since, as we saw, each lamp is connected

‘ directly to an extension of the power source terminals.

Return to Page 25 now, review it and then make another

choice.

i.)

«age 15

YOUR ANSWER: NO, R3

power 5 ourc e .

is not directly connected to the

You‘ re wrong, but at this stage of the game there's no
reason to feel bad about it. Let‘s look at point D of the

 

 

circuit. 8 C D
A A A ‘ 9.
—1: R1 «1 Q3
60V ": I2J2. ‘ 61’- 1J1-
E F 5

You' ll notice that there is no circuit element between
A and D. Since this is so, it must be simply an extension
of the positive pole of the power supply. Similarly, point
H is“ merely an extension of point B, the negative pole of

the power source .

It should not be difficult to conclude the same things

about R1 and R2 .

Return to Pa ge 8 and try again.

Page 16

YOUR ANSWER: 60 volts.

 

We know that the current through each resistor is the same as the
circuit current (3 amperes), since the current is the same at all points

in a series circuit.

a} i

We can use the E = IR form of Ohm's law to figure the voltage drOp
across each resistor. Thus the drop across R1 is 3 x 12 = 36 volts,

across R2 it is 3 x 6 = 18, and across R3 it is 3 x 2 = 6 volts.
Let‘s add these together:

E1+E2+E3 = 36+18+6 = 60volts.

In other words, in a series circuit the sum of the voltage drops is

equal to the total applied voltage.

Now return to Page 11. and select the correct answer to the

question.

Pa [1c 17

YOUR ANSWER: No amperes flow through Rl .
You're about as wrong as you could be here, but it

looks as if you may have been on the right path. Let's look at

the circuit again; but we' ll redraw it slightly so that you can

see more plainly what is happening.

  
 
 

R.

so.-:—

Apparently you remembered what we said about current --

in a parallel circuit, it divides . If you reasoned that since Rl
offered more resistance than R2, and R2 more than R3, then the
current would all go through R3, you were closer than the
answer indicates . It will turn out that R3 will have a larger
share in the total current, but don't rule out the other two

entirely .

Return to Page 19 and choose again.

Pa ge 18

YOUR ANSWER: The circuits are all electrically identical.

Well done. They are all circuits containing two resistors

connected in parallel across a voltage source.

In a circuit containing more than one resistor, if it is
possible to trace one end of each back to a single point on the
path from the voltage source's negative pole and to trace the
other end of each back to a common point on the path from the

positive pole, the resistors are in parallel.

We' ll investigate parallel circuits in more detail later on,

but for the moment it is important that you understand this:

In a series circuit, the voltage is divided between
the several resistors and the voltage drops across
the individual resistors add up to the total applied
voltage.

In a parallel circuit, the total voltage is applied

to each resistor .

And to help fix this in your mind, here's the diagram you

saw earlier: c p

5

VA) {1. r2,
~3— I2

”- 6f¢ 211.

”._..l

v v

F G H
Look at E.RZ , the resistor in the middle. What voltage is
applied across R2 ?

10 volts Page 20
20 volts Pane 23
60 volts P886 25

\fi

Page J. /‘ 1

YOUR ANSWER: Each lamp in the parallel string has 110 volts
applied, and there is a voltage drop of 11 volts
across each lamp in the series string.

Yes . One of the most important characteristics of the
parallel circuit is the fact that each device has the full voltage
applied to it. This is quite different from the series hook-up
where each device receives only a share of the voltage that left

the power source .

Parallel circuits differ from series circuits when it comes to
computing current, too. In a series circuit all of the current flows
through each device in the circuit; each component has the same
number of amperes through it. In a parallel circuit, the current
is divided up, part going through one branch and part going through
another.

Here is the parallel circuit we had before:

 

I Q. Q1 {23
:3.-
T

If a current of 45 amperes is leaving the battery, what

can we say about Rl ?

No amperes flow through Rl . Pafte 17
45 amperes flow through Rl . Page 22
Fewer than 45 amperes flow through R Page 26

1 8
More than 45 amperes flow through Rl . Page 27

Page 20

YOUR ANSWER: 10 volts.

 

No. If you recall, we agreed that each resistor is connected
directly to the power source, further, our last statement included:
. . the total voltage is applied to each resistor.

Return to Fe ge 18 recheck these points and then choose the

right answer.

YOUR ANSWER: The circuits are all electrically different.

Wrong. Let's look at the diagrams again:

 

 

 

“2:“ __L. 1.0. 3.0. _l__ 3.1L
.. fl .1— g.”
< A) _ t3) (0‘

The point we made about the earlier parallel circuits was

 

 

 

 

 

 

that each resistor was electrically identical to its neighbors

at the end connected to the negative pole' of the power source
and that, similarly, each was electrically identical to its
neighbors at the other end, which was connected to the positive

pole of the power source.

In the case of the above circuits, isn't it true that in each
one there are two resistors? Isn't it true that in each circuit,
each resistor is electrically identical at one end with the
negative pole of the battery and at the other end with the positive
pole of the battery? And doesn't this seem to make them fit the

definition of parallel circuits?

These are indeed parallel circuits that are electrically
identical. All that is different here is the way in which they are
drawn.

Return to Pa {’8 10 and select the correct answer.

YOUR ANSWER: 45 amperes flow through Rl .
You' re way off here. How much flows through R2 and
R3

a share.

? We said current divides, that is , each resistor gets

Return to and share the current in a more
democratic fashion.

YOUR ANSWER: 20 volts.

#_.x

I R. 31 Q3
60v — ‘M‘ 6.“- 2J-

_—
._.—.._

 

 

No. If you recall, we agreed that each resistor is
connected directly to the power source, further, out last
statement included: . . . the total voltage is applied to

each resistor .

Return to Page 18 recheck these points and then
choose the right answer.

Page 2h

YOUR ANSWER: Each lamp in either string has 110 volts
applied to it.

It sounds as though you were playing it cagey here,
and thought you'd be at least half right. The trouble is you're

also half wrong.

The same voltage is applied to each resistor in a
parallel circuit. In this case, each lamp in the string has
the full 110 volts applied to it.

In the series string, however, there will be a voltage
drop of 110 volts for the whole circuit. It will be shared
among all the components so that each lamp will have 1/10
of this total across it. Consider the second lamp in the
string. Will it get a full 110 volts?

Return to Page 25 review the material until you' re

sure of yourself and then make another choice.

3...!

YOUR ANSWER: There are 60 volts applied across R

1.]; Kl 1" f2;

éo V -— (UL 6.1L 211.

T

Good. The size of the resistor doesn't affect the voltage

2.

 

v

applied in a parallel circuit. R2 is connected directly to the
power source, and the full EMF of the power source is being applied
to it.

You could remove R1 and R3

electrons would continue. But in a series circuit there would be

from the circuit, and the flow of

no current if a device were removed, the circuit would be broken.

You have probably noticed this difference in strings of
Christmas tree lights . It used to be that almost all of these lights
were connected in series circuits . One burnt—out bulb would
interrupt the whole circuit. The more recent light strings are
connected in parallel circuits in which only a burned—out bulb will
not light.

Let's say you have two strings of Christmas tree lights, one
in parallel, the other in series. Each has 10 bulbs or lamps in it.

Which one of these statements is correct?

Each lamp in the series string has 110

volts applied to it, and there is a

voltage drop of 11 volts across each

lamp in the parallel string. Page 11*

Each lamp in the parallel string has 110

volts applied to it, and there is a page 19
voltage drop of 11 volts across each ‘
lamp in the series string.

Each lamp in either string has 110 Page 2L.
volts applied to it.

Pa ge 46

YOUR ANSWER: Fewer than 45 amperes flow through R1.

R: % KL 2:
6J1 .
'——‘ l2_'JL 1.0.

i
”T“ i

Correct. When the current arrives at the resistors, it
divides, some flowing through each resistor. You should expect
current through each resistor since you know that the full 60
volts is applied to each resistor. Ohm's Law will tell you the
current value. Of course, you would expect the current to prefer
R with just two ohms of resistance to the tougher going offered
by R and R . But some of the current will find its way through

R1 and R2 as well. We will take that up in more detail later.

 

60v

TURN TO PAGE 28 AND CONTINUE

b

YOUR ANSWER: There is more than 45 amperes flowing through R1

No. You're a bit confused here. When we said that the
current is divided in a parallel circuit, we meant that each
resistor would get only a portion of the total current. In a
sense, germs multiply by dividing. You can't say that about

current.

Go back to Page 19 and choose again.

 

CONTINUED FROM PAGE 26

In a simple parallel circuit the total voltage is applied to
each branch of the circuit. Ordinary house wiring is all in
parallel, and you can measure about 117 volts at every outlet.

In the circuit below, two resistors are connected in parallel

across a power supply. There are also three voltmeters in the

circuit. Are all the voltmeters parallel with‘the rest of the circuit?

 

 

iii

Page 29

YOUR ANSWER: Yes, the same current is found at all points in
a parallel circuit.

Electrical current in a circuit behaves in some ways very

like currents of water in a river system.

Would you say that the current, that is, the flow of water
in gallons per minute, was the same in all creeks, branches ,
' runs , and other tributaries? And that each of these was the

same as the current in the main stream?

 

/—"—-/\\\ f-\‘

Neither is it true of currents in parallel. electrical circuits.

Go back now to page 33 and find the better answer.

Page 30

YOUR ANSWER: Yes , I1 is larger than I2 .

Do you really believe that most of the current will go the
hardest way, and only a little of it will take the short—cut
through the easiest path?

Or, to go right back to basics , do you really believe that
the branch with the fewest free electrons will conduct better
than another path with many?

If you were driving from here to there (by yourself), would
you take the longest, narrowest, crookedest streets with the
most traffic on them, or would you prefer a freeway with no

traffic congestion?

Better return to Page 36 and use common sense to find

a better answer.

Page 31

YOUR ANSWER: No, the same current is not found at all points
in a parallel circuit.

Correct. The total current is divided between the various
branches of the parallel circuit. As you would expect, more
current will flow through a low-resistance branch than through a
branch with higher resistance. However, each branch will carry

some current.

In the circuit below we have two resistors in parallel:

 

I I R; R2
{— 6.0— 3.!)—

It' 5 true to say that R2 is in parallel with R1 or vice versa.

We may also say that each is in parallel with the power supply.

Sometimes we say that a component is ” in shunt" with another
component. This means about the same as being in parallel. A
swis an electrical bypath, whereby the current divides and flows
partly through the shunt and partly through the device that is shunted.
We could say, for instance, that a voltmeter is shunted across a

circuit for a voltage measurement.

In this circuit, then, R2 is shunted across R1 . Rl offers twice
as much resistance to current as does R2 . Is the same voltage

applied to each resistor?

Yes Page 36
No Pa ge 1.0

Page 32

YOUR ANSWER: No, 11 will not be larger than 12

“L R\ 2a.

:7 4:9. 3JL
I "
1. 3.3-.)

A
You are right. R1 is the larger resistor, but this means that it's

 

\Zv

 

 

easier for electrons to go through the other resistor. So you are
correct in assuming that more current passes through the lower
resistance, R2. The fact that Rl has twice the resistance doesn't
mean that it gets left out altogether, though. Wherever there is a
closed circuit available for electron flow, an EMF will cause s_oge

current, even if the resistance is very high.

Kirchhoff' 5 second law (for voltage) says that the algebraic sum
of all voltages around a closed circuit is zero. The circuit above
actually contains two closed circuits . That is , there are two different
paths that electrons may take on their way from the negative pole to

the positive pole of the battery.

12v—r

 

 

 

Starting from point P in each closed circuit, each circuit‘can be
analyzed the same way: 12 — 12 = 0. There is just one voltage drop
in each circuit. The current, too, can be figured for each circuit,

using Ohm's Law. What current is flowing in each circuit?

I1 = 1.33 amps 12 = 1.33 amps Page 31.

I,
I1 = 2 amps 12 = 4 amps Page 38
ll = 4 amps I2 = 2 amps Page h2

I1 = 6 amps 12 = 3 amps Page 14.3

 

3 ‘ Page 3'3

YOUR ANSWER: Yes, all the voltmeters are in parallel with

the rest of the circuit.

tilt

Correct. Each should give the same reading, for the total
voltage in a simple parallel circuit is applied across each

 

 

component. You can trace the wiring and see that each voltmeter

. is connected directly to the power supply.

So here‘ 3 one important thing you must remember about
parallel circuits: The voltage across each branch is equal to
the voltage across every other parallel branch and to the voltage

applied.

Is it also true that the same current is found in all parts

of a parallel circuit?

Y8 S Page 29

No Page 31

 

Page» 3h

YOURANSWER: I =l.33a,l =l.33a.

l 2

Does this mean that you believe that if they have a choice,

as many electrons will go the hard way as will take the easy

way?
Or did you figure that Ohm' 5 Law means that I = 16; has

the same answer as I =?-?

Or did you just glance at the diagram so carelessly that

you didn't really know what the question was ?

Better return to Page 32 and take a little more care.

YOUR ANSWER: No, not all the voltmeters are in parallel with
the rest of the circuit.

Sorry, but you didn't look at the diagram carefully enough.
Here's the same diagram repeated, then redrawn in a slightly
different fashion:

 

 

Page 35

 

“ __Lj -_I.
_lngir P3?

Look at the second and see whether all the meters are in

 

 

 

parallel with all the resistors. Then carefully compare the
second drawing with the first and make sure you see how they
are actually electrically identical. Now return to ' rage 2'8

and answer the question correctly.

Page 36

YOUR ANSWER: Yes , the same voltage is applied to each resistor.

Right. This is a characteristic of all parallel circuits. The
same voltage is applied to each resistor, but the current divides

some of it flowing through each resistor in the circuit.

If the total EMF is 12 volts , then this amount is applied to
each resistor. We can say: E,C = E1 = E2 . The voltage drop across
each resistor is equal to the total applied voltage. If you study
the circuit below, you will see that this is the same thing as
saying that the total voltage is applied to each branch.

It

————-snA

‘ ’ l

 

 

,3, 2: tan— 3.0.
__ L
“a

Kirchhoff's first law (for current) reminds us that there must
be as much current flowing away from a point as there is flowing
toward it. The total current, It, divides at point A. At point B,
the current (ll) flowing through Rl meets the current (I2) that has
come from R2 . They add together and form the total current again.
We may say: It =11 + 12.

Is Il larger than 12?

P 0
Yes age 3 '
No Page 32

I don't know Page 37

Page 37

YOUR ANSWER: I don't know whether I1 is larger or smaller
than IZ .

Well, let's figure it out by Ohm's Law, I =

70H?!

13 - 12v R1 = 6 ohms Therefore I1 =%, or 2 amps,

E = 12v R2 = 3 ohms Therefore I2 =-l§Z-, or 4 amps.

The only remaining problem to be solved is whether 2 is larger
or smaller than 41

We could put this another way: most of the electrons will
take the easier way. R2 must have more free electrons to carry
the current than R1; if it had fewer, its resistance would be higher,
not lower than Rl . If R2 has more free electrons , then the way is
easier for current to flow throughit than through Rl . If this is

the easier way, then more current will flow this way.

Now go back to Page 36 and find the right answer.

Page 38 C

YOUR ANSWER: I1 = 2 amperes, 12 = 4 amperes.

E
Yes, YOU can solve for current with the formula I = fi just
as though these were two separate circuits. There is nothing
"tricky" about it.

R1 R3
L L ext, 311.

>

You have seen that the total current in the circuit can be
found by adding together the currents through each branch, It =
I1 + 12. So, in this case, adding 2 and 4 gives the total current
of the parallel circuit, 6 amperes. There are 6 amperes of current
passing through the battery.

Now use Ohm's Law to find the total resistance of the parallel
circuit. What resistance does the battery encounter in moving 6
amperes of current?

2 ohms Page 1+1
1/2 ohm Page M.
3 ohms Page [.6

Page 39

YOUR ANSWER: The resistance would increase.

If part of the way is made easier, it certainly won't make
the total harder, will it?

Use the general rule you just learned: the total must now be
less than 2 ohms (the value of the smallest resistor), but not
less than 2/2 (the value of the smallest resistor divided by the

number of resistors in parallel).

The limits considered in the last example were: not more

than 3 ohms, not less than 3/2 ohms.

Now go back to Page hi. and choose the better answer.

Page 40

YOUR ANSWER: No, the same voltage is not applied to each

resistor.

Let's consider an ordinary composition resistor. This
device is made of graphite mixed with clay, then fired the
same as a brick would be. The result is many highly resistive
paths of graphite of very small cross sectional area, separated

by particles of non-conductive baked clay.

Here is a diagram of the internal electrical arrangement of

such a device, connected to a battery through metal caps on the

 

ends of the resistor:

r—Illllt—

Now if you'll look very closely at the diagram ,you will see
a large number of resistances in parallel all connected in series
with a source of EMF. Is the battery voltage applied to only one
of these resistive paths? To only a few? Or is the same battery
of the same voltage connected in the same way to all of the paths

in parallel?

Sorry; you made a boo-boo on that one. Return to Page 31
and try again!

 

3 ‘ Page [,1

YOUR ANSWER: The total resistance of the circuit is 2 ohms.
Eorrect. Use the formula R = T. In this case, you will
use A , and you may call your result Rt- Dividing 12 (volts) by

6 (amperes), you get 2 ohms for your answer.

 

 

It may be hard to believe that the total resistance of this
. parallel circuit is lower than e_it_hg of the resistors. This is ,
however, a universal rule: The total resistance of parallel
‘ resistors is ALWAYS less than that of the smallest resistor. It
is also NEVER less than the value of the smallest resistor divided
by the number of resistors in parallel. The total resistance here,
then, must be less than 3 ohms, but not less than 3/2 oth.

 

Resistance, you‘remember, depends upon the cross—sectional
area of the resistor (among other things). With two resistors in
parallel there is a larger total cross-sectional area, so there is
less resistance.

What would happen if you used a 2—ohm resistor instead
of 3 ohms for R2? ,
The resistance would increase. Page 39

The resistance would decrease. page [,5

V

Page [,2

YOUR ANSWER: I = 42., I = 2a

1 2

The form of Ohm's law to use here is I = E/R.

In both circuits you have to use the same value for E, the 12 volts

supplied by the battery.

To obtain 11 youdivide the voltage by 6 ohms. You arrive at

I"

I, by dividing the voltage by .3 ohms.

Z
Won’t this make "the value of I2 greater than the value of 11?
And doesn't your answer show just the reverse of this, with I1 greater
than I ?

2

Return to Page 32 and select your answer a little more carefully.

Pa ge [+3

YOUR ANSWER: I1 = 621, I2 = 3a.

The current in a circuit is NOT proportional to the resistance,
it's INVERSELY proportional.

That means that the circuit with the highest resistance will
have the lowest current.

Regardless of equations and arithmetic, plain common sense
should tell you that the 6—ohm resistor will not pass twice as
much current as the 3-ohm.

Please pay more attention to the question on Page 32
as you go over it again.

Page AL.

YOUR ANSWER: 1/2 ohm.

Oops I You slipped .

You divided current by voltage, instead of dividing voltage

by current, as you should have. Maybe you just didn't look at

the conditions of the problem carefully enough. Better take

another look at
E and which I.

Page 38 and make sure which number represents

Page L5

YOUR ANSWER: Putting in a Z—ohm resistor for R2 would decrease
the total resistance still more.

 

‘ _._L_ R1 R2 FORMERLY
12v. {3:— 6.0. 2.0. 3.0.

 

Yes, if all the current were forced through the 2—ohm resistor,
there would just be 2 ohms of resistance in the circuit. But one—
fourth of the total current goes through R1 . The effect is to lower
the total resistance of the circuit below 2 ohms.

We usually speak of this total resistance as the eguivalent
resistance of the circuit.

See if you can figure out the equivalent resistance of this
new circuit with its 6—ohm and 2~ohm resistors in parallel. You
already know that 2 amperes are flowing through the 6-ohm resistor,
R1 . That has not been changed, and the same current is still
flowing. But more current will now flow through (R2 , for it has less
resistance.

Calculate the new value of 12, It’ and Rt (the equivalent
resistance). '

12 = 4 amps, It = 6 amps, Rt = 2 ohms Page A7
12 = 8 amps. It = 10 amps, Rt = 1.2 ohms Page 1+9

12 = 6 amps, It = 8 amps, Rt = 1.5 ohms Page 52

YOUR ANSWER: 3 ohms.

Page [,6

N ow that' 3 too bad; you got an answer that would have been
right if the problem had applied only to the right-hand half of the

circuit .

. But you'll never get a good” grade by finding the right answer

to the wrong question. Better take another look at Page 38
see if you can see all the problem. '

and

it

Page L7

YOUR ANSWER: 12 = 4 amps, It = 6 amps, Rt = 2 ohms.
No.

It looks as if you used the same value, 3 ohms, for this
problem, as was given for the previous one. When we changed
the resistor, though, the problem was also changed.

Return to Page i. 5 and use the correct values to work out

a better answer.

Page [.8

YOUR ANSWER: 1/12 + 1/6.
No.

You've apparently added the reciprocal of the voltage
to the reciprocal of one resistance value.

If you added 1/12 of an orange to 1/6 of a dollar, how many
children would you have? Let's return to page 52 and stick
to one subject at a time. '

Page A9

YOURANSWER: I2 = 8 amps, It = 10 amps, Rt = 1.2 ohms.

Somehow you must have added the current through that 6—
ohm resistor twice. Watch your step a little more carefully
and you'll have a better chance of finding the right answer to
such problems as this on Page [,5

 

 

Page 50

YOUR ANSWER: You should add the fractions %+—12— to get i
Yes , the formula is; 11? = % + l
t 1 R2

This is the reciprocal formula for finding the equivalent
resistance of a parallel circuit.

As you remember, the reciprocal of a whole number is that
number divided into one. The reciprocal of 2 is 1/2, the
reciprocal of 4 is 1/4 , etc.

The reciprocal formula for finding the equivalent resistance
of a parallel circuit is: The reciprocal of the equivalent resistance
is equal to the sum of the reciprocals of all the resistances in

parallel.

 

..L R1 R2
lav. '1‘“:- 6-0— 8.0—

 

When you add the reciprocals of the two resistors in parallel
above, what is the result?

1/8 Page 53

»\:\\§~\\\\\\\w’f. 2/3 Page 55

YOUR ANSWER: 2

You must have worked the problem correctly, but you went

too far.

You were asked the sum of the reciprocals, not the total

resistance.

Try Page .‘53 again and stop when‘you have answered
the question asked.

Page 52

YOUR ANSWER: I2 = 6 amps, It = 8 amps, Rt = l .5 ohms.

Correct. The equivalent resistance is still lower than the
smaller of the two resistors , which had a value of 2 ohms.

There's a fairly simple way to find the equivalent resistance
of two known resistors in parallel -- even when you don't know
voltage or current. We come up With this formula by some
algebraic jiggling of Ohm's Law. Follow these equations carefully.

I :1 E. I = E. I = it.
1 R1 2 R2 t .Rt
And, since It =11 +12, then
E: ... £1. + £2
Rt R1 R2

- - _ _ .3. __3.
Now ina parallel Circuit, Et—El —E . So, R _R +

Divide by E and you get

1 l l l 1
—=-+-,orR --+- .
Rt R1 R2 t Rl R

 

So as you can see, you can find the total resistance by standing

each resistance value on its head, adding the resulting fractions,
1 1

E + 13:2 , and then standing the fractional sum back on its feet.
1

If R1 = 6 ohms, R2 = 2 ohms, and B = 12 volts , what fractions

should be added together to find fi- ?
t
1 l .
1—2 + -6— Page [48
1 1
._ _. P
6 + 2 age 50

Page 53

YOUR ANSWER: 1/8

Please, we don't add the numbers and then find their
reciprocal; we find each reciprocal and then add these reciprocals ,
which gives an entirely different sum.

Return to page 50 and try it the right way.

1 "f 1
.

 

l
Page 5h ‘

YOUR ANSWER: 37/ 63 ohms ,
Oh-Oh, you've slipped:

Your answer gives the sum of the reciprocals, not the total

resistance .

You've got to find the reciprocal of that sum of the
reciprocals to get the right answer to that problem on page 61

YOUR ANSWER: The sum of the reciprocals of the two resistors
18 2/3 0

Right. To add 1/6 and 1/2, you first change 1/2 into 3/6
and then your total will be 4/6 or 2/3 . This fraction is not the
answer to the problem, but it is equal to the reciprocal of”the
equivalent resistance.

We now have; 1%: = :2; . To solve for Rt we just invert our
two fractions: t . v

le

or 1.5

Rt:

Now use the reciprocal formula for Rt in this three-resistor
circuit.

" '2:— la 1,4 6
T _CL. .0. .0.
What is the sum of the three reciprocals?

2 Page
1/2 Page 57

7/1 2 Page 59

Page 56
YOUR‘ANSWER: Rt = 76.9 ohms.

, _Yes, the sum of the reciprocals is 13/1000, and when you
invert and divide 13 into 1000, you get the answer above.

When you have only two resistors in parallel, you can use
a simpler version of the reciprocal formula. The formula as we
have had it so far is:

l.__1_+.1_
Rt R1 R2

To simplify this, first, add the two reciprocals together, using

R1 x R2 for a common denominator:

 

,‘l- := R2 +. R1
R
1: R1 X R2 R1 X R2
' The result is:
..1. = Edit.

And when you invert both sides of the equation, you get this
simplified form:

R R1 XSR2

_.._ = t = ———-—fi
1 R1; + R2

As you can see, all you have to do is divide the sum of the
two resistances into their product.

Use this simplified formula to find the equivalent resistance

of two resistors in parallel, one 6 ohms and the other 4 ohms.
' Which equation below follows the formula?
Rt==lié Page 62
5 .
R = ..5_
1,2

t Page 65

Page 57

YOUR ANSWER: The sum of the reciprocals is 1/2.

 

_—_L_ 12 4 6
6V- E-

_n. n. ~0—

Correct. You use 12 for the common denominator, and you
can form the following fractions from the reciprocals: 1/12, 3/12
and 2/12.’ The total, 6/12, can be reduced'to 1/2. Any number-
of resistors in parallel may be handled this way.

To check your answers you can use this reciprocal of the
general rule for resistors in parallel: The sum of a number of
fractions is always larger than the largest fraction, but never
larger than that largest fraction multiplied by the number of fractions .
The answer above, then, must be more than 1/4 but not more than
3/4.

\ Now complete the solution for Rt in the circuit above, and
1 figure out the total current flow through the 6—volt battery.

3 amps Page 61

12 amps Pa ge 63

Page 581..» '

YOUR ANSWER: Rt is 1 . 7 ohms .

A

_L
E cut 7.Q 311

The sum of the reciprocals is 37/63. rFhis is equal

Correct.
to the reciprocal of Rt’ or, to put it mathematically:

(A)
\1

”at.
(D
O.)

Inverting both sides , we get:

\
Rt _ ._.

(.003
\103

And when 37 is divided into 63, we get 1 .7 ohms approximately.

The reciprocal formula will work for any number of resistors in

parallel. Figure out the equivalent resistance of. the four—resistor

circ uit-:sbelow:

 

 

, zoo 200

l k .500 .0. A
76.9 Ohms Page 56
Page on

130 ohms

Pa ge 59

YOUR ANSWER: 7/12

Handling three reciprocals at once is a little more confusing,
isn't it? You must have figured that 4/12 is the equivalent of
1/4.

Try Page 55 again and be a little more careful with your
fractions .

Pa ge 6C3

YOUR ANSWER: The equivalent resistance is 8. 77 ohms.
Ouch!

It looks as if you picked the answer that seemed to fall Within the
limits of the general rule.

But. . .

The resistors you were considering had values of 13K and 27K. That
is to say, they had values of 13, 000 ohms and 27,000 ohms.

K = 103 = 1000

That letter K makes a difference.

Return to Page 66 and this time figure the problem a little more

carefully, using powers of ten.

3:;

Page 61

YOUR ANSWER: There will be 3 amperes of current in the battery.

Yes. Rt (the equivalent resistance) is 2 ohms. Therefore, the
current in the battery (I = E/R) = 6/2 = 3 amps.

Now let's try one where the arithmetic is less simple.

 

._. 9 7 3
—— _n_ 11. 11-

By multiplying 9 and 7 together we get 63 which will serve
as a common denominator for the three reciprocals . We have:

What is the equivalent resistance (Rt) of the circuit above?
37/63 Ohm Page 51L

1 . 7 Ohms Pa {ca 58

 

-
LL

YOUR ANSWER: 12/5

Right. The correct equation is:

t = 2:: = 2% = ls—zxwhich is equal to 2.4 ohms.

 

R

If you ever get mixed up on which computation comes

first, just remember to

M ultiply
Add

and get

IE
. [3’
t2

then
Divide

Now try it with two larger resistors as in the circuit below. fl

 

J. 15 21

—-~ .0. ._fl_

What is the equivalent resistance of this circuit?

 

8.75 ohms Page 66
9.25 ohms Page 67
ll .5 ohms Page 7g

 

Pa ge 63

YOUR ANSWER: 12 amperes .

It looks as if you didn't set that 1/2 back on its feet again,
to get the total resistance, before you divided.

Remember, what you did to find that value of one-half was
simply to add all the individual reciprocals. That gave you l/R ,
not Rt itself. What is the reciprocal of l/Rt ?

When you think you know, try Page 57 again.

 

Fab; 614'

 

YOURXANSWER: 13 0 ohms .

(No.

r’
-‘ Let's get this thing straight: The total resistance of

resistors in parallel is the reciprocal of the sum of the

reciprocals .

 

You apparently remembered the general rule for resistors
in parallel, but you had to doctor your answer by shifting the

decimal point to make it stick.

The trouble was that you forgot to find that final RECIPROCAL

of the sum.

Now try Page 58 again, and do it right; then you won't
have to juggle a decimal point to be within the limits of the

general rule .

9)

 

Page 65

YOUR ANSWER: 5/1 2
N0.

It looks as if you tried to use the reciprocal formula

instead of this simplified form:

R1 XR2

R—__._.

t R1 + R2
Or did you add for the numerator and multiply to find the
denominator?

Whatever the reason, you had better return to rsC rib
and do it the right way.

:;i“i

“‘L. -

 

. Page 66
YOUR ANSWER: Rt is 8. 75 ohms.

. . 7 . _ 15 x 21
Correct. Put your numbers in this form. Rt — -———15 + 21 .

and the result is the equivalent resistance of these two resistors in

parallel.

This simplified formula will work for any number of resistors if you
work out the equivalent resistance for each pair, then use the same
formula again for pairs of totals, and again for pairs of pair equivalents,
etc. The job becomes rather complicated for more than.two resistors,

though.

With only four resistors in parallel, the complete equation looks

likethis: R xR R xR,
1 2 X 3 4
R1+R2 R3+R4

t
RlxR2 R3XR4

+
R1+R2 R3XR4

 

 

 

And for twice as many resistors the formula becomes four times as
large as that. Obviously, the simplified formula isn’t very practical in

such cases.

In the circuit below we have only two resistances; a 13K resistor

shunted across a 27K resistor.

/3/< 27K

Hillb—

What is the equivalent resistance in ohms?
8. 77 Ohms Page ()0
8780 ohms F9220 <38

877 ohms 15780 7’4

 

 

Page 67

YOUR ANSWER: 9. 25 ohms.

This is sheer guesswork, and not a lucky guess at that!

It's close, but not nearly close enough.

Return to Page 6.2 and actually do some arithmetic on this

prOblem.

Page 68

YOUR ANSWER: Rt is 8780 ohms.

Yes , this is the result rounded off to three significant figures .

There is another short cut available if all the resistors in
parallel are equal. To find the total resistance for any number of
equal resistors c0nnected in parallel, divide the resistance of
one resistor by the number of equal resistors .

i
N
of one resistor, and N is the number of resistors .

This can be expressed: Rt = , where R1 is the resistance

This is simple enough to figure out, if you consider what
happens in a parallel circuit containing N resistors all of equal

value. The basic formula for resistors in parallel gives you:

2P“

1
+ R....(Ntimes)

l l 1

+
pap—4H
:UI

._4

If you have five lO—ohm resistors , you may just divide 10 by

5, to get Rt = 2 ohms.

What would be the equivalent resistance of six 30-ohm

resistors in parallel?

5 ohms Page 1,.

_1_ ohms

Page ’39

YOUR ANSWER: Rt is 10 ohms . i
R. 30.0..

 

 

 

 

 

all!!! ‘
E Zé v

There are several ways to solve this problem. You

could have added the reciprocals, used an assumed voltage (60 volts

is the least common multiple), or you could have figured out the

actual current through each branch and added the currents to get It
so as to use Ohm‘s law.

Correct.

If you don‘t need to know the current through each branch, it' s
easiest to calculate the equivalent resistance first.
Using the reciprocal formula or the assumed voltage method, we

find that Rt is 10 ohms. Ohm‘s law (using I = E/R) then shows that
2.4 amperes will flow with 24 volts applied.

Now calculate the current through each branch. Which list
below is correct?

11 = 0.8 amp I1 = 0.8 amp

I2 = 0.4 amp I2 = 0.4 amp

I3 = 0.12 amp 13 = 1.2 amp
rage 73” fig, EC?)

Page ‘70

YOUR ANSWER: 11 . 5 ohms

You not only forgot the whole MAD idea, but you juggled
the decimal point to make your wrong answer come within the

limits of the general rule.

Maybe you'd better remember that two wrong procedures

don't make a right answer.

Now trY' Fag»: 5.; again.

Page 71 '

YOUR ANSWER: The equivalent resistance of six 30—ohm resistors
in parallel would be 5 ohms .

Right. To find the total resistance of any number of equal
resistors connected in parallel, you just divide the number of

resistors into the value of any one of them.

Many of our problems of finding the equivalent resistance have
included no reference to the power supply in the circuit. Sometimes
it's possible to solve a parallel circuit for resistance by assuming
that a certain voltage is applied. Then you figure out what current
would flow in each branch if such a voltage were applied, add the
currents to get the total current, and then divide the total current
into the assumed voltage to find Rt‘ All of this is much less
complicated than it may seem. Let's try an example.

 

LL R11 R2 0 R3
? {:— ZOJL 1011. 40.0.

T

For this circuit, let's say that 80 volts are applied across the
resistors . It doesn't matter whether this voltage is actually applied
or not. We will still be able to calculate Rt by using this (or any
other) value. If the EMF were 80 volts , what would be the current
through each branch?

I1 = 4 amps, I2 = 0.8 amp, 13 = 2 amps Page 75

I1 = 4 amps, 12 = 8 amps, I3 = 2 amps page 76

r
r .._. ——-l

Page 72

YOUR ANSWER: 17 . 4 ohms .

That's right. You could also have solved this by the reciprocal
formula , or you could have figured up the actual current from the
3 .9—volt battery.

Using the 3.9 volts shown would have made the problem much
more complicated than necessary, of course. The assumed voltage
method is much more convenient in a case like this.

If you assume a voltage of 400, you get 5 amperes through the
80-ohm resistor, 8 amperes through the 50—ohm resistor, and 10
amperes through the 40—ohm resistor. Added together, the total
current is 23 amperes. Divide this into 400 (the assumed voltage)
and you get 17.4 Ohms. NOW, TRY THESE PROBLEMS:

 

Rt, ..
“L l. 130 ohms
_ 10 [to 80 2. 8 ohms
3 0 2 t

I ..

 

2 ..
A 38

30V —— 2 p 68

I. . [ta

 

 

15 a, GO TO PAGE 77 AND CONTINUE

Page 73

YOUR ANSWER: 1/5 ohm

Sony; that's the result of dividing 6 by 30, but 6/30 is
not the correct quotient.

To find the total resistance of any number of equal resistors
connected in parallel, divide the resistance of one resistor by
the number of equal resistors . Rt = R1

N
Now return to page 68 and follow instructions more
carefully.

 

-
and. n

 

Page 74

YOUR ANSWER: 877 ohms°

Well, the digits are nearly right, but the decimal point is in the

wrong place.

Perhaps you worked out the problem almost correctly then fooled
yourself into thinking that, since 8770 seemed to be about right as far
as digits go, the only answer containing those figures, 877, was the
right one.

A little more arithmetic would have shown that the answer is actually
8775, not 877.0.

Try Page 66 again, and watch those decimals.

Page 75

YOUR ANSWER: I1 = 4 amps, I2 = 0.8 amps, I3 = 2 amps.

Sorry, that's not right. It can't possibly be right. You'll
find that R2 has only 1/4 the value of R3. That means R2 must
pass 4 times as much current as R3, and 4 x 2 is certainly not
0.8.

Return to Page 71 and try again. -

Page 76

YOUR ANSWER: 11 = 4 amps, I2 = 8 amps, 13 = 2 amps.

Correct. It doesn't matter whether these currents actually flow
or not, so long as you are trying to find total resistance. NOT

total current.

Adding these branch currents together, It = 14 amperes.
Using R = 91:" , dividing 14 into the " imaginary" 80 volts, gives
5. 71 ohms .
This is the actual value of Rt’ even though you may have used

an incorrect voltage in your calculations .

If you want to check this result using the reciprocal formula,
add your reciprocals and you will get 7/40 . Invert and divide
and you will still get 5. 71 ohms.

This is the assumed voltage method of finding the equivalent
resistance of a parallel circuit. When you use this method, it's
always best to choose a voltage higher than your largest number of

ohms , so that you can avoid current values of less than one.

 

.i. so so 40
:- (L {3 fi—
39 V. -:_—v

I

Pay no attention to the voltage given in this schematic , but
choose some other convenient voltage to figure Rt by the assumed
voltage method. What is Rt?

1 7 . 4 ohms Page 72

17.6 ohms Page 78

Page 77

PARALLE L C IRC U ITS

You should by now have these points about parallel circuits firmly
fixed in your mind:

(1) The applied voltage is the same for all branches
of the circuit.

(2) The total current is divided among the branches.

(3) The lower the resistance in a branch, the greater
the " share" of the current in that branch.

(4) The equivalent resistance of a parallel circuit is
always less than the value of the smallest resistor,
but not less than that value divided by the total
number in parallel.

Before we go further with parallel circuits, look at this diagram.

This is just another way of drawing a parallel circuit, of course.
R, 30 .0.

4| l I I}
E aéi‘v
Three resistors are in parallel with each other and the group is in
series with the power supply. Using any method you want, find the
value of Rt .

 

 

 

10 ohms Page 69
12 ohms Page 82

,q,

Page 78

YOUR ANSWER: l 7 . 6 ohms .
No, your arithmetic slipped somewhere, or you were guessing.

Perhaps you made the problem harder than necessary by your
choice of an assumed voltage.

The Least Common Multiple of the resistances is a good
number to use here; then you don't have any fractional amperes
to deal with.

Try Page 76 again, using that multiple.

Page 77

YOUR ANSWER: I - 0.8A, 12 = 0.4A, I

= I 2 a
l O l A

3

You were doing well until your arithmetic broke down. Check
your figures again and then return to Page 69 and get the correct

answer.

17’ 1
<

\‘ Page 80

 

R\ R2 J1. R3 R4
560v 20.0. IoJL

6012 3011
‘ _ T

YOUR ANSWER: Rt = 5 ohms.

Yes, the sum of the reciprocals is 12/60, and when you invert
this fraction, it becomes 5.

We'll say that the power supply is rated at 60 volts. Now you
can quickly figure the current through the power supply and
through the individual resistors .

Here are the same four resistors and the same power source,

but now they are connected in series:

R. 60.0. R2 3OJl

60V

 

Is the same amount of power expended in-each circuit?

Yes Page 88
No Page 91

YOUR ANSWER: P =19.2 watts PR .6 watts, P = 2.8.8 watts.

R1 2 A 33
_ Correct. Let's review the solution for R1,. 30 ohms. You could

have used P =31 (24 x 0.8), .9 -—~ :27»: {0.8 x. 0.8 x 30), or

2
4 24
E (2 X )‘I

P = R 30 All of these give you the same value for
PR1: 19.2 watts.

Adding together the power demands of all three resistors , we get
19.2 + 9.6 + 28.8 = 57.5 watts. Make a note of that figure.

MAN/Vat“
IWVVV-r

.._-“; . 'g .. ..-.

‘ - ’. 4.4-3!
You have seen that the iota? (Tiff-CC '2" f .r.. m The powe: su pply is 2.4
amperes. Using P= EI wire. is: the now-o“ drawn from ..rie power

supply?

Page 81'

 

Page 82

YOUR ANSWER: 12 ohms .

Wrong. Either you' re guessing or your arithmetic is simply
R, 30.0..

1.1111}
E 2.4V
You have three resistors in parallel with each other and with the

shocking .

 

 

 

power supply. You are asked to find the value of Rt ~- the total
resistance —— and you may use any method you want.

Perhaps the easiest approach is to use the reciprocal formula

R = l
t
l l 1
R1 R2 R3
But you can use the assumed voltage method, or, using the
values given, figure the current in each branch and then use Ohm' 3

law to get your answer.

That should be enough to put you back on the right track so that

you can return to Page 77 and select the correct answer.

fi1‘_l

, ,
I

l
Page 83 i

YOURANSWER: I1 = 0.8 amp, 12 = 0.4 amp, I3 = 1.2 amp.

Yes , the current through the three branches must add up to 2.4
amperes , the total current that should flow through the power supply.
You have already calculated the equivalent resistance of the circuit
at 10 ohms. R1 30“.

 

You may need to know the amount of power used up in each

' resistor. Use any form of the power formula (since you now know

the values of volts, ohms , and amperes for each resistor) and find

the amount of power expended in each resistor.

/ a 3

PR1 = 19.2 watts PR1 = 19. 4 watts PR1 = 19. 6 watts
PR = 9. 6 watts PR = 9. 4 watts PR = 9.2 watts
2 2 2
P = 28.8 watts P = 28.8 watts P = 28.8 watts
R R R
3 3 3
Page 81 €9.15 85 Page 9'0

 

 

Page 81.. 1

YOUR ANSWER: 56. 6 watts .

Oh, come, now; 2.4 (amperes) times 24 (volts) certainly does
not equal 56. 6 (watts).

This answer makes it look as if you're not bothering with calcula—'
tions. You should actually do this arithmetic. Then, unless it's
horribly bad, you'll get better answers than the one above.

Even more to the point, you'll get closer to understanding what
you're doing, and that is vital at this stage. If the foundations of
your learning are wobbly, you' re not going to be able to build on .1

them .

Return to Page 81 and use a piece of paper and a pencil this
time.

Page 85

YOURANSWER: P =19.4watts,P =9.4watts,P =28.8watts.
R1 R2 R3

Don't look now, but your arithmetic is slipping. Either that, or
you're just guessing. There shouldn't be any need to guess.

Surely
by this time you know the equations?

P = BI
P =12R
P = Ez/R

Return to Page» r33 and do a little accurate figuring.

Page 86

YOUR ANSWER: 3. 75 ohms .

You slipped a little in converting one of those reciprocals to
a common denominator. You know the principles: you've done it
right before; you can do it right again.

So do it!

Return to Page 89 and try.

a P5 ge 87

YOUR ANSWER: P = 60 W, P

=3ow, P =zow, P =1ow.
R1 R2 R3 R4

And your answer is wrong.

In a parallel circuit, the voltage is the same across each branch.
The smallest resistor will have the most current flowing through it.

And multiplying that current by the voltage will give the highest
power for the smallest resistor.

But you have the highest power listed with the biggest resistor.

) ‘ Return to Page 91. and check your calculations. You'should
' use the formula P = EZ/R.

 

Page 88

YOUR ANSWER: Yes , the same amount of power is expended in each
circuit.
That's just a guess , and not a very good one.

You don't even need to do the arithmetic on this one.
For instance: B 2

Bt’ and hence Btz, is the same in both circuits.

But in the series circuit, Rt is larger than the largest resistance ,
whereas in the parallel circuit it is smaller than the smallest.

So how can the two different answers , whatever they are , be the

same?

Think about it -- you' re going to have to think some more about

it in a minute , anyway -- and then return to Page #80- and try again.

 

0i

Page 89

YOUR ANSWER: The power supply produces 57. 6 watts.

Correct. This brings us to another characteristic of parallel
circuits: The total power used up in the circuit is equal to the sum
of the quantities of power expended in the branches .

A similar statement was made of series circuits, you may recall.
In series and parallel circuits alike, the total power expended is

equal to the sum of the amounts of power used in each component.

Below are four resistors in a circuit. This is another simple
parallel circuit —- each resistor has one end connected to one side
of the power source and the other end connected to the other side of

the power source. You may find parallel circuits shown in many

different ways, but the important thing is to see how the resistances

are connected to the power source.

A

R\ R?- 1 R3 R4

60!). 30.12

T - 20.“. )OJL
What is the equivalent resistance of this circuit?

5 ohms - Page 80

3.275 Chins Page 86

 

Page 90

= 9.2 watts, P = 28.8 watts.

YOUR ANSWER: P-R = 19. 6 watts , PR R3

1 2

A wild guess is no worse than this tame one. If it isn't just a
guess , you need to review your arithmetic, or your knowledge of
formulas, or your application of them. Here are the formulas again:

P=EI
P=IZR
P = EZ/R

Now, return to Page 83‘ and make a better try.

Page 91

YOUR ANSWER: No, the same amount of power is not expended in
each.

Right. The parallel circuit offers much less resistance, so there
is much more current than there is in the series circuit. Therefore,

there will be much more power expended in the parallel circuit.

R\ 60.0..

 

R1 R2 .1. R3 R4
39L T...“ 2011 ML

—._—

60V
The total amount of power expended in either circuit can be

found by adding together the amounts of power used in the various

5’
8

 

components. This is true for series and parallel circuits alike, but
it doesn't mean that the same total power is used whether the com-
ponents are in series or in parallel.

What amounts of power are used in each of the resistors connected
in series?

PR = 15 watts PR = 60 watts
l 1

PR = 7.5 watts PR = 30 watts
2 2

PR = 5 watts PR‘ = 20 watts
3 3

PR4 = Z . 5 watts PR4 = 10 watts

Page 91. Page 98

Page 92

YOUR ANSWER: The amounts of power expended by the four resistors
in the parallel circuit are (in numbered order)
60 watts , 120 watts, 180 watts , and 360 watts.

60.0— 30.17—

 

 

 

RI R; :1; R3 “9
:— 10
60 30 -— 3"
_rL _n.. Jk
R3 20.0. '0‘ <
60 v.
Right. 2
The best form of the power formula for this problem is P = F ,

since the voltage is the same across all branches of a parallel
circuit. Even though a lot more power is expended in the parallel
circuit, it is still true that the total power used by the circuit is
equal to the sum of the amounts of power used in each component in
the circuit.

Rt for this parallel circuit is 5 ohms. With an EMF of 60 volts
thereeshould be 12 amperes of current. Using P = E1, the total power
expended in the circuit is 60 X 12 = 720 _watts, and this is equal to
the sum of the values given in the answer.

Why is more power expended in resistors in parallel than in equal

resistors in series?

Because there is a voltage drop across each
resistor in the series circuit. Page 95

Because the combined resistance of the
parallel circuit is much lower. Page 102

Page 93

YOUR ANSWER: The largest resistance in a series circuit dissipates
the most power, but the largest resistance in a

parallel circuit dissipates the least power.

Correct. We can confirm this by using the formulas for power.

In the series circuit, the current is the same in all parts of the
circuit. In the formula P = 12R, since the value of I is constant
throughout the circuit, the value of 12 is constant, too. And if I2
doesn't change in value, then the larger the resistor, the greater
the value of P -- the more power used, if you like —— since 12 is

multiplied by a larger number.

In the parallel circuit, however, it is the voltage that doesn't
change from one part of the circuit to another. Voltage is the same
across each branch of the circuit. Using the formula P = Ez/R, E2
is the same for each branch of the circuit. And if E2 doesn't. change
in value, then the larger the resistor, the smaller the value of P —-
the smaller the amount of power used.

Let's say you want to find the power dissipated by each of the
three resistors below.

 

_ M
"Q—
R; ISI).
“N E R3 80.0.
R. ?

 

You'll need to know the value of R1. As a first step, find the
current through the 80—ohm resistor.

The current through the 80—ohm resistor is:
0. 2 amp Page 96f .,_,_

0.5 amp 5 54:» 3.00
0.8 amp , 103

Page 9L

YOUR ANSWER: The amounts of power expended by the four resistors
in the series circuit are (in numbered order) 15 watts,
7.5 watts, 5 watts, and 2.5 watts.
Correct. And the total power used in this circuit is 30 watts. Rt
for the series circuit, found by'adding the resistors together, is
120 ohms.- With 60 volts you know there 13 60/120 = 0.5 ampere of
current. You can then solve for each resistor using P = I R.

Note again: The total power used in the series circuit is equal to

the sum of the amounts of power expended by each component.

Rs 92

 

R2‘ 30313— ‘ I R4

1.." ’0 ‘0-
' R3 20.0.
65v

Now calculate the amount of power used in each branch of the

#2830

 

parallel circuit above. You will need a different form of the power
formula, sincelthe current will be different in each branch.

Which list below is correct?

PR1=60w\ PR1=60w

PR2 = 30 w . ‘ PR2 = 120 w _
Page 87 Page 92

P =20W ' ’ P =180w '

R3 R3

P =10w P = 360w

.Page 95

YOUR ANSWER: More power is expended in resistors in parallel

because there is a voltage drop across each resis—
tor in a series circuit.

There certainly is a voltage drOp across each resistor in a series
circuit.

But it is also true that there is a voltage drop across each

resistor in a parallel circuit, even though each of those voltage drops
is the same. . ‘

So the reason you have chosen isn't really a reason. It's a mis—
leading statement.

Now return to Pa ge 92 and find an accurate statement.

...—.._. ......

Page 96

YOUR ANSWER: 0.2 ampere flows through the 80—ohm resistor.

 

 

IA
R. :51
1E R3
\ .

 

Yes. I = E/R == l6/80 = 0.2 amps. This leaves 0.8 ampere to
flow through the other branch. Now you can use Kirchhoff‘s law
to find the value of R1. You know that there is a lG—voltdrop
across the inner branch with the two resistors , and that 0. 8
ampere is flowing through this branch of the circuit.

What is the value of R1?

20 ohms P8128 105
5 ohms l Page 108
0.2 ohm Page 112

\

Page 97

YOUR ANSWER: The largest resistor in a series circuit dissipates
the least power, but the largest resistor in a
parallel circuit dissipates the most power.

No. Let's go over it again.

'- The largest resistor in a parallel circuit will carry the
smallest current; the voltage is the same across each
resistor of the circuit.

- The largest resistor in a series circuit has the largest
voltage drop across it; but all the resistors in a series
circuit carry the same current.

The largest resistor in a parallel circuit carries the smallest

. current; the largest resistor in a series circuit carries the Same

current as all the other resistors.

A small number multiplied by a fixed number isn't larger than a

larger number multiplied by the fixed number, is it?

Yet that is what your answer seems to say. The fixed number is,
of course, the value of the resistance; the number that varies is the

value of current through the resistor.

Go over the points again and then return to Page 102 and

reason out the accurate answer.

age 98

YOUR ANSWER: P = 60 W, P =10 W.

= 30 W, P = 20 W, P
R1 R2 R3 R4

In a series circuit, the power used up by each resistor is propor—
tional to the resistance. The ratio of the resistances is l : 2 : 3 : 6.

So:
R3 must burn up twice as much power as R4
R2 must burn up three times as much as R4
and R1 must burn up twice as much as R2.
_ __E___ _ 60 =
It -— Rt —--120 0.5 amps.
So PR = IZR=O.5xO.5x10 = 2.5 watts.
4
And the rest are in proportion, as shown above.
Figure them out and then return to Page 91 with the correct
answer.

Page 99

YOUR ANSWER: There is a lZ—volt drop across the lS-ohrn re—
sistor.

Yes. E =IR; in this case, E = 0.8 x 15 or 12 volts.

l A
4—
{21 15.0.,
16v ——Z Rs 201;
I R. 51L
Now calculate the power used by each resistor in the circuit.

Which resistor dissipates the most power?

The S—ohm resistor Page 101*
The lS—ohm resistor Pa ge 110
The 80—ohm resistor Page 113

 

Page 100

YOUR ANSWER: The current through the 80-ohm resistor is
0. 5 amp.

No. Here's the way your calculation should have gone:

In a parallel circuit the voltage is the same across each branch.
In this case, since there are no complications, the voltage across
each branch is the voltage across the power source, 16 volts. The
resistance, R3, is given as 80 ohms.

13 = E/R

16/80 = ?

That's simple enough. Maybe your approach was more com-
'plicated? 0

Complete the calculation and then return to- _ Page 93 to

select the correct answer.

{5)

Page 101

YOUR ANSWER: The largest resistor in a circuit always dissipates
the most power.

Not so.

In a series circuit, all resistors carry the same current. So

2R will be largest for the largest R.

I
So far, your reasoning is okay.

In a parallel circuit the largest resistor carries the smallest
current. So 12R is smallest for the largest R.

Your answer just doesn't stand up. Return to Page 102 and
reason a little more carefully to find the right answer.

Page 102

YOUR ANSWER: More power is expended in resistors in parallel
than in the same resistors in series, because the
combined resistance of the parallel circuit is much
lower than that of the series circuit.

Correct. The lower resistance allows more current to flow in the
parallel circuit, Assuming the voltage is the same in each case, and
since P = E1, this will mean that more power is expended in the

parallel circuit .

 

 

R} GOJL _ I -
._.. Rim” Pig :2. _j; :23 R.
0 .22. ’2. . ._..
6 V — 031 p. 5011 305:. —--..___ 2041 \oJL G?
' P 10.0.
Temp“? < 1
60V

Now that you have calculated the amount of power used by each
of the resistors above , what could you say about the power expended

in R1, the largest resistor in each circuit?

The largest resistance in a series circuit dissipates
the most power; in a parallel circuit, the largest Page 93
resistance dissipates the least power.

The largest resistance in a series circuit dissi— '
pates the least power, but the largest resistance Page 97
in a parallel circuit dissipates the most power.

The largest resistance in a circuit always ' Page 101
dissipates the most power.

Page 103 I

YOUR ANSWER: The current through the 80-ohm resistor is
0. 8 amp.

No. Here's the way your calculation should have gone:

In a parallel circuit the voltage is the same across each branch.
In this case, since there are no complications, the voltage across
each branch is the voltage across the power source, 16 volts. The
resistance, R3, is given as 80 ohms.

I3 = E/R =16/80 = ?
That's simple enough. Maybe your approach was more com-
plicated?

Complete the calculation and then return to Page 93 to
select the correct answer.

 

Page 10A

YOUR ANSWER: The 5—ohm resistor dissipates the most power. Q

Did you guess that this must be right, since the smallest resistor

in a parallel circuit dissipates the most power?

You remembered the general rule, all right , but you overlooked

something. There's another resistor in series with that 5-ohm resistor.

t is possible to get the right answer to this problem without doing the
arithmetic, but not by missing the fact that this branch of the parallel

circuit is itself a series circuit.

Please go to Page -93 and read the reasoning there with a more

v ' ~ . 2 ’
enligntenea eye. 6 I

Page 105

YOUR ANSWER: 20 ohms .

No. You haven't gone far enough; you overlooked R2 , which
is in series with R1.

There was a hint that you would need Kirchhoff's law to help
find the value of R1. Kirchhoff 3 law for voltage says that the
algebraic sum of BMFs and voltage drops around a closed circuit

is zero.

Now let's round up the things known about the circuit. Here's
the diagram again: ‘ A

    

Rs 90.0..

In the center branch of the circuit, we know that 13t = 16 volts .

. We know that I in this branch is 0.8 amps. Using Ohm's law,

E = IR, we can figure that E2 (the voltage across R2) is 0.8 x 15

volts and that E1 is 0. 8 x Rl volts.

And Kirchhoff's law as quoted above comes to this:
Bt - E1 — E2 = 0.

If you substitute the values of E, you should now be able to
simplify your equation to get the value of R1. '
When you have it, return to Page 96 and select the correct

answer‘

Fa as 106

YOUR ANSWER: Circuit A will use the most power.

Wrong.
This is not really a problem in arithmetic, it's a test of recog—
nition.

Circuit A is a parallel circuit, drawn in conventional form. It
is quite true that a parallel circuit uses more power than a series
circuit across the same voltage, and both of these are certainly

across the same voltage.

But is circuit B a series circuit? ' .|

Return to Page 110 and look more closely this time before

selecting another answer.

3

. :3 Page 107

YOUR ANSWER: Half of the total power is dissipated in the
18, OOO—ohm resistor.
A A

1% y 54- K

qov g
T "q

v ‘

Yes , you can see this in the reciprocals. You get 3/54 as the
equivalent of the reciprocal of 18. The sum of the reciprocals is
6/54, and you can see that the 18K resistor accounts for half of
this sum.

In a parallel circuit, more current goes through a small resis-
tor than through a large resistor, as you've seen. The amount of
current, which is determined with Ohm' 5 law, varies inversely
with the size of the resistor. That is , a 5—ohm resistor carries
twice as much current as a lO—ohm resistor in parallel.

This relationship shows up in the reciprocal. You might like
to think of it this way: A smaller number has a bigger reciprocal.
For instance , .the reciprocal of 10 is 1/10. The reciprocal of 2 is
1/2 (or 0.5).

If you remember that the current and the amount of power
expended are directly proportional to the size of the reciprocal,
you may be able to save yourself some arithmetic in analyzing
parallel circuits.

You know that 0.45 watt is used in the 18K resistor above.
What power is expended in the 54,000-ohm resistor?

0.15 watt Page 116

l. 35 watt Page 123

 

”T
l
I
l
t

 

Page 108 ’

YOUR ANSWER: R1 is 5 ohms.
I A

R, 15.0.
\tvé F23 80.0.
T R. 5.0.

v

 

Correct. Using the formula E = IR, El can be written as 0.8 x R1
or 0. 8 R1. And using Kirchhoff' 5 law that the algebraic sum of the
EMFS and voltage drops around a closed circuit is zero, we can say:

Et - El - E2 = 0
16 -O.8Rl-(O.8X15) = 0.

That will simplify to:

0.8 R1 = 4
_ 4
R1 ‘ 0.8
R1 = Sohms

Now you know that the inner branch has a S—ohm and a 15—ohm
resistor. This is the same as a ZO—ohm resistor in shunt with the
80—ohm resistor.

The'two resistors of the inner branch are actually in series with
each other, so it is proper to say that there is a partial voltage drop

across each one.

What is the voltage drop across the 15~ohm resistor?
12 volts , Page 99“
15 volts Page 111

, ' Page 109"
:J 1

YOUR ANSWER: O. 9 watt is used in this circuit.

Correct. Adding the reciprocals of these resistors, you have
1/ Rt = 6/54, and this gives you 914 for Rt' This will allow 0.01
ampere or 10 milliamperes to flow through the power supply, and
the circuit will use 0.9 watt of power.

 

54K

 

How much of the total power is dissipated in the 18, OOO—Ohm

resistor?
Half, of the power. Page 107
More than half of the power. Pn ge 115
Less than half of the power. Page 121 .4

an,

Page 7

YOUR ANSWER: The lS—ohm resistor dissipates the most power.

 

[A—sa—fi
J R; R3
__—— [5.0. 30;).
/6V ‘2':
M
5.0.

Correct .

 

You have learned that the smallest resistor in a simple parallel
circuit uses the most power. And, as you may have seen for your—
self already, this problem does not really contradict the general
rule.

You must think of the 5—ohm and the lS—ohm resistor as joined to
make a ZO—ohm resistor, since they are in series in this branch of
the circuit. Now the .20—ohm branch dissipates the most power:
12.8 watts against the 3.2 watts used by the 80—ohm resistor.

R2 is 15. or g of the total resistance in the series branch.
Therefore, 3/4 of the power in this branch is dissipated by R2.
3/4 of 12.8 is greater than 3.2 watts.

Which circuit below will dissipate the most power?

 

 

 

 

 

 

E8 K

5 4K

«VV‘ 9WM_L

27 K ‘2:— [8 K 4V.

._-~__ 7x
S‘IK

Add
W1 9m “

(A) ' (a)
Circuit A will use the most power. ' Page 106
Circuit B will use the most power. Page 11h
Both will use the same amount of power.

. Page 117
I don t know.
Page 122

Page 1111

YOUR ANSWER: 15 volts.
Now this is sheer guesswork, and poor guesses don't pay off.
You'll get along much faster if you take time to use the figures

you are given . 1

Now return to Page 108 and stop trying to cut corners.

 

Eflh \

Page 112

YOUR ANSWER: 0.2 ohm.

No. Decimal points cause trouble when they get in the wrong

place, as in your answer.

But more important still, you apparently forgot that there was
more than one resistor in this branch of the circuit.

There was a hint that you would need Kirchhoff's law to help
find the value of R1. Kirchhoff‘s law for voltage says that the
algebraic sum of EMPs and voltage drops around a closed circuit

is zero.
Now let‘ s round up the things known about the circuit. Here‘ 5

 
 
  

7U
- '0

the diagram again: i A
i «1“ s.
,3 g l
e: «1.1541 a a.
WV .2. .3 ‘1 80.0.
T 'n-
i E
i E

In the center branch of the circuit, we know that Et = 16 volts.
We know that I in this branch is O. 8 amps. Using Ohm‘ s law,
13 = IR, we can figure that 132 (the voltage across R2) is 0.8 x 15
volts and that E is 0. 8 x Rl volts.

1
And Kirchhoff‘s law as quoted above comes to this:

Et - El - 32 = 0.
If you substitute the values of E, you should now be able to
simplify your equation to get the value of R1.
When you have it, return to Page 96 and select the correct

answer.

‘

Page 113

‘1)
YOUR ANSWER: The 80-ohm resistor dissipates the most power.

No.

You seem to have forgotten the general rule that in a parallel circuit,
the SMALLEST resistance dissipates the most power. That should have
given you a hint that your answer was wrong. 'Please notice that we said
the smallest resistance, not the smallest resistor.

. Begin again with the reasoning on Page 93

5 ”*4

 

 

Page 111. ‘X

YOUR ANSWER: Circuit B will use the most power.

How can you arrive at that conclusion?

Circuit A is obviously a parallel circuit. The values of the
individual resistors are the same in both circuits, the voltage is

the same in both.

If circuit B were a series circuit (which it isn' t), then

 

circuit A would use the most power.

Return to Page 3.10 and study circuit B carefully before

picking another answer.

\,

‘\

Page 115 ,

YOUR ANSWER: More than half the power is dissipated in the

18 , 000 ohm resistor.

You remembered, correctly, that in a parallel circuit, the
least resistance dissipates the most power. But that doesn't
necessarily mean that the smallest resistor will dissipate more

than all the other resistors put together.

Suppose you had one Z—ohm resistor in parallel with eight
8—ohm resistors. The equivalent resistance of the eight in paral—
lel would be one ohm, wouldn't it? And one ohm of resistance
would dissipate more power than the two ohms . The smallest
single resistor certainly wouldn‘t dissipate more than the combina—

tion of all the others in that case.

So return to Page 109 and try again.

 

 

.1 AA W “"' \-

l

Page 116

YOUR ANSWER: Knowing that 0.45 watt is used in the 18, 000-

ohm resistor ,

you can see that 0.15 watt is

used in the 54, COO—ohm resistor.

Correct. Because 54 is three times as large as 18, the recip-

rocal of 18 is three times as large as the reciprocal of 54.

Therefore, the 18K resistor carries three times as much current

and expends three times as much power as the 54K resistor.

If you thought the diagonal resistor set—up looked odd, take a

look at this:

 

 

 

 

--——~———-O C}
0v

 

 

6

What is the current shown on the ammeter?

O. 004 ampere
0 . 002 ampere

0. 006 ampere

Page 12h

Page 125
Page 126

\L

Page 11?

YOUR ANSWER: Both circuits will use the same amount of power.

Correct. Both are simple parallel circuits with identical com—
ponents . , 3 K

 

 

 

 

A. W a.

Circuit B may look a little peculiar, but it's still just a

parallel circuit. If the 27K-ohm resistor were connected from points
‘ P to Q, it would have the same effect it has now, and it would.
look more like the circuits you have seen.

Now figure out how much power is used by either of these
circuits.

0. 9 watt Page 109

900 watts Page 120

Page 118

YOUR ANSWER: 0.12 watt is the total power.

No. You have figured the amount of power dissipated by

one resistor.

You can't find a total of any kind by ignoring some of the
things that make up that total.

Return to Page .125 and consider those other things too.

 

 

 

 

 

 

‘47 Page 119
' YOUR ANSWER: 0. 36 watt is dissipated in the circuit.
so K ' 30K 30K
sovo_
Correct. Rt is found by dividing 30,000 by 3, since they
are equal resistors, and, using I =%—, It = 10% = 0.006
ampere.
P 1. 5mw
R h.
l a
1;}. ‘2’ Z) Fajk, faé? L 2 10w
2%, 40032. 1119ng ' '
3. lOmw
A
Le .OOSmW
, Mme .
“ ‘5?

In the Circuit above, if R3 is removed, It will
1. increase
2. decrease

3. remain the same

GO TO PAGE 131

 

Pa ge 120

YOUR ANSWER: 900 watts .

Watch those decimals! Your answer might have been right if
you had kept in mind that those resistors are NOT 18, 27, and
54 ohms .

Please look carefully at the diagram when you return to
Page 117 and see for yourself.

9

I“ ., Page 121

YOUR ANSWER: Less than half the power is dissipated in the
18 , OOO-ohm resistor.

You should know that can't be so, even Without using arith-
metic.

This is a parallel circuit. Stop a moment and remember the
general rule for currents in the branches of a parallel circuit -—

the smallest resistance draws the most current.

That should be enough to make you suspicious about this answer.

Now return to Page 109 and try arithmetic -y- it really works!

Page 122 ‘

YOUR ANSWER: I don't know. 1

At least that' s honest.

The trouble, probably, is that you didn't recognize circuit B.
The circuit certainly isn‘t drawn in conventional form. But you' 11
find many circuits in electronics which are not. So let's fix
this firmly in mind:

Whether a circuit is "this" or "that" depends not on how the

circuit is drawn, but on how it is connected.

Electrons don't flow through lines on pieces of paper, but

within conductors in electronic equipment. ~

Circuit diagrams rarely show the actual or even the relative
locations of the parts; they merely show how the parts are
connected together. Usually they are drawn with two ideas in 9'
mind: to show the circuit most clearly, and/or to draw the diagram
in the least time and the least space.

So whenever you study any circuit diagram, look to see what

kind of circuit it is, before you do anything else.

Consider this one, for instance:

WW WM

That's a parallel circuit, even though it' s drawn more nearly

 

 

 

 

like a conventional diagram of a series circuit. Study it for a bit,
and see why. . I

Then return to Page 110 and try again.

7W

Page 123

YOUR ANSWER: 1.35 watt is used in the 54,000-ohm resistor.
Where were you when that last image was on the screen?

The amount of power expended is DIRECTLY PROPORTIONAL
TO THE SIZE OF THE RECIPROCAL OF THE RESISTANCE.

The bigger the resistor in a parallel circuit, the smaller the

power consumed .

The largest resistor does NOT dissipate the most power in
a parallel circuit. To repeat, the larger the resistor; the smaller

the power cons umed .

If you are comparing two resistors in parallel and one is
three times bigger than the other, the bigger one uses not three
times the power but the reciprocal, one—third the power used in

the smaller resistor.

Remembering that, return to Page 107 and figure out a more

accurate answer.

 

Page 121. .

YOUR ANSWER: 0. 004 ampere will be shown by the ammeter.

When a circuit is drawn in what seems to be unconventional
form, or whenever you don't recognize it, try redrawing it. This
tip will help you not only in this course , but in your actual

work on any kind of electronic device .

This drawing, for instance, does NOT show two resistors
in series:

   

.._..
._..—._.
_—
v

If you redraw the circuit of image 3818 in more conventional
form, you' ll see at once that the meter does not measure the

current through two of the resistors in parallel.

Now return to‘ page 116 and try redrawing that circuit.

Page 125

YOUR ANSWER: The ammeter should show 0.002 ampere or
2 milliamperes. A R‘ 8

    

 

 

 

 

 

.. +
-———O 50v O——————-
A. E3.

Correct. No matter what your first reaction, this is just a
simple parallel circuit. At right is the same circuit drawn in a
more usual fashion. In figure A we have indicated the polarity
of each end of the resistors and lettered them as well. In
figure B are the same resistors , and you can see that they ”are

really connected in the same way.

Note that the positive ends of the resistors are all connected
together, and that they are all connected directly to the positive
side of the power source. Similarly, the negative ends of the
resistors are all joined directly to the negative side of the power

source .

The ammeter is in series with one resistor only, and does not
indicate total current, but only the current through that one

resistor.

Now calculate the total current and from that determine the

total power dissipated in the circuit.

0.12 watt 311,120 171.8
0.36 watt {‘19
1.2 watt .

3.6 watts .3,

if,"

Fage 126

YOUR ANSWER: 0. 006 ampere will be shown on the ammeter.

When a circuit is drawn in what seems to be unconventional
form, or whenever you don‘t recognize it, try redrawing it. This
tip will help you not only in this course , but in your actual work
on any kind of electronic circuit.

This drawing, for instance, does NOT indicate two resistors
in parallel:

If you redraw the circuit of image 3818 in more conventional
form, you' ll see at once that the meter does not measure the

current through all the resistors in parallel.

Now return to Pa. we lie and try redrawing it.

Page 127

YOUR ANSWER: l. 2 watts is the total power dissipated in the
circuit.

That's a double error. You've got the decimal point in the

wrong place , and you've also neglected two—thirds of the circuit.
That' s no way to find a total.

Return to Page 125 and consider all parts of the circuit,
then watch that decimal point to see that it doesn't wander.

 

7‘.“ - k

Page 128

YOUR ANSWER: 3.6 watts is the total power dissipated in the
circuit.

And your answer would be right if only the decimal point didn't ‘

matter.
But the decimal point does matter.

Check your calculations and then return to Page 12:5 and

don't let that decimal point get away from’you again.

Page 129 }

YOUR ANSWER: When one branch of the circuit is removed the

generator current decreases .

Yes , the equivalent resistance of the circuit increases when you ‘

remove one branch . ‘

 

551, 11.0. .24 2.4
.20., @ _. .

 

The sum of the reciprocals is 8/24, and you can quickly see
that Rt = 3 ohms. .
Or you can find Rt even faster by combining like resistors: The
two 24—ohm resistors at right are equivalent to 12 ohms; combine
' this with the other lZ—ohm resistor and you get 6 ohms , andcombine
(that with the 6—ohm resistor and you get 3 ohms.
I Remove the 6—ohm resistor and Rt becomes 6 ohms; remove a
24—ohm resistor and it becomes about 3.4 ohms . There is an in—

crease in total resistance when you remove any of them.

‘

Well done. You have come to the end of tne lesson. On your answer

sheet write down the length of time it took you to complete this lesson. W

 

Page 130

YOUR ANSWER: The generator current would remain the same if one

branch of the parallel circuit were removed.

But the generator is supplying the total current through the cir—
cuit, isn‘t it?

When the applied voltage is constant, as here, the total current

depends only on the total resistance .

Your answer thus means that the total resistance remains the

same when you remove part of the resistancei

So try image Page 131

Page 131

CONTINUED FROM PAGE 119

Remember that the voltage across each branch is the same.
Assuming that voltage from the generator remains constant. Ohm's
law determines the current in each branch independent of the others .

 

Izov 6.0. n 2341 2.4—IL

 

What will happen to the generator current when one branch is
removed?

Decrease . Page 129

Remain the same. Page 130

APPENDIX D

ANSWER SHEETS USED WITH SKINNER PROGRAM

635

 

 

‘ 1' :A - F "1
.'.'Jnl'L£11

CLASS

'—_ -—--

tmsorx l-S

RATE

SECT.

 

\O

@ASU REE-{ENE
current
pressure
reelstance

 

AN SW ans

snraOL
K 5
(
(

‘s/ \. -1

(3
2
H
F3

ANSWER SHEET

SERVICE NO.

DATE

 

WORK

16.

17.

18.

31.

32.
33-
3h.

E:

most R ( )
most emf ( )

resistance
volts
current

current in A
current in B

".._—.._.-

35.
36.
37.
38.
39.

AC.

42.

#3.

45.
A6.
A7.
A8.
A9.
50.

51.

most current ( )

(_)

most emf

 

52. 1.
2.
3.

53. l.
2.

SA.

55.

56.

57.

58.

59.

6o.

61.

62.

63.

61.

65.

66. 1. 103
2°
3.

67.

68.

5.

3.

69.

70.

71.

72.

73.

7b..

75.

76.

77.

78.

79.

80.

81.

C, J
K; 3

5.

"‘9‘: '11:» a..._.t.i..'_w’. ..g'. ‘

.. . . —._-v-—_-- kv ...... ..
"1PBL_! p—‘m‘ ' ..‘. ”gag; ‘t. _. _-_ .» . .

88.

90.

91.

92.

93.

9A.

95.

96.

97.

98.

99.

100.

101.

102.

103.
101..
105.
106.

107.

 

108.

109.

110.

111 .

112.

113» .

111v, .

115.

I._.:
[...]
O\
O

117.

3-13.

ll9.

128.
129.
130.
131.
132.
.133.
134.
135.
136.

137.

Time

to complete lesson

 

LESSON 2-S ANSWER SHEET

NAME RATE SERVICE NO.' DATE

 

CLASS SECT.

 

ANSWERS . WORK

10°

11. EDE

( )

()
EAE=<>

t1}

BC

l2.
13.
ll».

159

16.

17. EAB = ( )v
EBe= ( )v
ECD = ( )v

18. EAB = ( )
EBC = ( )
ECD = (_ )

19.

2o.

21.

22.

23.‘

21+.

25.

26.

27. EAB = ( )
EBC =,.;( )
ECD = ( )

28.

29.

 

 

 

30., 31., 32., 33., 34., and 35.

36.

38.

40.

1+2.

A3.

A5.
1+6.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

AB BC UJ TOTAL
R
_I
E
AB, BC CD TOTAL
R
I
E
AB BC CD TOTAL
R
I
E .
l. 3.
2. A.

2.

. im....-..*_.r-.:wx.~:w.a- ~— - W x;’ '

37;

39.

 

CD

TOTAL

 

 

 

 

 

 

 

 

 

 

A7.

1:8.

49.

50.

51.

52.

53.

51+.

55.

56.

57.

58.

59.

60.

61.

62.

63.

61..

 

 

65.

66..

6’7.

68.

69.

70.

71.

72.

73.

7A.

75.

76.

77.

78.

' 79.

80.

82.
83.
81+.

85.

86.

88.

89 .
90‘.
91 .
92 .

93.

9h.

 

' IGH

IIJ
EKL

 

 

95. IAB

 

 

 

 

 

 

 

 

 

 

 

ECD =
PEF =
96. 1. EAB = 5.
2. ICD = 6-
3. REF = 7.
1. EGH = 8.
9'7. ( ) ( )
98., 99., 100.
AB BC CD TOTAL
R
I
E 1
101.
102.
103.
101+.
105.
1060
107.

108..

1W" ' 7'
_ .7 n.._-_-..m..-.m-AM-—~

109.

110.

111.

112.

113.

114.

115.

116.

117.

118.

119.

Time

AD =

to complete lesson‘

 

 

 

 

 

LESSON 3-S ANSWER SHEET

NAME RATE SERVICE N O . DATE

 

CLASS SECT .

 

ANSWERS WORK

9.
10.

110 I:

12.

13.

 

 

 

 

 

 

 

 

 

 

11.. 1. ( )
2. ( )
3. ( )
15.
16. 1:24 =(
IZ-i R=(
17.
18.
19.
20.
21.
22.
23.” .6 6. l 6666'
R 24L 1.11
I 28 ( ) l 3a
E l < > < ) I Av
21.. AB CD TOTAL
R
I
1'? 11
25.

26.

 

 

 

 

 

 

27.

28.

29.

30.

31.

32.

33.

3A.

35.

36.

37.

38.

39.

1+0.

112.

A3.

AA.

3.

A.

A5.

1+6.

47.

’ A8.

A9.

50.

51.

52.

53.

51+.

55.

56.

57.

58.

59.

60,

 

CD

TOTAL

 

()

 

23.

ha

63.

 

 

12v

 

 

12v

 

12v

 

 

H
H
II

(

2=('

V

61.

62.

63.

61+.

’65.

66.

67.

68.

69.

70.

. 71.

72.

73.

71+.

75.

76.

77.

78.

79.

80,

“O.

86.

87.

88.

89:

90.

91.

92.

93.

9A.

95.

96.

97.

98.

99.

100.

101.

102.

103.

10A.

105.

106.

107.

108.

109.

110.

111.

112.

113,

118.

119.

111, 115, 116, 117.

 

AB CD

TOTAL

 

 

 

 

"Ut‘JHFU

 

 

 

 

 

 

 

AB BC

CD

TOTAL

 

 

 

 

 

 

 

 

 

 

 

Time to complete lesson

APPENDTX E

ANSWER SHEETS USED WITH CROWDER PROGRAM

660

-.

-,'~ .\..—v '1‘,-

 

V:~_-.-_ _ U-v-Iv rm-
— . _.. .

LESSON 1-0, 2-0, 3-0, (Circle one)

Rate Service No. Date

 

 

 

Section

 

Page Answer Work

 

 

 

 

 

 

 

 

 

 

 

 

 

Lw-c-

 

v-h

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.1 11114 .ifif ‘|‘J

 

APPENDIX F

CRITERION TESTS

 

LOCDVOV

664

FORM 81

INS TRUC TTONS TO TRAINEES

 

Check the number of pages to make certain the test is complete and
legible.

Do not mark this test in any manner. Do not use the answer sheet,
front or back, for scratch paper.

Fill in the data section of both Answer Sheets as the instructor indi—
cates.

Completion and fill-in questions: Use the separate answer sheet
marked "Completion Questions. " Write the word or words that make
the statement complete and true. Do not write any completion answers
on the IBM Answer Sheets.

Multiple Choice questions: Choose the most complete and correct
answer and black in the proper space on the IBM Answer Sheet. Be
sure to black in the complete space and use only the IBM pencil which
is provided.

If you wish to change an answer, erase your first answer completely.
Be sure your answer number corresponds to the question number.

Turn in all scratch paper. Do not make notes.

Any necessary additional instructions will be given by the instructor.

HANDLE IN A CONFIDENTIAL MANNER

665

Ohm's Law states that current is
proportional to resistance and
proportional to voltage.

 

 

An electromotive force of is required to push
. S amperes of current through a resistance of 8 ohms.

 

20 picounits times 10 kilounits divided by 5 microunits equals

10 kinoamps is equal to milliamps.

6) (4 X 103) = X 10:-2

 

(2.5 x10-

In the circuit below, current equals 4 amperes. The resistance of
R215

 

  

R1=15(l R2

Illli

100v

 

The current flow through a 4 ohm resistor which is dissipating 100
watts is

 

In the circuit below, 1R1 is equal to

 

RZZZOCl

 

. R3=10fi

 

R42300{l

66.6

 

9. In the circuit below, ER4 is equal to

 

 

 

R5=100f1 R4

10. In the circuit below, the total power dissipated is

 

 

 

 

 

L
2
——— 240v
RZZZOQ
R4=3Ofl R3=3OO
11. In the circuit below, amperes of current is flowing from
the battery.
R2=ZOQ __JL_
R1=IOQ E:— 80 V

667

12. In the circuit below, the power dissipated by R3 is

 

 

 

__ R1=ISQ R2=3SQ R3=4Ofl
28V?

 

 

13. In the circuit below, when 51 is closed, IRI will

 

 

'81

R1 R2 R3

 

ll'l'i

 

 

14. In the circuit below, the total resistance equals

 

 

Ri= R2: R3: l

62.50 83.39 12581 :ZSOV

Q a) 2,. l

15. Adding a resistor in parallel with another resistor will cause a/an
in total current.

 

 

16. INCREASING the voltage applied to a resistor will cause the

. current to decrease.

. current to increase.

. resistance to increase.
. power to decrease.

whooNl-d

17.

l8.

19.

20.

 

668

In the following circuit, the current through R1 equals

1
2.
3.
4

(4 x10-

Convert 22 KO to no.

Abomr—a

324,000,000 X 8900 = ?

n-D-OONH

l
2
3.
4

1.5amps.
1 amps.
.Samps.
2 amps.

5

1.2 X 102
1.2 x10—
12x10”1

12 X 102

)(3X10)=?

2

22 X 109 HQ
22 X 106 HQ

9

22 x10— no

6

22 x10‘ mm

2.70 X10
27.00 X10
2.70 X 10

.27 X10

ll
12
12
ll

 

 

::::100v

 

 

Rl=SOQ

R2=150Q

21.

22.

23.

24.

669

 

 

 

Find ERZ in the following circuit.

1. 56 volts. <:;:>

2. l4 VOltS. ‘ R :13 KO I

3. 76 VOltS. :: ma 3 >R2

4. 108 volts. €3184v E
’VWVVVj

R3=13 Kn

In the circuit below, the value of R2 is

l. 75 KO.

2. 50 Krt L

3. 30 KO. ‘2 Rl=30 KO

4. 20 KKI. g:

_—_160v

J

@60v

1

—J\/\/\/\/ —

 

In a series circuit containing three resistors, which of the statements
concerning current is correct?

rho 00 [\D I-‘
H
H
II
I-4 ~
._.:
+
r-q
N
+
r- «I
00

IT - 11 - 12 — I3 = 0

In the circuit below, Pt is

l 81 milliwatts.
2. 270 milliwatts.
3. 750 milliwatts.
4 810 milliwatts.

 

 

<
R3=30KO% V 90v

 

670

25. In the circuit below, the voltage at point A with reference to point B

    

 

is

l. -1 volts. 50.
2. 1 volt. ‘

3. -5 volts.

4. 5 volts.

 

26. The power dissipated in each branch of the circuit below will

be 6 watts .

vary directly with the resistor wattage rating.
be 18 watts.

vary inversely with branch resistance.

$.me

 

30 KO 20 KC. 10 KO.

illit—
w—

 

 

27. In the circuit below, IRZ is

 

 

1. .5a .

2. 1.0a

3. 2.9a =

4. 5.5a R1 100m 3
:_ >
—1_—“:—E=100v R3=500 §>R4=1000

V

A
V

 

Ra=1oon ]

 

 

28.

29.

30.

671

In the circuit below, current through R1 is

 

1. 2 amps. , f

2. 4 amps.

3. 6 amps.

4. . 5 amps.
——__:E=100v Rl=50Q 2R2=25Q
”— T

 

 

Four resistors are connected in parallel directly across a voltage
source. If a lead to one of the resistors should be opened the effect
would be to

increase the amount of current drawn from the source.
increase the voltage across the remaining resistors.
decrease the total current drawn from the source.
decrease the voltage across the remaining resistors.

.5me

In the circuit below, It =

 

 

1 l7 ma. @ j

2 10 ma. ‘

3. 9 ma. _ =

4 7 ma. R1— R2 R3
30 KO. 15 KO.

 

 

 

 

 

 

 

 

 

LOGDVOU

672

FORM 82
INSTRUCTIONS TO TRAINEES

Check the number of pages to make certain the test is complete and
legible.

Do not mark this test in any manner. Do not use the answer sheet ,
front or back, for scratch paper.

Fill in the data section of both Answer Sheets as the instructor indie
cates.

Completion and fill—in questions; Use the separate answer sheet
marked "Completion Questions. " Write the word or words that make
the statement complete and true. Do not write any completion answers
on the IBM Answer Sheets.

Multiple Choice questions: Choose the most complete and correct
answer and black in the proper space on the IBM Answer Sheet. Be
sure to black in the complete space and use only the IBM pencil which
is provided.

If you wish to change an answer, erase your first answer completely.
Be sure your answer number corresponds to the question number.

Turn in all scratch paper. Do not make notes.

Any necessary additional instructions will be given by the instructor.

HANDLE IN A CONFIDENTIAL MANNER

 

 

673

1. If the resistance in a circuit is increased, it is necessary to
the EMF in order to maintain the same current

 

flow.

2. If the current doubles in a circuit when the resistance is halved, the
EMF has

 

3. 165 x 10“-10 expressed in scientific notation is

 

4. 100 ii units equals basic units.

 

5. L25 X 10_8) (2 X 103) _

 

 

 

 

 

 

 

5 x 10—4

6. In the circuit below, watts of power is dissipated by R3.
: 100V 60v
—l———/\/\/\/\/\/\/————P

7. The mathematical relationship between total current and the current
flow through three resistors in a series circuit can be eXpressed as

 

8. In the circuit below, R3 equals

 

 

 

 

 

 

 

 

 

 

 

674

 

9. In the circuit below, R1 equals

 

<R3=300§L

{I l

10 A formula that shows the relationship between individual votage drops
and the total voltage in a series circuit containing three resistors is

 

In the circuit below, if R3 opens, the current through R2 W111

it

In the circuit below, the total resistance is

11.

 

«tutti——
WW

 

12.

 

1

ZOKQ ZOKQ lOKQg

i
I

 

 

 

 

 

13.

675

In the circuit below, the power dissipated by R3 is

 

)
)
I

l
1

14. In the circuit below, when 31 is closed, PR1 will

15.

In the circuit below, the current through R1 is

1

l
2
3.
4

4,111]:

 

——Iv\/\/»

 

 

1

R2: R3: l

 

 

R1:
125 500 750 2
. 5a
In the circuit below, It is
1. 00 ma. 1
1.72 ma. >
10.00 ma. %,5 < Rl=2.5K
17.20 ma. 1: 1
—‘_ <: R =4K
g 2

 

 

 

17. In the circuit below, Et =

1.5V

1. . ‘
2. 16.0 v.

3. 1500.0v.

4. ——

 

16,000.0 v. 20 “a

  

R = 800KQ

18. If 4 basic units are divided by 2 micro units the answer would be

1 2 kilo units.
2. 2 milli units.
3. 2 meg units.
4 2 basic units.

19. 155 picoamps = kiloamps.
1. 1.55 x 10'10
2. 1.55 x 1015
3. 1.55 x 10‘7
4. 1.55 x 10‘13

20. 20 microvolts times 1012 is equal to

2 x 107 volts.
20 x 107 volts.

2 x 1019 megavolts.

4&0on

2 x 10_7 microvolts.

21. In the circuit below, the voltage across R3 is equal to

 

l. 72 volts.
2. 45 volts. i R1=3K
3. 27 volts. :— 45v
4. 3 volts. ‘
R2=12K j
\I

 

R3 = 9K§L

 

 

 

 

 

 

 

 

677

22. What is the relationship between power, current and resistance?

23.

24.

25.

1. Power is inversely proportional to current and resistance squared

Power is directly proportional to current squared and resistance

3. Power is directly proportional to resistance and inversely
proportional to current

4. Power is directly proportional to current and inversely proportional
to resistance

[\3

In the circuit below, IRI is equal to

1R2.
IR3.

IR4.
all the above.

A m N H

 

 

R4=3OO

How much power is dissipated by a 5000 resistor which drops 30
volts ?

l 60.00 mw
2 .18w
3. 1.80w
4 15.00 kw

In the circuit below, PR2 is

10 watts.
15 watts.
25 watts. l_

50 watts. g 100v RZ=6OQ <>

R1=40§1

(DWNH

 

R3=100

 

 

 

 

 

678
26. In the circuit below, R1 equals
50 I
83. 3

l
2
3. 125 :250V
4 186.6 —

 

R1 R2=83. 3

27. In the circuit below, It =

 

Rl=8OKQ

l
I

IN

R2=240KQ§ R3=60K§1

l
l

>

fi
3ma

 

 

 

28. Adding a resistor in parallel with another resistor will cause a
decrease in

1 Rt.
2 Pt.
3. Et‘
4 It.

29. In the circuit below, what is PR1?

 

1 2 watts
2 100 watts <>
3 400 watts <’
' —— > =4 00
4 20,000 watts % Rlloon <, R2 0
4

\I

V

 

(A) l

.5a

 

 

 

 

 

 

679

30. In the circuit below, if SI is closed

IRZ decreases. /'
ERZ decreases. S1

ERl decreases .

l
l

R1 R2 R3

.5 co N 9—-
. . . .

[M

It increa ses.

 

 

 

 

APPENDIX G

RAW DATA: SCORES ON ALL VARIABLES FOR ALL SUBIECTS

680

 

 

 

681

_ —
000 GM 00 OM —F fie NM NM ON NM 0— MM N~ We MF mF V—— — ~
000 mm NF Cm F0 00 0m OF 00 0¢ NF HQ NM Nm 00 m0 OO— — _
000 00 QF Fm N0 N0 00 00 0F Nm NO ON NM 0. OF HF Q—~ _ H
000 mm 00 —€ N0 OM OM 0m 00 0€ NF 0m NM 00 NF OF ¢—— — _
COO NV Q0 0Q ~F NM NM NM 00 OM Nm flN N~ NM 00 nF 0O~ — —
000 "F OF Fm mm 00 0m NF 00 06 NF MG N— NF ”F ”F F__ a _
000 00 cc om m0 HO Nm NF mm 00 0m 00 NH 0Q <0 N0 ON~ — n
000 mm N0 Fm _Q 00 00 00 06 0? NM on 0N NM F0 00 man _ ~
000 FF OF Fm _0 OF 00 00 00 Nm 00 00 00 Nm NF QF mN_ _ a
000 FF mm FW 00 0F 0F NF 00 0m NF H0 Nm NF HF F0 ®- — ~
000 cm OF Fm Um 0F NF 00 00 0m 00 HO 0m 00 F0 00 0—_ _ A
OOO NO OF _G Km 00 0m 0m 0F OF OF NO Nm NF m0 Ow QN_ — _
000 DF QF Fr mm 00 NF NO 0m 00 NF 0Q 00 04 F0 00 Cg— — —
000 FF FF Fm —F 00 Nm 00 00 0” NF 0m 0” NF F0 m0 HN— — —
OOO mF 0Q FW —0 OF NF 0F MF 00 0F 00 NF 00 F0 00 DO_ _ ~
000 00 NF 0m N0 00 Nm 0F HF NF NF 00 00 NH 00 ~F O- — fi
000 mm 00 FW _m 0m 0H Om mm 0m 00 ”a N0 Nm _F F0 0—~ _ d
000 00 N0 Fm F0 0F NF 0E 0F NF 00 ”F 0F 00 OF 0F Oflx ~ _
000 NO OF Ffi mm ”F 00 0F 0m 00 0G HO ON 00 #F Q0 _u— _ g
000 —F NF 0m _F 0F 0F NF NF 00 0F HQ 0N 0m 0F 00 0—_ a #
000 um 00 Fm N0 0F NF NF 0m 0Q NM 0N N— On 00 me 00— H _
OOO HF QF Gm NO OF NF 00 00 0m 0m 0‘ ma 0m 00 m0 O—H ~ —
uOu «u 0... you an 0... you «u 05 a M
H n: n K: N: H I! H.“ a.“ .._.“ Em Eh .._.-Hm gm Bflm Bflm mm UN— OH «IL .W
n z
a you at

 

mHOmHm—Dm A13» mOm mmqm<Hm<> Add. 20 mmmOOm ”<H<Q >><m
mu mam/.2.

 

 

 

 

 

682

00

cm mp
F0 00
«0 mo
v0 me
he do
VF 0F
mo 00
3
00 «r
Omtlflm
mm mm
FtFttth
m» mm
NH an
0F 00
.NH .NN
mo 00
.00 m0
FF NF
OF} 3
~F NF
00 NH
F0 00
00 0F
>0 on
00 m0
F0 00
OF 00
~O n0
9. ch
0F 0F
NF ~F
00 mo
F0 00
ms or
0F «F
Q. :
mF 0F
9. or
0F F0
E. mp
mF mF
F0 00
FF «F
0F HF
0F m.F
me do
0F ad.
on E.
lmw oh
.flh. .9.
NF —F

 

 

 

 

 

683

mmnnnmmmmnmnmmmnmmm

NVNNNNNN’VNVNVNNNNNNNVVM

NNNNN’UMMMNNNNNNNMNN

 

 

 

684

mu

~F
Vb
00
m0

mmmmu‘imm
mmnmnnm

O
N
NNNNNVN’VNNNNNNNNVNNNNN

0
H
NNNNNNNNNNNNNNN’UNVNNNNNN

 

 

 

 

685

\D
H
nnnnmmmmnnnmmnmnnnnmmm

<-
v-l
nflmmn‘ififlnmmnm

an

FHFHFHFNFFF
nflndnnmwmd

Mi

 

 

 

686

00¢

N”; «F1
0.” F0
.mwflrerh.
NF 00
gig
0F mF
Mmt QF
0. .0
ON QF
00 «F
NF 0F
Om FF
0Q FF
0m FF
oq "F
Nm ~F
Om ~F
N0 N0
00 Nm
on mo
NF _F
00 0F
NF mF
0F 00
N0 ~F
0m 0F
00 FF
Nm 0F
N0 00
0F OF
00 no
00 FF
NH NF
0F 0F
00 OF
Nm NF
mm 00
0” OF
No OF
NF 0F
00 mF
00 mF
Nm om
wfl mF
0m —F
NF mm
0N QF
0? F0
0m GF
00 «MF
0‘ MN
OH. 00
NM QF
0.0. ..N
0— 0F
01h .OH.

0'9

cicficde

mmmmcncomannmmcomtnmmmmmn

____~~_4_q_~~~~~u_~~

—1~—+—-~|—-—<—-4~

 

 

 

 

687

mm

mammmma‘mmmmo‘m

00000000000000

NNNNNNNNNNNNN

NNNNNNNNNNNNNN

—.<_.._.__‘~_._._._._....._..._._—a—

 

 

688

NF

11')
H
NNNNN

m
H
NNNNNNNNNNNNNNNNNNMNNINN

 

 

 

689

O0N
mum
DDN
nFN
ONN
muN
€0—
OQN
ODN
mom
OuN
O0N
mmn
ncN
.nN
Nmn
F0—
OWN
0ON
«mm
mNN
WON
¢0N
NO—
MF—
OON
0¢N
WHN
0N0
owN

mFN

”mm

00
—F
00
O0
Q0
mm
mm
0F
00
«0
nm
«0
F0
mF
N0
00
cm
FF
F0

00
F0
cc
F0
mF
«F
um
00
Q0
00

Nv

F0
N0
00
O0
00
he
00
FQ
O0
F0
"F
N0
00

N0
N0

00
00
0F
c0
00
00
0F
00
00
_F
00
N0
OF
NO
c0
~F
um
00
O0
00
00
«F
00
NF
N0
00
«0

v0
F0
F0
on
0?

on
N0
0c
cm
N0
F0
on
00
F0
00
00
0c
om
00
N0
Fm
fin
om
N0
Fm
00
F0
on
N0

0n
mm
mm
mm
mm
mm
am
no
F0
00
—0
F0
0?
mm
_0

N0
0O
_F
~F
N0

0F:

—0
0F
F0

—F

.0
F0

—0
mo
F0
0O
0O
m0
0F

—0
mm

—F
0F

~F
Fm
0O
N0
0F
0O

0N
0F
F0
mm
mm
F0
MF
.0

F0
N0
NF
HF
00
—m

0m

0m
00
00
00
00
0M
0N

mm
00

0F
O0

00
00
NW
On
0c

OM

OM

Nm

Om
Om

mm
00
Om
00
Om
OF
00
On
0N
Nm
0N
Nm
O0
00
00

Nm

00
0c

NF
OF
NF
00.
00
Nm
10¢

00

NF
00
O0
O0
00
O0
00
0c
On
00
00
O0
O0
Nm
NO

mm

Nm
00
Nm
N0
00
N0
0N
OM
OM
OF
O0
00
NF

.-00

F0
ON
00
Ge
0c
0c
O0
0Q
00
no
00
O0

On
00
Om
00
00
Nm
O0
06
NM
NM
03
On
OF
00
00

0N
O0
OH

.NH

Na
9.
2.
NF
N»
om
co
Nn
on
00
2.
N...
mm
ms
2.
3

00
OF
Nm
NF
N0
NF
0c
00
00

00

OF
NO
om
0Q
00
NH
0c
00
OH
NH
NF
00
OF

0N

.Om

0c
O—
0N
NF
N0
00
mm
N—
On
mm

on.

00
O0
00
NF
Nn

Om.

Nn
O0
O0
N—
Om
OF
On
O”
NH
O”
Nm

0N
NM
Om
Nm
Om
0Q

OH
N0
N0
N0
N0
00
OF
N0

N0
NF

3
on
Ms
Nm
«m
on
om
on
on
on
9.
om
or
2.
00
mm
mm
cc
on
em
mm
mm
06
3
«.6
on
up

Nn
OF
00
O0
NF
00

OF
0N

N0
Om
Nm
Nm
N0

00
0Q

00
NF
¢F
0F
-00
00
flm

.0H.

OF
OF
0F
0F
m0
0F
00
.ON
QF
vF
OF
«H
QF
0F

QF
OF
00
00

HF
HF
00
00
O0
mF
F0
HF
~0
NF
O0
00
OF
00
«F

.0”
QF
OF
O0
¢F
flF
an
F0

O0
0F
0F
00
00
0F

uF
OF
¢F
c0
fiH

”N

F.—
NM—
FN—
«Na
Ou—
ANA
{~—
.44—
NN—
FNu
«Nu
«fl—
FH—
~0—
0—fi
OO—
Nflé
ONu

nNn-

«nu
0N“
-—
F04
FH—
cu"
F0“
Nqfl
Gou
0‘4
N0—

N0—
Oc—
cNd
Fm—
mm“
NN—
Ofia
uu.
Odn
nau
Ofiu
F‘—
0-
O0—
”Au

NCUNCVNCUNCUNOUNCUNfuN

NFTMF10F1MFIHFHWflfWMIWN(“MFG”(WHCONIWHIOHFOH

fifQHFQMFOMFOMIOHCWHIWM

 

 

690

nauIOMUIn .uocuuxmnw

.ousuuoqIHH
o__ MN m
qw. mw n
mg. mu m
m._ nN n
0N. mm m
mj MNA.
om_ nN n
m~_ mu m
a—_ mu m
N: MN m
¢N~ MN fl
0N. MN m
.nfid m~,fl
on. MN m
(Qua. m~.d
«N. mm m
.01 .34.
of mm m
can add
o__ man
fir: adn
0N_ ma n
ma; mflm
~N_ man
«1. afi‘n
-_ ma n
towta ad...“
>__ man
m_q may”
owg mam
dNa afld
oe_ mu m
w~_ de
Fm: m." n
mun adyn
No. my. n
Fad an m
o: m." n
F: ad 0
em. ma 0
QMJ man”
m__ ma m
.N4~ afl 0
cm. ma n
Nu— mH .n

 

 

 

BIBLIOGRAPHY

Angell, G. W. ”Effect of Immediate Knowledge of Quiz Results on
Final Examination Scores in Freshman Chemistry, ” I. Edu. Res. ,
_4__g (1949), pp. 391-394.

 

Bendig, A. W. Analysis of Variance with Unequal Subclass Fre=
guencies. University of Pittsburgh, March 1960 (Mimeo).

 

Briggs, L. I. "The DevelOpment and Appraisal of Special Procedures
for Superior Students and an Analysis of the Effects of :Knowledge
of Results,‘ Abstracts of Doctoral Dissertations, No. 58. The
Ohio State University Press , 1949.

 

Briggs, L. I. Teaching Machines: History and Possible'Applioation-s
to Air Force Education and Training Prggrams. Paper prepared for
presentation at a Symposium on Training Media, Washington,

D. C., August 18 and 19, 1959, sponsored by National Academy
of Sciences, National Research Council.

 

 

Briggs, L. I. , Goldbeck, R. A., Campbell, V. N., and Nichols, D. G.
"Experimental Results Regarding Form of Response, Size of Step,
and Individual Differences in Automated Programs. " In I. E.
Coulson (Ed. ), Programmed Learning and Computer-Based Instrqu
”ti—on. New York: John Wiley & Sons, 1962° pp. 86—-98.

 

Bryan, G. L., and Rigney, I. W. Current Trends in Automated Tutoring
and their Implications for Naval Technical Training. Contr Nonr-
228(02), Office of Naval Research, December 1959.

 

 

Buros, O. K. (Ed.). The Fifth Mental Measurements Yearbook.
Highland Park, New Jersey: The Cryphon Press, 1959.

 

Campbell, V. N. "Adjusting Self—Instruction Programs to Individual
Differences: Studies of Cueing, Responding and Bypassing, “
AV Comm. Rev. , 1962, 1_O (6), A—124 (abstract).

 

Cassel, R. N. , and Ullom, W. L. "A Preliminary Evaluation of
Programmed Ingruction with Students of High Ability, " Psychol.
Rpts, 1962,1_O, pp. 223—228.

691

 

 

 

10.

ll.

12.

l3.

14.

15.

16.

l7.

l8.

19.

20.

21.

692

Cave, R. T. , and Zachert, V. Programed Learning with Extremes of
Student Ability. Paper read at meeting of American Psychol.
Assn. , St. Louis, Missouri, 1962.

 

 

Center for Programed Instruction. Programed Instruction and Special
Students. Programed Instruction, 1963, g, p. 3.

 

 

 

Chesire, L. , Saffir, M. , and Thurstone, L. L. Computing Diagrams
for the Tetrachoric Correlation Coefficient . Chicago: University
of Chicago Bookstore, 1933.

 

 

Coulson, I. E. , and Silberman, H. F. ”Effects of Three Variables in
a Teaching Machine " L Educ. Psychol. , 1960,. §_1_, 135-443.

 

Crawford, M. P. ”Concepts of Training. " In R. M. Gagne (Edfl.
Psycholggical Principles in System Develogament. New York:
Holt, Rinehart and Winston, 1962, pp. 301—341.

 

Cronbach, L. 1., and Gleser, G. C. Psychological Tests and Person—
nel Decisions. Urbana: University of Illinois Press, 1957.

 

 

Crowder, N. A. "Automatic tutoring by Intrinsic Programming. ‘ In
A. A. Lumsdaine and R. Glaser (Eds.), Teaching Machines and
Programmed Learning. Washington, D. C.: National Education
Assn., 1960, pp. 286—298.

 

 

Crowder, N. A. , and Zachert, V. Use of the Autotutor at Keesler A13
to Train in Fundamentals of Electronics. Memorandum Report No.
1, United States Air Force and Western Design and Electronics,
1960.

 

Davis, F. B., and Davis, C. C. Davis Reading Test. The Psycho—
logical Corporation, 1958.

 

Detambel, M. H. , and Stolurow, L. M. "Stimulus Sequence and
Concept: Learning, " LExper. Psg/Chol. _. 1956, i1, 34~-40.

 

Dite, D. , Woodruff, A. B., and Faltz, C. The Relationship Betweeg
Level of ReadingAbility and Performance on Programmed Learning.
Paper read at meeting of Midwestern Psychol. Assn. _. Chicago
Illinois, 1962.

 

Evans, J. L. , Glaser, R. , and Homme, L. E. "A Preliminary Investiga-
tion of Variation in the Properties of Verbal Learning Sequences
of the "teaching Machines" type. " In A. A. Lumsdaine and R.
Glaser (Eds. ), Teaching Machines and Programmed Learning.
Washington, D. C.:; National Education Assn., 1960, pp. 446*451.

 

22.

23.

28.

29.

31.

32.

33.

693

Faltz, C., Woodruff, A. B., and Dite, D. The Relationship Betwegn
Learning Ability and Performance on Programmed Learninc. Paper
read at meeting of Midwestern Psychol. Assn. Chicago, Illinois,
1962.

Fattu, N. A. "Training Devices. " In C. W. Harris (Edg‘l, Encyclopedia
of Educational Research. New York: Macmillan Co. , 1960, pp.
1529i, 35.

Ferster, C. B. , and Sapon, S. M. "An Application of Recent Develop~
ments in Psychology to the Teaching of German, " Harv. Educ.
Rev. , 1958, as, pp. 58—69.

Finn, I. D. , and Perrin, D. (3. Teaching Machines and Programed
Learning, 1962:: A Survey of the Industry. Washington, D. C. r‘
National Education Association, 1962.

 

Flanagan, V. C. "Units, Scales, and Norms.“ In E. F. Lindquist
(Ed.), Educational Measurement. Washington: American Council
on Education, 1951. Pp 752-760.

Freeman, V. T. "The Effects of Reinforced Practice on Conventional
Multiple Choice Tests, " Automated Teaching Bulletin, 1959, 1
(September), pp. 19—20.

Froehlich, H. P. “Training Conditions, Ability, and Academic Outs

comes," I. Appl. Psychol. , 1962, fl, pp. 67w71.

Froehlich. H. P. "A Note on Under— and Overachievement Measure~
ment," Pers. Guid. 1. (March 1963), pp. 6214323,.

 

Fry, E. B. "A Study of Teaching Machine Response Modes. " In A. A.
Lumsdaine and. R. Glaser (Eds.), Teaching Machines and Pro—
grammed Learning. Washington, D. C.: National Education
Assn., 1960. Pp. 469—474.

Fry, E. B. , Bryan, G. L. , and Rigney, I. W. _Teaching Machines:
An Annotated Bibliography. Los Angeles:: University of Southern
California, Electronics Pers. Res. Group , November 1959 (Tech.
Rep. 28).

Gilpin, I. "Design and Evaluation of Instructional Systems , “ A;
Comm. Rev. , 1962,1_0, pp. 75184.

Glaser, R. "Learning and the Technology of Instrucuon, " A1." Comm.
Rev. , 1961, 9, pp. 42455.

 

 

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

694

Hedberg, R. D. Automated 'I'eachin : Research Problems. Paper read
at meeting of Amer. Psychol. Assn, New York, New York, 1961.

 

Homme, L. E. , Willey, R. E. , and McMahan, W. H. A Study in the

Application of Teaching Machines. U. S. Naval Training Device
Center. Technical Report: NAVTRADEVCEN 1000*1. 1962.

Hughes, I. L. , and McNamara, W. I. "An Experiment Comparing the
Effectiveness of Programmed Textbooks and Conventional Classe
room Instruction in an Industrial Training Situation, " Amer.
Psychol., 1961,16, 438. (Abstract.)

Iensen, B. T. "An Independent Study Laboratory Using SelfeScoring
Tests," 1'. Educ. Res., 1949, 43, 134.1327.

Keislar, E. R. "The Development of Understanding in Arithmetic by a
Teaching Machine,“l I. Educ. Psychol., 1959, i9, pp 247—253.

Keislar, E. R. , and McNeil D. ”A Comparison of Two Response
Modes in an Autoinstructionai Program with Children in the
Primary Grades," I. Educ. Psychol., 1962, 53, pp. 127-1310

Kendler, H. H. "Stimulus=Response Psychology and Audiovisual
Education, " AV Comm. Rev. , 1961, 9, pp. 33411.

Krumboltz, I. D. , and Weisman, R. G. “The Effect of Intermittent
Confirmation in Programed Instruction, ” “I. Educ. Psychol. , 1962:
5_3, pp. 250-253.

Lambert, P. Schedules of Reinforcement. Effects on Programed
Learning. Paper read at meeting of American Psychol. Assn. _, St.
Louis, Missouri, 1962.

Levin, G. R. and Baker, B. L. ”Item Scrambling in a Selfelnstructional
Program," i. Educ. Psychol., 1963, 53, pp. 138e143.

Lindquist, E. F. Design and Analysis of Experiments in Psychology
and Education. Boston: Houghton Mifflin Co. , 1953.

Little, I. K. “Results of Use of Machines for Testing and for Drill,

Upon Learning in Educational Psychology, " ‘v. Exper. Educ. , 1934,
3, pp. 45—9.

Luchins, A. S. “Implications of Gestalt Psychology for AV Learning,
AV Comm. Rev... 1961, 9, pp. 7:31.

 

 

 

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

695

MacPherson, S. I. , Dees, V., and Grindley, G. C. ”The Effect of
Knowledge of Results on Learning and Performance, " uart.
Exp. Psychol., 1949, I, pp. 167—174.

Mager, R. P. "On the Sequence of Instructional Content, " Psychol.
Rpts., 1961, 9, pp. 405—413.

Mager, R. F. , and Clark, C. ”Explorations in Student-Controlled
Instruction,” Psychol. Reports, 1963, l_3, pp. 71—76.

Mayer, S. R. Use of a Teaching Machine for Air Force On—The—[ob
Training in the Page System. Air Force Command Control Devel-
opment Division, Bedford, Massachusetts. APCCDD—TN-GO-Sl.
1960.

Mosier, C. I. , and McQuitty, I. V. “Methods of Item Validation and
Abacs for Item—Test Correlation and Critical Ratio of Upper—Lower
Difference," Psychometrika, 1940, g, pp. 57—65.

Mowry, H. W. , and Webb, W. B. “Remedial School and the Needs
of Failing Trainees. ” In H. W. Mowry, W. B. Webb, and E. A.

Garvin, Studies in Air Technical Training. Washington Univer_
sity, Technical Report No. 2, 1955.

Porter, D. "A Critical Review of a Portion of the Literature on Teaching
Devices," Harvard Educ. Rev., 1957, _2_7, 126147.

Porter, D. "Some Effects of Year Long Teaching Machine Instruction. "
In E. Galanter (Ed. ), Automatic Teaching: The State of the Art.
New York: Iohn Wiley & Sons, 1959, pp. 85—90.

 

Roe, A. "Automated Teaching Methods Using Linear Programs, " I_.
Appl. Psychol. , 1962, Q, pp. 198—201.

Roe, K. V. , Case, H. W. , and Roe, A. "Scrambled Versus Ordered
Sequence in Instructional Programs, " I. Educ. Psychol. , 1962,
5.3:, pp. 101-104.

Scharf, E. S. "A Study of the Effects of Partial Reinforcement on
Behavior in a Programmed Learning Situation. " In R. Glaser and
I. R. Taber, Investigation ofthe Characteristics of Programmed

 

Learning Sequences. University of Pittsburgs, October 1961 ,
pp. 83—96.

 

 

 

58.

59.

60.

61.

62.

63.

64.

65.

66.

67.

68.

69.

696

Shay, C. B. "Relationship of Intelligence to Step Size on a Teaching
Machine Program," I. Educ. Psychol. , 1961, Q, pp. 98—103.

Siegel, L.,and Siegel, L. C. "The Instructional Gestalt: Acon-
ceptual Framework and Design for Educational Research, " AV
Comm. Rev., 1964, 1_2, pp. 16—45.

Silberman, H. F. Characteristics of Some Recent Studies of Instruc—
tional Methods. Paper read at Conference on Application of
Digital Computers to Automated Instruction, co—sponsored by the
Office of Naval Research, and System Development Corporation,
Washington, D. C. , October 1961.

Silberman, H. F., Coulson, I., Crinn, E., and Melaragno, R.
Development and Evaluation of Self—Instructional Materials for

Underachieving and Overachieving Students. Technical Memoran—
dum, TM-727, System Development Corporation, 1962.

Silverman, H. F., Melaragno, R. I., and Coulson, I. E. "Confirma—
tion and Prompting with Connected Discourse Material, " Psychol.
Rpts., 1961, 9, pp. 235—238.

Skinner, B. F. "The Science of Learning and the Art of Teaching, "
Harv. Educ. Rev. , 1954, 94, pp. 86-97.

Skinner, B. F. "Teaching Machines. " In A. A. Lumsdaine and R.
Glaser, Teaching Machines and Programmed Learning. Wash—
ington, D. C.: National Education Association, 1960. Pp. 137—
158.

Skinner, B. F. "Why We Need Teaching Machines," Harv. Educ.
Rev., 1961, Q, pp. 377—398.

Stolurow, L. M. Teaching By Machine. Cooperative Research Mono—
graph No. 6, U. S. Department of Health, Education, and Wei-—
fare, OE—34010, 1961.

Thorndike, R. L. Personnel Selection. New York: Iohn Wiley & Sons,
Inc. , 1949.

Trittipoe, W. I., Trittipoe, T. G., and Halen, C. P. "The Effective—
ness of Three Programed—Learning Frame Styles, " I. Programed
Instr., 1963, 9, pp. 35—43.

Walker, H. M., and Lev, V. Statistical Inference. New York:
Henry Holt and Co., 1953.

 

70.

71.

697

Winer, B. V. Statistical Principles in Experimental Design. New
York: McGraw—Hill Book Co. , 1962.

Wong, W. , and Douglass, W. B. A Comparative Evaluation of Stu—
dent Performance Under Automated and Non—Automated Learning
Conditions. International Business Machines Corporation,
Owego, New York, August 1963.

 

 

 

 

 

 

e ”111111111

 

   

        

    

 

 

 

 

ATE UNIVERSIT), L B

mum H I ll ill Ill will“

93 03056 7204