ch1 problems for electrical engineering

7/30/2019 CH1 Problems for electrical engineering

1/15

Parl I

HOMEWORK PROBLEMSSect ion 2.1: Def in i t ions2. 1 An isolatedlc c electron! lravelioghrough nclectr ic clcl rclnsolDenit ialpointwher-etscoulontbic otenlit lc eru) perunil charge rr) lrf l ,q()sl7 kJ/( ' an dvclocit) = 9.1Vnr/s o some inal point$ here tscoul(nnbic otcnt it lenerrype runitchirue s6 kJ/C.Dere rine lhc charue n velocir\ .ofrheclectron. e! lectgritVilut ionalbrccs.2. 2 I hc unituscd or volt tge 5 hc \olt . Io r cuf ienl hc

a rp.rc. nd lor tesislrrrcehc ohlt).Using hedehnil ions l volt t lc. cufrenl. nd c\ ist i1nce.xpressi . r . , , r , : r ' r l i l \n J -Ln . l : ' r . r r , r t . r ,lKs ur r t \ .2. 3 'Ihecepiicir lol a crf barler\ s usually pccil ieclnlumpere olrrs. bil l tery t led tt . sit \ . 10 0A h shouldbe ablc o supflv I{X)A tbr I h. 50 A tbr. h. 25 A ior-l h. I A 1br10 0h or an) olher onrbinalion'ieldingprorluct f 10 0A, h

x. I low nt itn)coulorlb\ol chitrce houldwe bc tbleto dra$ fr(nrr full\ chlrsed 100A,h blt ter). ,1h, Hou mirn\elcctronslocs oLu-l[ ts\\ 'ero pit f l lre q re./

2 . 4 Thechargec l c iehou I n F igu | e 2 .1i sane ran rp leal it t \ ' t )-t-at(1tur,. !r.hc current s heldconslantrt 50rD A i)f 5 h. Jhcn l rs srl lched o l0 rnA lirr th enexr5 h . F indil . I hc ot i l chargcritnslcnedo thcbullery.b. The energ! rn|\ len.cd o he blt lcr), .

Hil, / Rccall hr t lrterq! r/ js lhc le-srxl l po$er.or2. 5 Bltrclies .. .g.. cad-JciLir ef ies) k)re hemrcil lcnerg!n (l conrcrl t Io clcctf ic nelg)on dentand.Bnllcries o nol slorcelectf ic hargc I cl largc il l . I iers.Charge arricr\ electrons)nleron c cf t inrl ol the

battcry.tcquire lcclf ic potential ncrg,-.nde\iLIronr heolher efI ] t int l t a owervolta-qe.crrenberl l l ( l e L l r o ' t. r . , r c r ' . r r t r c i l r . r r ' . . r ' lt s . , , r c r r r e r , ,lhinkof posir i lecarricI1\iowing n he oppotile(lrrcclion.hrt is.convcntionululTent. ndexil ingat ahr-shefolt i lge.All currenl\ n lhis course. nlcssollter\ \ \L' t i t led. lr e onrell jonal ul enl. Benjantjnl fanklin auscdhis me\s ) Fo rd baller), i lh a raled\ollage ll V tnd r f l led capuctlv l-50A h.alctcrnt iner. Thc rated hclnical nefg\ '! loredn thcbi ter\ , .h. Thc tolalchrrge hl l clt l lbe upplied t lhe raredvoll iue.

2. 6 What dcternincs he bllowinga. Horv nuch urrcnl s supplied al a constantvollage)byan deil lvollage ource.b. Ho$ muchvoltages supplied at a constantcurrenl) y an dealcllt -rcntoulce.

2.7 An automorive aflery s r iLred t l l0 A h. Thisrreanshilt utdet cef lajn eslcondit ionsl cirnoutpulI A rt l l V lbr 120h (under rher ell condit ions.hcbattcry )a) have ther lt in{s).iL How tnuch otalenergy s rtored n th ebattery.)b. I f the headlighls re cf i or )o\ernight ll h). ho$nuch energywill sl i l l bc sroredn rhcbtlter. \ ,-n lhen)Lnrlng. 'Assume 150-W tXtl power ating orbothheacll igbrsogether.

2. 8 A cal batter 'Vept n slora-qen theDrsemenl eedsrech{rging.l thevoltagend thecufent provided ]th ccharger uringa charue vclcar esho$n n FigureP2.u.a. Find the Iotarl harge ranstered I() the balter),.b. Find he otalenergy fansfirrcdo thcbatery.

2, 9 Supposehecurrent lo$, ing hfough wire rssivenb! thecurveshoun n l- i-eure 1.9.a. Find heantouul l charge./. l lal lous throughth e$,ifebct\ ,reenr= 0 anci : = I s.h . Repea t a r t fb r1 .= 2 . . .1 .5 ,6 ,7 .8 .9 . and 0 ! .c S k e t c h q ( / ) l b r 0 : : 1 0 r .

-l0 nrA

?'6

t . 7 5t . 5

l : 5 vI V

0

Figure P2.4


2/15

Chatler2 Fundamenratst Etecr. iu . i rcurts

1 2

1 0

F igu r e P 2 .9

2,1O The char-l i lgschcnle se d n Fil lurepl . l0 is ane\anlplcol a constanl-volla-qehat.gc rl l)currentIlnlr l. he chrrger oltages such hal he curenl intolne ba ef) 'does ol c\ceed10 0nl^. as hown nFr !L r c , . . f l r e ,h . r r . - e r\ , , , 1d .1 ( .r \ c . r \ ( \ , , r r ( .

maximum f 9 V ts shown n Figure 2.I0.Th ebat(erys chargedbr (r h. Find:a. The olaicharqe eliveredo Ihebaltery.b- Thc energv ranslerred o thc batlervdurinq hc,cnargtng ycte.

/ / i / r / :Recall ha t he enel-s!. ,. ! thc ntcgrar r po\,rer.rt a 9

E

a.=

,=6ad

2E; 0

-2F igu r e P2- l O

2, 11 ' fhecharging chemc se


3/15

Pan I

a

t . l0 7 i Y

.:

Figure P2.12F igu r e P 2 . l 1

2.12 Ihc char-l ingschenrcuscd n Fi ,r :urcP2 . l is cal iecln tul t tr tr l t t trrcttt t l tatqa l rr /r , . I he curre t \ l i t f l \ lr tlhe hiehcst e\ c l ancl hen clecreases| i th l iotc for thcentrrc chi rrgecvcle. ls shown. l hc brt lefY is chi -trgedl o r 1 l h . F r n d l.1. The tolal chxfgc dcl i \ered to lhc bxl tef\.b. I he encrg) trrnsl i ITc-d o th! ' bnttery during rhc

chtrginq r\ c lc./1rr l : Rccal l th i l l lbc enr-fgv.u. i\ thc' i | tegral ol powcf. of

Sect ions 2.2,2.3: KCL, KVL2 .13 t j \e K i f c hho s c u en t w lo de te r n inche

unk no r n u r fen l sI Ihe i r c u i t fF iqu r cP l . l J .A \ \ u m c t h n l1 l A . i = l A . / \ : 8 A . l n d

2 .14 lpp l y K C I - o i nc l hcc unc n l i n hec i r c u i l fl - r S u r e,1 .14 .2.15 r\pply KCL lo l incl hc currcnr in lte circuiro1 'F jgu l cP l . l 5 .

F igu r e P 2 . l 3

Figure P2.14


4/15

Chrp.er FuDdarnenratsl Electr icCircurl \

Section 2.4: Sign Convention2.18 In the circuits f Figurep2.18.thc direcrions fcurrent ndpoJariticsf voltagc av e lreadv cc n

dehned. ind hc aclual alues f lhe rnt_rrcareoLrrrcotan dvollages.

F igu r e P 2 .15

2.16 Appll KC L ro hlcl rhevollages iLnd . i r FigureP t . 6 .

- t 0 \ ' +

\. ) t / r r \I| , - ' r fl r , I t 0 ( )F igu r e P 2 .16 +

r h )

2.17 t.Jse hm t-rw rn d KCL to (lctcrntinehc currenl/ i I l hcc i r c u i l ' F i { u r ep l .17 .

1 ) o

( c )F igu r e P 2 . l 8

2,19 I,inci hcpo"cr delivercd 1 cact) ource n rhccifcuilsof I. ' i8urc 2.19.

3 0 o

l 0 o

/ . I

F igu r e P 2 .17


5/15

Paa I

l 0

( a )

- l o v +

5 O

Figure P2.212.22 For the circdt shown n Figure P2.22:

a. Determinewhich componenlsare absorbing owerand which are deliveringpower.

b. Is conservation f power satisficd'lExplain vouranswer,

>..1/\( b )

F i g u r e P 2 . 1 9

2.20 Dctelminewhichelenrentsn thc circuitof FiguleP2.20 rcsupplying oweran dwhichar cdissipatingporvcr. lso delerminehc amount f powerdissipxtedancl upplied.

Figure P2.222.23 For the circuit shorvn n Figure P2.23.determine

the powcr absorbed y lhc 5 f,2 csislor.+ 1 5

Figure P2-2O

2.21 In thecircuitof FigureP2.21. crcrr lrnene powerab\orbed y lhe csistorR and hepowerdelivered ylhc cuffentsourcc.

Figure P2.232.24 For hecircuitshown n FigureP2.2il. etcrminc

whichcomponenls rcsupplying oweran dwhichar edissipaling ower AIsodetermineheamount fpowerdissipatecJnclsupplied.


6/15

-l i:^.,,,ll r I+ -

Chapter f(nd rnentats f E lcc t r icCircurrs

+ l l ) \ '

t00 '

.' t-']]ll{) ' t)L-- r

Figure P2.24

2.25 Fo r hccilcuir houlr n Figurcpl. l- i. clelcminc\\ 'I rchcorlt fo|entsfe supfl! ing powcrun d$hich ar cdissipating o$cr. \ iso delcmtinc he mount o1 .powcf lr\srprted nd upllted.

2.26 l t . . tn eicclr ic hct ler fcquircs l A l r I I0 V.LL' tcfmner . Thc power i t . i i \s ip i t les ts heat or o lhc,r o\ses.b. l h e encr!\ 'd is\ l la tcd b],the hculef i jt i t l {_h

r . Thc c( ) \ t o i lhc l r tc fg \ i l lh ! , po\ \ .c lcontpluvcht|11c\ t t l te r i i le 6 r c ls/kWh.

Figute P2-27

2 , 28 A l+ - ro l t au t om o l i ven l l e r \ . i \ on | ec l cdo $ohc'adliehts.uc h hat he \\o Ioitds re n prrallcl:eacof theherdli-shrss nlcndecjo be a 75-\\ ' old.horvever. I{)( lW he|dli-chls nri\rakcrrl\ rshllcd.Whal s fhc esistancel eachhcadlighl. nduhlt isth e otal esislancece nb) lhe bulIery. l2.29 What s hc equivil lenre\ isla|ce ccrl v th ebiit lcry f Problen2.18 f two l5-W lail l ighls re

addeclin parallel) lo hc rwo7-i-W cacrrrheadlighrsl2.3O Fo r hecircuitshown n Firufe pl. i0. dercrl incthc po$ef absofbcd rv hevafiable csistof . rancir)l . ,Dr rl . ' n O. Pr , , rhe n$ , . . r b i , | ? l r , ,. r \ . rlunct iolt f R.

2.27 In th ecircuir hown n Figurep:.11.dctemrineIcmtinal oltage f thesoIrce. hc powcrsupplieclothc circuit or oad), nd he ethciencv f thecircuiAssumc ha t heonly oss s due o lhe nler-nalresistancef th csource. lf icie cy is clct ined s hefat iool loadpower o sourcc ovvcr

r = i t \ R . 5 k O / i , ( a

t l ) \

i ! )

9 ^ *t tFigure P2.3O

2 . 3 ' l Rc tc r o F i cu re t . l t .i l . t r ind hc olnlpoNer upplicd ) lhc ( leal ourcc.b. l i ind he po$erdirr ipated ncl ost \ , lrnlntenoniclea]oLrrce.


7/15

Prfl I

c. Whll is he po\lersupplied v thc source() hecircuitas ntodelcd v he oad c'si\ l iucc'.)d. Plol he ernli !l voltalcan dpowc.sLtpplied) hecifcuila\ it i tnct ion t culTe .Repert / : i ) 5. 10. (1. 0 A.

l r - 1 2 V / l \ = ( 1 . . r Q

\ o n ( l e , ' l \ o r r . .Figure p2"312,32 In thecrlcuitof Figlrre :.11. i r , : r '-1 n,1 1,"po\\! 'rdeli\ered \ rhcsourccs -1 0 t\V. ind ?. ,. r.and . C i i r cn :11 31 , 52 . /1 ,= l { )kO. 1 l i = l l 1 ( ) .

Figure p2.322.33 , \ C t r So l iWh i l c. o r r ! l i l i ' l i s h rbL r lb\ r l l c ( l l \iol lot!s:

/'r = frlcd po\\c'f = (r0 W/'orr ratcd pt icalpo*er. = ll l0 luntcns lrn) u\,0rage)I lLrr: tcn: m \V( lpe r r l i nL l i t c I 5 (X ) ( r \ e rL rge )\ r : l t l c r l opc f i l l i n ro l l i r ge l l 5 \

fh c resistancelthe li l irntcnl l lhc hulb.nleitsurcd$ it h l sl lndurd tull ir lctcr. \ 16.7e When he bulbisconreclecinl o u circuit nd is opcrtt in-rtt hc rr le(jval! le\-giyenbo\ e. dctcrninclu. hc resi\ t i lnce lthe li lanlcnt.b. The ctl icieno ol thebulb.

2.34 ,f n rncandesccnlighrhulb.r lcd r 100W $il ldi\sif l le l(X) \ t \ hell l l ld l i-ght t hci conncctcdilcfo\\ a I l0 V i( leul olt i lse oufca. l lbrecoflhe\e

bulbs rc corrnecledn series crosshe same o0rcr.dclc.nrlnehepoweretchbulbwil l di\siprte2.35 ,,fn ncandescenrightbulb ared t 60 W wil ldis\ ipatc ()W.ls heat nd igh(whenconnecledncro\s l0(lV idcalvoll i tse ource. l(XlW bulb\\ i l l dissipte 10 0W whenconncctcalucrosshe sxntc\ot lrce. l thc bI lb\ at econncclecln series cfosshrsanre ourcc. lcler 'nt inehcfo$,er ht l cither-onc l thelwo bu lhs l l d i \ s i p i r l e .2.36 Fir| hecilc it shown n FigureP2..16.n( l'l he eqLrivalcnte\ ist i incc ce n rYhc soufcc.'fhe currcnt ,'fhe powcrdclivelecl l the ource.

d. The volluges and ..c. The minintLrntoucr rat ing equiredin R .G i r e n :' : l l V R r : S O . R = 1 0 C . . : I A

b.

R

F igu re P2 . 36

2.37 Fol t ltecilcuirshoun in frgureP2.37. indlr. Thc cLrrrents iLndt,b. ' l 'hepo$cr delivercd y thc l A curcnl soufcc n(lb! l l)e ll V !oltage ourcc.. Th , t " t . r ln . $ f l , l \ , r l . r l ( ' l h_ \l . , r f i J i r .L e r i = 1 5 0 . / ? . = l 0 A R , = 5 ' } . , R r = 7 Q . l n dc \ p f c ' \ s r nd . s i r nc l i o | s l r , . H / / r .App l ) 'K ( ' l _ l hcnodcb! 'tweenR un dRr.)

F igu re P2 . 372.38 Delerminchc pou,er eliverc(l l rhc lcpcr)clenr. , ' r ' t . c' r l l ( . i . u . t , t r ! r t r ( : . { | )


8/15

Chapier l.undamenrals f Elecldc Circurrs

F igu r e P 2 .382,39 Consider iMH hobbyist arlerieshown n thecrrcuilof FigureP2.J9.

a . I f l ' r = 12 .0 Rr :0 .15 A andR r : 2 .55Q.iind the oadcunent / and hcpowcrdissipaled ythe oacl.b. lf we conncct second allcry n parallelwith

battefy thatha svoltage 'r = 12V andR: = 0.28 O. wil l thc oa dcurrenl 1 increase rdecreasc? i l l the power issipiitedy lhe oadincrease r decrease? y how much

alrrcry .1 tl! .r} rlF igu r e P 2 .39

2.40 With no load aftacheci.hc voirirgcal lhe rerninalsof i particular ower upply s 50.8V Whcna l0 Wload s altachcd, he voltagedrops o zl9 Va. DeIenrioeu5and Rs br rhisnoniderlsource_b. Whatvoltagewouldbemeasuredl lhe enninals

in thepresencef a l5-Q load esistor '.)c. Holv muchcurent could be drawn iont this powersupplyunder hon-circuit ondir ions

2.41 a 220-Vetecrr ic ealer as wohcaring oilswhrch anbc switched uch ha ieithcr oil canbe usedindependentlyr the wo ca nbeconnecledn series rparallel, ielding tolalof fourpossible onligurations.lf the warmestsetlingcorrcsponclso 2,000-Wpowerdissipation nd hecoolest onespondso J00 W ilnd

a. The aesislance f eachol the two coils.b. l'he powcr dissipalion or eachof Ihe orher wo

possiblearlangenlcnls.Sections 2.5, 2.61 Resistance andOhm's Law2.42 For hecircuils f FigureP2.42. etemtinehere\i.turvir lur:. nrludinp he pouer :rrrn!, e.e..irr tto rchic\( lh( inJi(Jledvollaee\.Rrsr.r r. uI

r v r i l r h l e n r . . r . r - . - - . . r nd . W r . r l i r E \ .

t 5 ( 2 | 7 ! ru . s r r r l l a v

I5 o

( n r = l 1 5 \

1 1 0I o r r - 1 6 l vf r = l 7 t l 2

l ( ) t r = l 0 V

R r = l t O


9/15

2.43 Fo | hc cir-cuil how in Figure ' : ll hn di l The cquivalenlesislanccec n v th c sourccb. l_hccul crll Ic. Th epo\er delivcfcd y th esourcc'( 1 .The o l t x [ e : ' 1 . -c. The minimunlpo*'cl rxt ing eqll iredbf Rr '

6 t )

/ t = 1 ( )

Figure P2.43

2.44 Fin(l heequi!r l lcnl csi\ lancc ee n r1'the oulce nl: ieurcPl. '1'1.nd u\ e re\[ lt 10 in d - II xnd I"r i )

Figure P2.44

2,45 Find hee'cluivil lentesisll lnceoc nb\ thesourccitn(i hecLtfrerrli1 l he circuitol FigurePl '1 5

r ! ) r ! )

Parr Circurts

2,46 ln thccircuitof FigLrrc 2'16' he powerahsorbedbY he i5'Q resislors I5 W li ind R

R 4C)

Figure P2.46

2.47 Fin r l the equivalcnl rcsista ice b"' t$eclr ten l l ln i i ls1lan d b in the c i rcu i t o l Figurc I )l 17

t o

t o

Figure P2.47

2.48 fb r th ecircuilsho''u n FigureP2'48 lnd heequivalcnleristancece n y lhc sourceHo N ruchpo\\'er s dcliveredbY he soLlrce

l c ) l l l l l l

i ' ( "/, | ] ! ). tQ

:l ll .l !2

l o

Figure P2.45 Figure P2.48


10/15

Chrrprer Fund.r cnt ir ls l Elecl c Cirru it \

2.49 For hecircuitshoun n FigureP2.-19.incl hccquilalcnl esisrancc.hcfc Rr = 5 Q. R. =I k O . / ? r R : : I (X)O. R ; - 9 .1O ar t dR 6 = t k o .

F igu re P2 . 492.5O ('help rcsisror\ rc lhf icntedb\ dcposil ing lhin

h)ef of crfbononk) nonconLiuct in! l) lncjr ici l lsubslrr le \e eFigure '1.50). l \uch r cylindef rsf irdius rnLl cnglh / .dclenninc he hickncss f th ct i lnr cquircd in n re\ i\(drceR i l

Assllne he esist lnce f n 1u\e l: igureP2.-5) is gi \enb \ t he xp res r i on : R r r l l+ 1 (7 7 i j ) l$ i t hT T . - A P : I , = l s ( - : 1 = ( ) . - l - lt = 0 . 35 'C / WlRl : 0 . I I g : andP i s heloue fLlissipdleLln he esislive lcmcntof the usc.Determinehe aled urrenl l which he cifcuil wil lmeltxnd open. hal s. blo$ (1/ irrt : he useblou$hcn R beconrcsnfinilc.)

R : l l k et s: 1 . 9 M J : 9 r r n r/ ) n r\cglect heenLlurt. lce\ l lhec!l i l ]derrnd i\sun1elha l thc lh icknc\s is nruch \n lx l lc f thrn thc fadi \ .

F igu re P2 . 5O2.5'l The rcsist ivc'clcntcntsl fuscs-ighlbulbs. ealefs.

c l c . . r e ig i r i hc rn l l \ on l i ea r i . e . .he es i s l r nccsdcpcnclcnln he currerl l)rcu!:h hc elemenl).

_rl [f t , .F igu re P2 . 51

2.52 U\e Kirchholl s currenlLr$ rnd Ohnr's r$ todelernlinehc cllrrcnt r eachol lhe resistors?, . Rt .i l ndR6 n he i r cu i l f F ieu rcP l . 5 l . \ r = I0 V .R - l ( ) o . R r = . + 0 Q .R , = 1 0 a . R r = R . =R 6 = t 5 O .

Figure P2.522.53 With relerenceo Problenr. lJ. useKirchholl r

cuffer1tn\\ andOh r'\ ] i lw () ind t lrc csislnncesr.Rr . /1 , .R r . and t . i l R , : 2 O Assu r re r : :R. r r r , l . R .

2 . 54 Assu r l l i n -er :2 Q. R : - 5 f ) . R, : :1 C.R . ' : I Q . R : = I Q . : = . 1 A . a n r l l , - 5 . 1V i r h ecircuilol FigureP2.ll. u:e Kirchholl s currelrh\ \rnd Ohm hw to l i ldn . / r . / r . / r . and \ . b . Rr ' .

2-55 Assunring , : I a. /? : I Q. Rr = ,1,11 .Rr :6 s2 . nd ' s = i l V in he i r cu i l l F rgu rcP1.55. seKirchh(t l \ \oll l lre ln$ ruxlOInn l ir \r oI ln t la . i , , . r , . nd , .b. Thecurrenl hrough ilch e\i\ tancc.


11/15

Prr I Cr r lu r (s

Figure P2.55

2.56 Assurn inql ) I O. Rr- I s? . / l : = 5 O.Rr . 1 A. und r : l .+V in hc i rcu i l i F igurcP1.55.use Kirchhol] s vo l l i le l i lu and Ohm s la$ tol l r ( l11 . , / . / i , . t l r l ( lb . Thc voJl l l l lc lc r l )s \e i lch fesist lnce.

2,57 Assume thlt thc vol laqe rource in the c ircLt i tol 'Figure '1.55 it norv eplaccd br a curfcnl \oUfce. DdR l j= l Q . / t = I O . R : = I Q . . I t r : . 1 . A . r n d/s : l l A. L,sc Kifchholl \ vol l lge l i r * anclOhnr'sl l l \ \ f t ) a le lcfn ine he \ o l l lL l ]c c ro \s each rc\ is lnDce.

2.58 Thc ro lr ir l le r l i \ i ( le r ncl$u.k ol FigLrr.e l .5 l i iscxpecled o ptoyrdc 5 V it l thc oulpul. The fc\ iskn.s.no\re\er. ntx\ no l bc e\ l tc t l) thc santc: hat is . the i fto lef i lnccs Ie sUch halt he fesist l lncesmll\ nol bee x i r c t l \ l S llL. I l the rcsistorshuvc +l{} percenl o lcrrnce. l ind thc

worsl cr\e output \1) l t l l l l cs.h. F ind thesc !l l i lges l in lo lcrtr)ces of 15 percenl.( j i r c n :r - l 0 V . R : 5 k e . R . : 5 k A

F igu re P2 . 58

2.59 Finci heequilt lent c\ i\ t i lnce l_thc ircuitolFicurc '1.59b\ cornbinin{ csistorsn scf ies nd inp l ' 1 r l l e l ., : , 1 O . R r l l A . R . = 8 A . R , : 2 e .i ( r = 1 6 Q . R , - 5 Q

Figure P2.592.6O Find hecqullalent csistanceec nnl l ie sollrcernd thecurreDt in thecircuitof FigureP2.60.Given:1 . , I 2 : R l : : l Q . R r 1 Q . R r = 5 0 O .R r i l g . R r l 0 Q . R : - l l 5 ) . R 6 6 O .

F igu re P2 . 6O2.61 I n t hcc i r cu i t t ) f- i gu reP2 . 6L . rhcpowerso rbedby the 0 Q resislor\ l0 W Find R Civcn: , : 5l )V .R L 2 0 5 2 . R : = 5 O . R r = 2 f 2 R j : l i O .R 5 l l O . R 6 3 0 A .

F igu re P2 . 612.62 Determinehc equrvalenlcsistancel the Dtinitcnctwork l rcsisl(n-\n lhe circuitol FigurePl.6l

Figure P2.622.63 F'or hecircuitsho$n n FieureP2.61 in. l

a. Th eequivalcnlL'\ islnncece nh),the our-ce.b. The cuffcot htuu{haDd hc powerabsofb(l b\ ' the90-Q esist irncc.iren: I r = ll0 V. R = 90 A.R: = 5 f )Q . R r :10 A. Rr = 20 Q. Rs= .10O.R6= l0 A . R r :60 Q. Rs l l 0 O .


12/15

2.66 In thebridse ircuit n FigureP2.66.fnodcs1o rternlinals)a arrd D arc shorledand

R ,= 2 2 k A: .1.7 S)

l t {k03.3 f )

resist lrnce retwecn hc nodcseler'milc'he equi!r lentor lcnrinals 4 nd B.

F igu re P2 . 632.64 ln thc ci|cuit ) l FigurcP].6.1.ind hecquivll lent

fe\ islrnce ookinq n xt . f 'ninxls / itnd if lerlninulsiincl l i lreopen n(l grrn i l lemrnrls r 'and a/ lfeshorledo-getlter.\ lso. ind heequivalcnl csistaIcelooling in i l terninals and11 flcnninal\ (/ and reopen nd il lef lr lnal\ d an(l ) rrc shofrcd r)gethe'r.

F igu re P2 . 642.65 ,qr in en:Lineeringit cwhich oLr re super'r ' isrn-u.

l-hor\efo\\ 'er lok)rnrust c iIed dj\ t ince / l ioln uponrble-sencrrt(r '(Fi!ufe l.6-5). ssuntelt elanerii lorcurlbc mocilr lcds n idcrl sourcc r ' i lh hevoltrge ivcn. 1he narneplalen he noI(r-! ivc\ helollo\r ' in! aLrd oila-qesrd th ecorf! '\ lon(linglu l l l ond un-cn t :

l ; ; = l t 0 vI r / , , . , l 0 5 V - 1 , / r r 7 . 1 0 AI u . ^ - l 1 7 V - - / v r r - 6 . - 1 7 A

Il, i : l5(l nr rnd the nolornru\tclclivcf ts ir l l-rale(lpo\\,cf. ctcfrr l inelte minimunt \W(; condLtclors$ hrchnrust c used n a ruhhcr n\ult led cublc.A \ \ume ha l l r c n l \ l ossc \n hcc i r cL r i tc cL r r - i nhc$ il-cs.

F igu re P2 . 662.67 Dctcrnrinchc oltarge el $eer) odes 1antl B in

th ccifcuil sho\!n n Figulc- 1.67.

R = l l k oR. = 220 A

R, = 6.llks )Rr= 0.21MA

Figure P2-67

2.68 Detcrnrinchc vohagc cl\ \cennodc'sl antlB ilt he i f cL r i lho$n n F igo re ] . 67 .

R : : l l l k QR r - l l k O

2.69 Detelminc hcvc)ll luc cross , ir Fi! :urcP].69.: 1 . 7m Q: l0 kr2

R : t . l k o

1 = 1 l V R LR : 3 k O R r

( r b l cF i g u r e P 2 . 6 5 F igu re P2 . 69


13/15

Pall I

Sections 2"7, 2.82 Practical Sourcesand Measuring Devices2.7O i\ thtrtnistot s a nonlincar evicewhichchinSesit s cfninal rcsistancealue ds Is surrolrndll ls

lcrnperiitufchitnges.he rc\ islancelld lcmpcfaturcserrcf ir l lyi: t \ 'c cll l ion i lhc t itn ol '

R , , { 1 ) : R 1 , t '' / / '\ \hercR, , : fe\ isfdncc 1 enuerilturc . !2

R,) rcri isl l lrrccl t tcrnperit lurc = 298 K. AI : nll leri i l constlnl.K

7_ . r = ahsoluteernperaturc.l. l f Rr = 101)Q lnd I : 0.01 K '. plol /1,r,(7asr l lnclion ol-the rrrroundingenperatturc lir-

.r50 7 : 750.b. I l lhe irernisk)f s n pit l ir l lcl \ i th a 150-Q c\ i\1or.

l ind hc erpre\sionor'1hc quivnlentcsist itnceLn dplol /?,, 7- )on the lntc griPh 1i)f ar t o.

2,71 ,\ rnoving oil rnclcrnrovenrenl l\ n nlclefresistuce '17 l0t) Q. and ir l l-scalc eilcctronscauscd ) a mctetculrenl , , , l0 /1i\ .Thc novclncnlntuslhe used o indicatc rcsstt fe lcasureclY hercnsor p ct rnarimumoi l00 kPiL- ec - igufePl.7l.ru .Drarv circuil equired odo hi\ . \ho$ing l l l

irppropfi lr leonnectionsel\ \ 'ccnhe ernlrndl\ tth! 'scnsof nclntelefnto!cntenl.

b. Deter inc haviLlue l each ontlonenl n lhcc fcur

r. Whll l \ the ineuf enge.lhrt s. he mlnt lnut l l lndnrlximulnpressurehall anaccurt lel)bcntcit \ufell, '

2.72 Thecircuitof FigureP2.72 s used o measureheinternal mpedance f a battery.The batlery beingtesleds a NiMll battery ell.

. A lreshbaftcry s being ested,and t i s fbund thatIh evoltage ,, , , i .s 2.28V with theswitchopenan d2.27V wilh theswitch loseci. ind he nternalresistancef thcbattery.

b. The sanlebatlery s testedone year ater. and yourislbund o bc 2.2 V with thc switchopenbu t0.31Vwith th cswitchclosed. ind he nlemtl resistanceof th eballen.

Figure P2.72

2.73 Considerhepract ical mmeter. escribednFigure '2.73. onsist ing f an dcalanlmeler n serrcsu' i lh a l-kQ resistor. hemclersees full-scalcdcllect ionwhen hecurrent hrough l is 30 A. Il wcdesire o conslrucl mull irange mneter eacjlnglir l l scale alues f l0 mA, 100mA , and I A.dcpending lh c setl ing f a rotar)switch. etemrineappropriatealues f Rr. R:. andRr .

Figure P2.732.74 A circuil hatmelsurcshc nternal esistancei a

pliicticxl ammeler s shown n Figure P2.7'1,whereRs = 50.000Q. l 'r = 12V an dR, , s a variablercsistorhal canbc adjusled l wil l .lu. A\slrnle hat , ,


14/15

7l l r l r , r t ' t , I u n . l i , I ( . . . . . t t - , r L C f - U . r ,

Figure P2.74

2.75 A plactrcal1)ltnrclera\ n intcrrraresrsrance,,, .Whrt i\ lhc vlluc ol r, , , l rhcD-rctcfc.r( ls LllI V$hcncL )nnec t edssho$n n F i su r c ] . 75 .

/ l \ = 1 5 l l )l r = lFigure p2.75

2-76 Lis ing Ie c i rc ! i l o l l r i !u re p2.75. l ind rhe yo l l igeth l i t hc ntc le l rcarls l i , - l -+ V in d R\ h i l r lhclb l lorv in! I ' r lues:R\ : 0.1r,, , . .-1f, , 0.6r,, , . )r , , , . 1r, , , . r , . rnd 10r . , , , .l lo$ l i l f !c (o r \n t i t l l ) \hould the rntcrnul csj\ luncc ol 'thc mctrr bc rc l l t l i \c to R\ ,

2.77 A rcltnrcte l i \ usc( l o deternt ire rnc !ot i iLqlaLcro\si l f r . \ r \ t r !e c le l l lcn t i t) the c ifcuit ol l ; i !u re p1.77. Thein\trLrn lcnt s nlodcle( l br eI idcul \o l lmcler in p i l r i t l le lwt lh l l l l0 kQ tcsi\ lor. s \hown. l -he reier is p l i lcc( llO nlersUte hc voll i lge tctos\ /?r. Assurne Rr : 8 kA .R : - l l k Q . R . - 5 0 k O . / i . = t t 5 k e . l r d/ r : I l 0 n r , \ . F i r r ( l h c \ o l l u l c i r c r o \ sR r $ i l h t r ) d|\ i lhoul lhe vojl lnc lcf in lhc c ircLt i l i r r . the bl lowing\ a l u e s :t r . R r : i 0 0 Qb . l r - l l s )c . R r - l 0 k Od . f t r : I 0 0k O

2.78 .{ l lnrnr.-ters usct l r shoun n Ligufcpl.78. Th eilDtI ltL'tetodclconsislsJ1 It dcil l mrllcler n \c es

Figure P2.77

\r r lh r re\t \ l iucc. The antr l1cter]todcl i \ l l lucd i rr hebrrnch as sho$n it) thc t jgurc. Fi l)d lhc cuftcnl Ihfou-qhR,bolh | \ ' i th and without th e umntetel n the c ifrut l tot lhclb l lo$ in - ! :v i l lues.us\Ul]t ing ht t R. = l0 O. R :t l ( x )Q . / i : = m0 o. R i : 1 . 1 o . , R , : 150O. andl r = l - l V .

b. /1. = I00 Qc . R ; = l 0 Q( t . R5: I S)

Anrmerer.rcnrl r)x\lfl

t i o

Figure P2.78

2.79 Shorvnn Firurc I)1.79s an alLrnrnuntcilnll lc\crcd ! ' i tm oadecl\ the ir |cc . SIfuin rulc.RL .Rt. R,. un dRr rrc i l l luchccltr irc Jt_. ir lnt\ \no[ l]in Fi. ! :urc 2.79 rr( l orr)ccte.!rl o lhc circLrit horvn.The orcc iru\es tcn\ ionsltcss tr he ()p l lh ebc:lnl ha tcruse\ hc len! lh (rnd hcr. l ir .e hafesislnncc)l n iud Rt k) ltcrcasc|( l r contPrcrsion\ lre\s on th ehottont l thcbeunt hit cau\e\ h. lc| lcth(lud thc.elbrchc elistnnce)olR. itnd R: lo dccral\c.Thiscauscst voltage i 50 rnV ut no(leB rr th r.sp.,ctlo rode .1.Dctcrnt ineh. ' i)rcc 1'/ 1 , I k Q I , : l t V, : 15 rrrnr /r - I00 ntf i f : 69 C\ /ml


15/15

Pr|t I

Figure p2-79

2.80 Shosl in i j isurePl.E() s I srr 'ucluratteelclnli lc\credbcarn radcd r u tbrce . Strtin gitu{esRt. Rt. /1,.u d 1itruc L rchf( l f() he bc|t it ssh{)wnan dconn(-clednk r hecr|cuitshoun. The irrcc uuseslu c-nsiorrlrr 'rsJn hc (rpof thebcitm hll crusc\ helenglh xlrd hefctolehc rc\i\t i lncc-rl Ri itnd lr to

incr'casencl compressiontrcss n thebottollt l thebeam hatcauscslte en-ethan d hereti)reheresrslaDce)f R t i lnd Rr to decrcase.hi \ gencrllesvollage etween odesB endA. Dclenninc hi sv o l tngci F : 1 . 1MN and

R , , = l k c l ' r : 1 , 1 L : t . ] t ntt, = -l cnr /r = 7 crn f : 10 0GN/rnl

F igu r e P 2 .80

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I

. / r \ . 1I t r j +I_- T-----_l1 L l

:-

ch1 problems for electrical engineering

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