Direct Current Circuits


 Edith Barker
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1 8 Direct Current Circuits CHAPTE OUTLNE 8 Electromotive Force 8 esistors in Series nd Prllel 8 Kirchho s ules 84 C Circuits 85 Electricl Meters 86 Household Wiring nd Electricl Sety ANSWES TO QUESTONS Q8 The lod resistnce in circuit is the eective resistnce o ll o the circuit elements excluding the em source n energy terms, it cn be used to determine the energy delivered to the lod by electricl trnsmission nd there ppering s internl energy to rise the temperture o the resistor The internl resistnce o bttery represents the limittion on the eiciency o the chemicl rection tht tkes plce in the bttery to supply current to the lod The em o the bttery represents its conversion o chemicl energy into energy which it puts out by electric trnsmission; the bttery lso cretes internl energy within itsel, in n mount tht cn be computed rom its internl resistnce We model the internl resistnce s constnt or given bttery, but it my increse gretly s the bttery ges t my increse somewht with incresing current demnd by the lod For lod described by Ohm s lw, the lod resistnce is precisely ixed vlue Q8 The potentil dierence between the terminls o bttery will equl the em o the bttery when there is no current in the bttery At this time, the current though, nd hence the potentil drop cross the internl resistnce is zero This only hppens when there is no lod plced on the bttery tht includes mesuring the potentil dierence with voltmeter! The terminl voltge will exceed the em o the bttery when current is driven bckwrd through the bttery, in t its positive terminl nd out t its negtive terminl Q8 No there is one bttery in circuit, the current inside it will be rom its negtive terminl to its positive terminl Whenever bttery is delivering energy to circuit, it will crry current in this direction On the other hnd, when nother source o em is chrging the bttery in question, it will hve current pushed through it rom its positive terminl to its negtive terminl Q84 Connect the resistors in series esistors o 5 kω, 75 kω nd kω connected in series present equivlent resistnce 47 kω Q85 Connect the resistors in prllel esistors o 5 kω, 75 kω nd kω connected in prllel present equivlent resistnce kω 9
2 Direct Current Circuits Q86 Q87 n series, the current is the sme through ech resistor Without knowing individul resistnces, nothing cn be determined bout potentil dierences or power Q88 n prllel, the potentil dierence is the sme cross ech resistor Without knowing individul resistnces, nothing cn be determined bout current or power Q89 n this conigurtion, the power delivered to one individul resistor is signiicntly less thn i only one equivlent resistor were used This decreses the possibility o component ilure, nd possible electricl disster to some more expensive circuit component thn resistor Q8 Ech o the two conductors in the extension cord itsel hs smll resistnce The longer the extension cord, the lrger the resistnce Tken into ccount in the circuit, the extension cord will reduce the current rom the power supply, nd lso will bsorb energy itsel in the orm o internl energy, leving less power vilble to the light bulb Q8 The whole wire is very nerly t one uniorm potentil There is essentilly zero dierence in potentil between the bird s eet Then negligible current goes through the bird The resistnce through the bird s body between its eet is much lrger thn the resistnce through the wire between the sme two points Q8 The potentil dierence cross resistor is positive when it is mesured ginst the direction o the current in the resistor Q8 The bulb will light up or while immeditely ter the switch is closed As the cpcitor chrges, the bulb gets progressively dimmer When the cpcitor is ully chrged the current in the circuit is zero nd the bulb does not glow t ll the vlue o C is smll, this whole process might occupy very short time intervl Q84 An idel mmeter hs zero resistnce An idel voltmeter hs ininite resistnce el meters cnnot ttin these vlues, but do pproch these vlues to the degree tht they do not lter the current or potentil dierence tht is being mesured within the ccurcy o the meter Hoory or experimentl uncertinty! Q85 A short circuit cn develop when the lst bit o insultion rys wy between the two conductors in lmp cord Then the two conductors touch ech other, opening lowresistnce brnch in prllel with the lmp The lmp will immeditely go out, crrying no current nd presenting no dnger A very lrge current exists in the power supply, the house wiring, nd the rest o the lmp cord up to the contct point Beore it blows the use or pops the circuit breker, the lrge current cn quickly rise the temperture in the shortcircuit pth
3 Chpter 8 Q86 A wire or cble in trnsmission line is thick nd mde o mteril with very low resistivity Only when its length is very lrge does its resistnce become signiicnt To trnsmit power over long distnce it is most eicient to use low current t high voltge, minimizing the power loss in the trnsmission line Alternting current, s opposed to the direct current we study irst, cn be stepped up in voltge nd then down gin, with higheiciency trnsormers t both ends o the power line Q87 Cr hedlights re in prllel they were in series, both would go out when the ilment o one iled An importnt sety ctor would be lost Q88 Kirchho s junction rule expresses conservtion o electric chrge the totl current into point were dierent rom the totl current out, then chrge would be continuously creted or nnihilted t tht point Kirchho s loop rule expresses conservtion o energy For singleloop circle with two resistors, the loop rule reds + ε This is lgebriclly equivlent to qε q + q, where q t is the chrge pssing point in the loop in the time intervl t The equivlent eqution sttes tht the power supply injects energy into the circuit equl in mount to tht which the resistors degrde into internl energy Q89 At their norml operting tempertures, rom P, the bulbs present resistnces 4 Ω, nd 9 Ω, nd 7 Ω The nominl 6 W lmp P 6 W 75 W W hs gretest resistnce When they re connected in series, they ll crry the sme smll current Here the highestresistnce bulb glows most brightly nd the one with lowest resistnce is intest This is just the reverse o their order o intensity i they were connected in prllel, s they re designed to be Q8 Answer their question with chllenge the student is just looking t digrm, provide the mterils to build the circuit you re looking t circuit where the second bulb relly is inter, get the student to unscrew them both nd interchnge them But check tht the student s understnding o potentil hs not been impired: i you ptch pst the irst bulb to short it out, the second gets brighter Q8 Series, becuse the circuit breker trips nd opens the circuit when the current in tht circuit loop exceeds certin preset vlue The circuit breker must be in series to sense the pproprite current (see Fig 8) Q8 The hospitl mintennce worker is right A hospitl room is ull o electricl grounds, including the bed rme your grndmother touched the ulty knob nd the bed rme t the sme time, she could receive quite jolt, s there would be potentil dierence o cross her the is DC, the shock could send her into ventriculr ibrilltion, nd the hospitl st could use the deibrilltor you red bout in Section 64 the is AC, which is most likely, the current could produce externl nd internl burns long the pth o conduction Likely no one got shock rom the rdio bck t home becuse her bedroom contined no electricl grounds no conductors connected to zero volts Just like the bird in Question 8, grnny could touch the hot knob without getting shock so long s there ws no pth to ground to supply potentil dierence cross her A new pplince in the bedroom or lood could mke the rdio lethl epir it or discrd it Enjoy the news rom Lke Wobegon on the new plstic rdio
4 Direct Current Circuits Q8 So long s you only grb one wire, nd you do not touch nything tht is grounded, you re se (see Question 8) the wire breks, let go! you continue to hold on to the wire, there will be lrge nd rther lethl potentil dierence between the wire nd your eet when you hit the ground Since your body cn hve resistnce o bout kω, the current in you would be suicient to ruin your dy Q84 Both  nd 4 lines cn deliver injurious or lethl shocks, but there is somewht better sety ctor with the lower voltge To sy it dierent wy, the insultion on  line cn be thinner On the other hnd, 4 device crries less current to operte device with the sme power, so the conductor itsel cn be thinner Finlly, s we will see in Chpter, the lst stepdown trnsormer cn lso be somewht smller i it hs to go down only to 4 volts rom the high voltge o the min power line Q85 As Luigi Glvni showed with his experiment with rogs legs, muscles contrct when electric current exists in them n electricin contcts live wire, the muscles in his hnds nd ingers will contrct, mking his hnd clench he touches the wire with the ront o his hnd, his hnd will clench round the wire, nd he my not be ble to let go Also, the bck o his hnd my be drier thn his plm, so n ctul shock my be much weker Q86 Grb n insultor, like stick or bsebll bt, nd bt or home run Hit the wire wy rom the person or hit them wy rom the wire you grb the person, you will lern very quickly bout electricl circuits by becoming prt o one Q87 A high voltge cn led to high current when plced in circuit A device cnnot supply high current or ny current unless connected to lod A more ccurte sign sying potentilly high current would just conuse the poor physics student who lredy hs problems distinguishing between electricl potentil nd current Q88 The two gretest ctors re the potentil dierence between the wire nd your eet, nd the conductivity o the kite string This is why Ben Frnklin s experiment with lightning nd lying kite ws so dngerous Severl scientists died trying to reproduce Frnklin s results Q89 Suppose ε nd ech lmp hs Ω Beore the switch is closed the current is A 6 Ω The potentil dierence cross ech lmp is A Ω 4 The power o ech lmp is A 4 8 W, totling 4 W or the circuit Closing the switch mkes the switch nd the wires connected to it zeroresistnce brnch All o the current through A nd B will go through the switch nd lmp C goes out, with zero voltge cross it With less totl resistnce, the (c) current in the bttery A becomes lrger thn beore nd () lmps A nd B get brighter (d) The 4 Ω voltge cross ech o A nd B is A Ω 6, lrger thn beore Ech converts power A 6 8 W, totling 6 W, which is (e) n increse Q8 The strter motor drws signiicnt mount o current rom the bttery while it is strting the cr This, coupled with the internl resistnce o the bttery, decreses the output voltge o the bttery below its the nominl em Then the current in the hedlights decreses
5 Chpter 8 Q8 Two runs in series: Three runs in prllel: Junction o one lit nd two runs: Gustv obert Kirchho, Proessor o Physics t Heidelberg nd Berlin, ws mster o the obvious A junction rule: The number o skiers coming into ny junction must be equl to the number o skiers leving A loop rule: the totl chnge in ltitude must be zero or ny skier completing closed pth SOLUTONS TO POBLEMS Section 8 Electromotive Force P8 () P becomes W 6 so 67 Ω so Ω nd 7 A ε + r so Ar FG P8 r 97 Ω P8 () term becomes 5 6 Ω so 79 A ε r term becomes 79 A Ω so ε 4 ε P8 The totl resistnce is 5 Ω 6 A () lmp rbtteries 5 Ω 48 Ω 4 59 Ω P P btteries totl 48 Ω 86 86% 5 Ω FG P8
6 4 Direct Current Circuits ε P84 () Here ε + r, so + r 6 5 Ω+ 8 Ω b 48 A 5 Ω 4 Then, g 48 A Let nd be the currents lowing through the bttery nd the hedlights, respectively Then, + 5 A, nd ε r r so ε b + 5 A gb 8 Ωg+ 5 Ω 6 giving 9 A Thus, 9 A 5 Ω 965 FG P84 Section 8 esistors in Series nd Prllel P85 A nd b + g 6 A b + Ωg A A b + Ω g or Ω Thereore, 6 P86 () p 7 Ω + Ω b g b g 4 Ω s Ω 4 7 Ω 99 A or 4 Ω, 9 Ω resistors Applying, 99 A 4 Ω Ω so 7 A or 7 Ω resistor 88 Ω so 88 A or Ω resistor FG P86 P87 For the bulb in use s intended, P 75 W 65 A nd 9 Ω 65 A Now, presuming the bulb resistnce is unchnged, 6 A 96 Ω Across the bulb is 9 Ω 6 A 9 A9 W so its power is P 6 78 FG P87
7 F HG ρ ρ ρ ρ P88 eq + + +, or ρ A A A A 4 ρ A A A + A + A + A e 4 j KJ 9 5 e A A A A4 j Chpter 8 5 P89 we turn the given digrm on its side, we ind tht it is the sme s igure () The Ω nd 5 Ω resistors re in series, so the irst reduction is shown in n ddition, since the Ω, 5 Ω, nd 5 Ω resistors re then in prllel, we cn solve or their equivlent resistnce s: eq 94 Ω c Ω + 5 Ω + 5 Ωh This is shown in igure (c), which in turn reduces to the circuit shown in igure (d) Next, we work bckwrds through the digrms pplying nd lterntely to every resistor, rel nd equivlent The 94 Ω resistor is connected cross 5, so the current through the bttery in every digrm is 5 Ω 9 A 94 n igure (c), this 9 A goes through the 94 Ω equivlent resistor to give potentil dierence o: 9 A 94 Ω 568 From igure, we see tht this potentil dierence is the sme cross b, the Ω resistor, nd the 5 Ω resistor Thereore, b 568 () Since the current through the Ω resistor is lso the current through the 5 Ω line b, b 568 Ω 7 A 7mA 5 b FG P89 *P8 We ssume tht the metl wnd mkes lowresistnce contct with the person s hnd nd tht the resistnce through the person s body is negligible compred to the resistnce shoes o the shoe soles The equivlent resistnce seen by the power supply is MΩ+ shoes The current through both 5 resistors is The voltmeter displys MΩ+shoes 5 Ω MΩ M MΩ + () We solve to obtin 5 M Ω MΩ + shoes MΩ5 shoes shoes b g With shoes, the current through the person s body is 5 5 µ A The current will never exceed 5 A MΩ µ
8 6 Direct Current Circuits P8 () Since ll the current in the circuit must pss through the series Ω resistor, P so eq P mx mx mx P 5 W 5 A Ω Ω+ + Ω 5 Ω mx mx 75 eq F HG K J FG P8 totl power P 5 A W P P P Ω 5 A 6 5 W 5 W P8 Using Ω, Ω, 4Ω resistors, there re 7 series, 4 prllel, nd 6 mixed combintions: Series Ω 6 Ω Ω 7 Ω 4 Ω 9 Ω 5 Ω Prllel Mixed 9 Ω 56 Ω Ω Ω Ω Ω 7 Ω 7 Ω 4 Ω 5 Ω The resistors my be rrnged in ptterns: P8 The potentil dierence is the sme cross either combintion + 5 c + 5 h F HG K J so + 5 nd Ω kω FG P8 *P84 When S is open,,, re in series with the bttery Thus: kω () A When S is closed in position, the prllel combintion o the two s is in series with,, nd the bttery Thus: kω () A When S is closed in position, nd re in series with the bttery is shorted Thus: 6 + kω () A From () nd (): kω Subtrct () rom (): kω From (): kω Answers: k Ω, k Ω, kω
9 P85 p s F + HG K J 75 Ω Ω 6 75 Ω bttery 8 67 A 675 Ω s P : P 67 A Ω P 4 W in Ω P 67 A 4 A 8 4 W in 4 Ω 4 67 A Ω 5, 4 67 A 4 Ω 67 8 p 4 b g P W in Ω Ω bg P 4 W in Ω Ω b g Chpter 8 7 FG P85 P86 Denoting the two resistors s x nd y, x+ y69, nd 5 y + x x + x x x69 x x 69x ± x x 47 Ω y Ω x *P87 A certin quntity o energy E int Ptime is required to rise the temperture o the wter to C For the power delivered to the heters we hve P where is constnt t t Thus compring coils nd, we hve or the energy Then () When connected in prllel, the coils present equivlent resistnce p + + Now For the series connection, s + + nd ts t t tp t tp t t s
10 8 Direct Current Circuits P88 () : Ω Ω A 5 A P : P A Ω P 5 A Ω P 99 W P 49 5 W The  Ω resistor uses more power P + P 48 W P 45 48W (c) s + Ω+ Ω Ω : Ω, so A P : P A Ω P A Ω P W P W FG P88() FG P88(c) The  Ω resistor uses more power (d) P + P + A Ω W b g P A W (e) The prllel conigurtion uses more power *P89 () The resistors,, nd 4 cn be combined to single resistor This is in series with resistor, with resistnce, so the equivlent resistnce o the whole circuit is n series, potentil dierence is shred in proportion to the resistnce, so resistor gets o the bttery voltge nd the 4 prllel combintion get o the bttery voltge This is the potentil dierence cross resistor 4, but resistors nd must shre this voltge goes to nd to The rnking by potentil dierence is > > > 4 (c) (d) Bsed on the resoning bove the potentil dierences re ε ε 4ε ε,,, All the current goes through resistor, so it gets the most The current then splits t the prllel combintion esistor 4 gets more thn hl, becuse the resistnce in tht brnch is less thn in the other brnch esistors nd hve equl currents becuse they re in series The rnking by current is > 4 > esistor hs current o Becuse the resistnce o nd in series is twice tht o resistor 4, twice s much current goes through 4 s through nd The current through the resistors re,, 4 continued on next pge
11 Chpter 8 9 (e) () ncresing resistor increses the equivlent resistnce o the entire circuit The current in the circuit, which is the current through resistor, decreses This decreses the potentil dierence cross resistor, incresing the potentil dierence cross the prllel combintion With lrger potentil dierence the current through resistor 4 is incresed With more current through 4, nd less in the circuit to strt with, the current through resistors nd must decrese To summrize, increses nd,, nd decrese 4 resistor hs n ininite resistnce it blocks ny current rom pssing through tht brnch, nd the circuit eectively is just resistor nd resistor 4 in series with the bttery The circuit now hs n equivlent resistnce o 4 The current in the circuit drops to o the originl 4 current becuse the resistnce hs incresed by 4 All this current psses through resistors nd 4, nd none psses through or Thereore,, Section 8 Kirchho s ules P so 74 A + A 74 + so 9 A ε 6 + ε 9 5 FG P8 P8 We nme currents,, nd s shown From Kirchho s current rule, + Applying Kirchho s voltge rule to the loop contining nd, Applying Kirchho s voltge rule to the loop contining nd, Solving the bove liner system, we proceed to the pir o simultneous equtions: S or S T nd to the single eqution T A Then 846 A Ω nd + give 846 ma, 46 ma, A All currents re in the directions indicted by the rrows in the circuit digrm FG P8
12 4 Direct Current Circuits P8 The solution igure is shown to the right P8 We use the results o Problem 8 A s kj FG P8 () By the 4 bttery: U t 4 46 A s J By the  bttery: 88 A Ω s J A Ω s J A Ω s J A Ω s J By the 8Ω resistor: t 846 A 8 Ω s 687J By the 5Ω resistor: By the Ω resistor: By the Ω resistor: 66 By the Ω resistor: 5 (c) J + 88 kj 66 kj rom chemicl to electricl 687 J + 8 J J + 66 J + 5 J 66 kj rom electricl to internl P84 We nme the currents,, nd s shown k Ω kω [] 7 6 k Ω kω [] [] + () Substituting or nd solving the resulting simultneous equtions yields 85 ma through 69 ma through b b b 8 ma through c 6 8 ma k Ω 69 g g g Point c is t higher potentil FG P84
13 Chpter 8 4 P85 Lbel the currents in the brnches s shown in the irst igure educe the circuit by combining the two prllel resistors s shown in the second igure Apply Kirchho s loop rule to both loops in Figure to obtin: nd With Ω, simultneous solution o these equtions yields: ma nd ma b g From Figure, c c 4 Thus, rom Figure (), 4 6 ma 4 4 Ω Finlly, pplying Kirchho s point rule t point in Figure () gives: 4 6 ma ma + 5 ma, () FG P85 or 5 ma rom point to point e P86 Nme the currents s shown in the igure to the right Then w+ x+ z y Loop equtions re w x 8 x y + 6 y 7 z+ 8 Eliminte y by substitution Eliminte x S T S T x 5 w+ 5 4 x w z 44 w x 9 z 5 7w z 4 7 w 9 z Eliminte z 7 5 5w to obtin 4 7 w w FG P86 7 w 7 A upwrd in Ω Now z 4 A upwrd in 7 Ω x A upwrd in 8 Ω y 8 A downwrd in Ω nd or the Ω, A Ω
14 4 Direct Current Circuits P87 Using Kirchho s rules, b g b 6 g + 6 nd FG P87 b g Solving simultneously, 8 A downwrd in the ded bttery nd 7 A downwrd in the strter The currents re orwrd in the live bttery nd in the strter, reltive to norml strting opertion The current is bckwrd in the ded bttery, tending to chrge it up P88 b + b b b g b g b g b g Let A, x, nd y Then, the three equtions become: b xy, or y x b b 4 x+ 6 y+ 5 nd b 8 8 x+ 5 y Substituting the irst into the lst two gives: 7 b 8 x+ 5 nd 6 b x+ 8 Solving these simultneously yields b 7 7 Then, b 7 b 7 A or b 7 7 Ω P89 We nme the currents,, nd s shown FG P88 () + Counterclockwise round the top loop, Ω 4 Ω Trversing the bottom loop, 8 6 Ω + Ω 4, +, nd 99 ma 99 A Ω b 8 b FG P89
15 Chpter 8 4 P8 We pply Kirchho s rules to the second digrm 5 () + () + () Substitute () into (), nd solve or,, nd A; 5 A; 5 A Then pply P to ech resistor: : P Ω Ω Ω A 8W F 4 Ω: P H G 5 A K J 4 Ω 5 W (Hl o goes through ech) : P Ω 5 A Ω 45W Ω FG P8 Section 84 C Circuits 6 6 e je j 6 e j P8 () C Ω 5 F 5 s Q Cε 5 C 5µ C (c) t ε e tc P8 () t e tc t Q C F L HG K J O exp e je NM P jq 6 5 C 9 b Ωge Fj 96 A exp qt Qe tc L NM b5 Cg 96 A O P Q L 6 O 8 s e NM P 9 Ω FjQ 6 9 s 9 b Ωge Fj 6 6 P ma P 46 µ A µ exp P 5 µc FG P8 (c) The mgnitude o the mximum current is 96 A P8 U C nd Q C Thereore, U Q nd when the chrge decreses to hl its originl vlue, the stored energy is onequrter its originl vlue: U U C 4
16 44 Direct Current Circuits P84 qt Q e tc so 6 e 9 C 9 ln 4 or qt e Q 9 C e tc thus C C ln s *P85 We re to clculte z tc tc tc e z C dt dt e C P86 () τ C 5 Ω F 5 s 5 6 e je j 5 6 e je j τ Ω F s F HG K J + C C e C C e e (c) The bttery crries current 5 µ A Ω The kω crries current o mgnitude e Ω e So the switch crries downwrd current s µ A + b µ Age F HG K J t s P87 () Cll the potentil t the let junction L nd t the right Ater long time, the cpcitor is ully chrged L 8 becuse o voltge divider: L Likewise, L A 5 Ω A Ω 8 F Ω HG + K J Ω 8 Ω or A Ω 8 Ω A Thereore, 8 6 edrw the circuit L 9 Ω + 6 Ω b g b g C 6 6 s nd e tc so t Cln 8 9 µ s 6 Ω FG P87() FG P87
17 *P88 () We model the person s body nd street shoes s shown For the dischrge to rech, tc tc qt Qe Ct Ce e tc e + tc t C t ln F H G K J F H G K J e F j H G K J 6 Cln 5 Ω F ln 9 s 6 t A e C jln 78 µ s 6 6 e je j P89 () τ C 4 Ω F s Chpter pf 8 pf 5 MΩ FG P88() ε e tc e t 4 6 s tc 6 t q Cε e e t t q 6 µ C e µ Ae FG P89 P84 Q C Then, i q() t nd Qe tc When t () Thus, b t () e tc t () b g, then e tc g e b tc g t C ln ln F H G K J t C ln Section 85 Electricl Meters P84 grg egj p, or p e r g g g j e g g j 6 Ω Thereore, to hve A ma when g 5 ma: 5 ma 6 Ω p 99 5 ma Ω FG P84
18 46 Direct Current Circuits e j P84 Applying Kirchho s loop rule, 75 Ω + Thereore, i A when g 5 ma, p e 75 Ω 5 A 75 Ω A 5 A g g j e j g g p Ω FG P84 P84 Series esistor oltmeter s : Solving, s 6 6 kω b g FG P84 P844 () n Figure (), the em sees n equivlent resistnce o Ω 6 Ω Ω Ω A A 6 Ω A 8 Ω 8 Ω 8 Ω () (c) FG P844 The terminl potentil dierence is A 8 Ω 5 4 F HG b g n Figure, eq Ω 8 Ω ΩKJ The equivlent resistnce cross the em is 78 9 Ω+ 5 Ω+ Ω Ω ε 6 The mmeter reds 67 A 9889 Ω nd the voltmeter reds 67 A 78 9 Ω (c) F HG b g KJ n Figure (c), Ω Ω Ω Thereore, the em sends current through tot Ω+ Ω Ω The current through the bttery is 6 68 A 9889 Ω but not ll o this goes through the mmeter b g The voltmeter reds 68 A Ω The mmeter mesures current 5966 Ω 9898 A 85 The connection shown in Figure (c) is better thn tht shown in Figure or ccurte redings
19 Chpter 8 47 P845 Consider the circuit digrm shown, relizing tht g ma For the 5 ma scle: 4 ma b + + g ma 5 Ω F FG P845 or HG K J Ω () 4 For the 5 ma scle: 49 ma + ma 5 Ω + b g b g b g Ω () b g or For the ma scle: 99 ma ma 5 Ω + + or 99 5 Ω + + () Solving (), (), nd () simultneously yields 6 Ω, 6 Ω, 5 Ω P846 e jb g () A + 6 Ω Ω 9 94 kω e jb g 5 A Ω kω e jb g (c) A Ω 5 kω P847 Ammeter: r 5 A Ω g e gj () gbr+ Ωg () or g r+ Ω oltmeter: 5 Solve () nd () simultneously to ind: g 756 ma nd r 45 Ω FG P846 FG P847 Section 86 Household Wiring nd Electricl Sety 8 F P848 () P H G ρ A K J e Ω m j t b m tg A πe5 mj P Ω W W
20 48 Direct Current Circuits P849 () P : So or the Heter, P 5 W 5 A 75 W For the Toster, 65 A W And or the Grill, 8 A A The current drw is greter thn 5 mps, so this circuit breker would not be suicient Al P85 so Al Cu Cu Al Cu ρcu Cu ρ 8 Al Al Cu A P85 () Suppose tht the insultion between either o your ingers nd the conductor djcent is chunk o rubber with contct re 4 mm nd thickness mm ts resistnce is ρ e Ω m je mj 5 Ω A 6 4 m The current will be driven by through totl resistnce (series) Ω+ Ω+ 5 Ω 5 Ω t is: ~ ~ 4 A 5 5 Ω The resistors orm voltge divider, with the center o your hnd t potentil h, where h is the potentil o the hot wire The potentil dierence between your inger nd thumb is ~ e 4 A je 4 Ωj ~ So the points where the rubber meets your ingers re t potentils o ~ h + nd ~ h Additionl Problems P85 The set o our btteries boosts the electric potentil o ech bit o chrge tht goes through them by The chemicl energy they store is U q 4 C b6 J C g 44 J The rdio drws current 6 Ω A So, its power is P 6 b A g 8 W 8 J s Then or the time the energy lsts, we hve P E t : t E 44J 8 s P 8 J s Q Q 4 C We could lso compute this rom : t 8 s h t A
21 F + H G K J ε ε ε P85, so P or r + r + r P Let x ε, then r x P + or + rx r With r Ω, this becomes + 4 x 44 which hs solutions o Chpter 8 49, 4 x 4 x 576 ± () With ε 9 nd P 8 W, x 66 : Ω or ± Ω For ε 9 nd P W, x ε P 99 + ± ± The eqution or the lod resistnce yields complex number, so there is no resistnce tht will extrct W rom this bttery The mximum power output occurs when r Ω, nd tht mximum is: P mx 7 6 W 4r P854 Using Kirchho s loop rule or the closed loop, + 4, so A b + 4 A 4 Ω Ω 4 Thus, b 4 nd point is t the higher potentil ε P855 () eq ε P series ε ε eq + + b g b g b g ε P prllel ε ε (c) Nine times more power is converted in the prllel connection *P856 () We model the genertor s constntvoltge power supply Connect two light bulbs cross it in series Ech bulb is designed to crry current P W 8 A Ech hs resistnce 44 Ω n the 4 circuit the equivlent 8 A resistnce is 44 Ω+ 44 Ω 88 Ω The current is 4 88 Ω 8 A nd the genertor delivers power P 8 A 4 W FG P856() continued on next pge
22 5 Direct Current Circuits The hot pot is designed to crry current P 5 W 47 A t hs resistnce 8 8 Ω 47 A 88 Ω 4 44 Ω FG P856 n terms o current, since 47 A 5, we cn plce ive light bulbs in prllel nd the hot 8 A pot in series with their combintion The current in the genertor is then 47 A nd it delivers power P 4 7 A 4 W ε P857 The current in the simple loop circuit will be + r () ter r ε ε + r nd ter ε s ε + r nd ε s r (c) P ε + r dp ε ε + d + r + r FG P857 Then + r nd r e j L s 6 s O e NM Ωj e QP P858 The potentil dierence cross the cpcitor t e tc mx Using Frd s Ω, 4 Thereore, 4 e e 5 Ωj Or Tking the nturl logrithm o both sides, e e j 5 Ω 6 5 Ω ln 6 Ω 5 nd Ω 587kΩ ln 6 5
23 Chpter 8 5 P859 Let the two resistnces be x nd y Ps 5 W Then, s x+ y 9 Ω y 9 Ω x 5 A nd p xy Pp 5 W Ω x + y 5 A x 9 Ω x so Ω x 9 x+ 8 x+ 9 Ω x Fctoring the second eqution, x6 x so x 6 Ω or x Ω Then, y 9 Ω x gives y Ω or y 6 Ω x y x y FG P859 The two resistnces re ound to be 6 Ω nd Ω P86 Let the two resistnces be x nd y s Then, s x+ y P nd xy p p P x + y Ps From the irst eqution, y x, nd the second e e s j j x Ps x Pp becomes or x x+ P x P PP s s p x 4 F H G K J + Ps ± Ps 4PP s p Using the qudrtic ormul, x x y x y FG P86 P P s s Ps 4PP s p Then, y x gives y The two resistnces re c h b g P86 () ε ε + ε P + P 4PP s s s p nd s s s p P P 4P P 4 4 A Ω ; so 44 Ω nside the supply, P 4 A Ω W nside both btteries together, P 4 A 6 Ω 9 6 W For the limiting resistor, P A Ω W b g (c) P ε + ε 4 A W
24 5 Direct Current Circuits *P86 () FG P86() The power delivered to the pir is P + + we wnt to ind such tht d P d dp + d + b g + This is the sme condition s tht ound in prt () b g For minimum power P86 Let m mesured vlue, ctul vlue, current through the resistor current mesured by the mmeter () When using circuit (), b g or But since nd, we hve m m L N M O Q P FG P86 () nd b g m () m b mg 5 When > m, we require Thereore, m b g nd rom () we ind 5 Ω When using circuit, 5 Ω But since m 5 + () m When m >, we require bm g 5 From () we ind Ω
25 P864 The bttery supplies energy t chnging rte Then the totl energy put out by the bttery is z z ε t de C C de dt z F Chpter 8 5 e C H G P ε ε ε K J z ε t de exp C dt F exphg K J exp The power delivered to the resistor is t F dt t C C C HG C K J F ε HG C K J ε ε So the totl internl energy ppering in the resistor is z z F HG K J F HG de dt z F HG ε t P exp C z ε t de exp C dt K J F HG K J F HG K J ε C t dt ε C t ε C ε C de exp exp C C C F HG The energy inlly stored in the cpcitor is U C Cε Thus, energy o the circuit is conserved ε C ε C+ ε C nd resistor nd cpcitor shre eqully in the energy rom the bttery P865 () q C e tc (c) e j L O 6 NM 6 6 e j e je j QP q F e 99 µ C dq dt F H G K J du dt F e tc 5 8 e HG K J 7 A 7 na 6 Ω d F dt H G F HG q C q dq KJ F H G C K J dt 6 e q C F H G K J du dt 99 C A 4 W 4 nw 6 C KJ e j 8 7 (d) P bttery ε 7 A 7 W 7nW j K J K J F HG K J
26 54 Direct Current Circuits P866 Strt t the point when the voltge hs just reched nd the switch hs just closed The voltge is nd is decying towrds with time constnt C tc C t e QP We wnt to know when C t will rech L O Thereore, e tc NM L NM O QP or e tc or t Cln Ater the switch opens, the voltge is b g C t e L NM O oltge controlled switch C FG P866 c, incresing towrd with time constnt + C t C QP + When C t e t + C So t + C b g or e t b + g C b g ln nd T t + t b + gcln P867 () First determine the resistnce o ech light bulb: P + b g : 4 Ω P 6 W We obtin the equivlent resistnce eq o the network o light bulbs by identiying series nd prllel equivlent resistnces: FG P867 eq + 4 Ω+ Ω 6 Ω + b g b g The totl power dissipted in the 6 Ω is P 6 Ω eq 4 W The current through the network is given by P eq : P 4 W A 6 Ω F The potentil dierence cross is 4 Ω 8 H G A K J The potentil dierence cross the prllel combintion o nd is F 4 H G A K J 4Ω + 4Ω F HG b g b gkj eq
27 Chpter 8 55 *P868 () With the switch closed, current exists in simple series circuit s shown The cpcitors crry no current For we hve P P 4 A 8 5 ma 7 A The potentil dierence cross nd C is e jb g 85 A 4 A 74 The chrge on C 6 e j µ Q C C 74 C The potentil dierence cross nd C is 85 A 7 Ω The chrge on C e jb g 6 e j µ Q C 6 C 778 C The bttery em is eq + 85 A A 4 b g b g FG P868() n equilibrium ter the switch hs been opened, no current exists The potentil dierence cross ech resistor is zero The ull 4 ppers cross both cpcitors The new chrge C 6 e j µ Q C 6 C 4 C or chnge o µ C 778 µ C 444 µ C FG P868 *P869 The bttery current is ma ma () (c) The resistor with highest resistnce is tht crrying 4 ma Doubling its resistnce will reduce the current it crries to ma Then the totl current is FG P ma ma, nerly the sme s beore The rtio is 99 The resistor with lest resistnce crries 5 ma Doubling its resistnce chnges this current to 75 ma nd chnges the totl to ma 8 ma The rtio is 8 reduction (5% insted o 9%) 648, representing much lrger This problem is precisely nlogous As bttery mintined potentil dierence in prts () nd, urnce mintins temperture dierence here Energy low by het is nlogous to current nd tkes plce through therml resistnces in prllel Ech resistnce cn hve its vlue incresed by dding insultion Doubling the therml resistnce o the ttic door will produce only negligible (9%) sving in uel Doubling the therml resistnce o the ceiling will produce much lrger sving The ceiling originlly hs the smllest therml resistnce
28 56 Direct Current Circuits *P87 From the hint, the equivlent resistnce o Tht is, T + + L eq Leq T + eq L + eq eq Only the + sign is physicl: T L T eq L eq L eq eq eq T eq T L eq b T T T L ± For exmple, i T Ω F H eq T L T T And L Ω, eq 5 Ω P87 () Ater stedystte conditions hve been reched, there is no DC current through the cpcitor Thus, or : stedystte K g b g For the other two resistors, the stedystte current is simply determined by the 9 em cross the kω nd 5kΩ resistors in series: For nd : ε 9 b+ g µ A bstedystteg k Ω kω Ater the trnsient currents hve cesed, the potentil dierence cross C is the sme s the potentil dierence cross b g becuse there is no voltge drop cross Thereore, the chrge Q on C is b g b gb g Q C C µ F µ A 5 kω 5 µ C FG P87 continued on next pge
29 Chpter 8 57 (c) When the switch is opened, the brnch contining is no longer prt o the circuit The cpcitor dischrges through b + g with time constnt o b + gc 5 k Ω+ k Ω b µ F g 8 s The initil current i in this dischrge circuit is determined by the initil potentil dierence cross the cpcitor pplied to b + g in series: C b µ A g5 kω i 78 µ A k Ω+ kω b g b g FG P87(c) Thus, when the switch is opened, the current through chnges instntneously rom µ A (downwrd) to 78 µ A (downwrd) s shown in the grph Thereter, it decys ccording to e 78 µ A e or t> b g b g t + C t 8 s i (d) The chrge q on the cpcitor decys rom Q i to Q i 5 b t + C q Qie Qi t Qe b 8 sg i 5 t 8 s 5 e t ln5 8 ms t 8 s ln5 9 ms g ccording to *P87 () First let us ltten the circuit on D plne s shown; then reorgnize it to ormt esier to red Notice tht the ive resistors on the top re in the sme connection s those in Exmple 85; the sme rgument tells us tht the middle resistor cn be removed without ecting the circuit The remining resistors over the three prllel brnches hve equivlent resistnce F HG K J eq Ω So the current through the bttery is 4 A eq 5 Ω FG P87()
30 58 Direct Current Circuits P87 ε e tc so ln ε A plot o ln F HG K J F H G C K J t F ε HG K J versus t should be stright line with slope equl to C Using the given dt vlues: () A lestsqure it to this dt yields the grph bove 4 x i 8, x i 86, xy i i 44, y i 4, N 8 c h c hc h e ij c ih e xijc yihc xihc xiyih Ne xijc xih N xiyi xi yi Slope 8 N x x ntercept 88 FG P87 F HG K J b g + ε The eqution o the best it line is: ln 8 t 88 b g ts ln ε Thus, the time constnt is τ C 84 7 s slope 8 τ 84 7 s nd the cpcitnce is C 6 Ω 847 µ F P874 () For the irst mesurement, the equivlent circuit is s shown in Figure + b y y y so y () For the second mesurement, the equivlent circuit is shown in Figure Thus, c y + x () Substitute () into () to obtin: F x H G K J +, or x 4 Ω nd 6 Ω, then x 75 Ω y y c x Figure y Figure FG P874 The ntenn is indequtely grounded since this exceeds the limit o Ω c x b y
31 Chpter 8 59 P875 The totl resistnce between points b nd c is: k Ω kω kω k Ω+ kω The totl cpcitnce between points d nd e is: C µ F + µ F 5 µ F The potentil dierence between point d nd e in this series C circuit t ny time is: e tc t 6 ε e Thereore, the chrge on ech cpcitor between points d nd e is: t 6 t 6 q C µ F e 4µ C e b g b g t b F g b Cg 6 t 6 nd q C µ e 6µ e kω b c d e kω S C µf C µf +  FG P875 *P876 () Let i represent the current in the bttery nd i c the current chrging the cpcitor Then i i c is the current in the voltmeter The loop rule pplied to the inner loop is + ε i The loop C dq rule or the outer perimeter is ε i bi icgr With ic, this becomes ε + dt dq dt r ε q Between the two loop equtions we eliminte i C by substitution to obtin F ε q dq ε + r HG K J + C dt r + r + r ε ε + + C q dq dt r r q r dq ε r C + r dt This is the dierentil eqution required To solve we ollow the sme steps s on pge 875 dq ε + r dt rc q + rf rc q ε rc + r K J z HG z q t q t dq + r q rc r rc dt q ε rc + r ln + HG + r K J ε rc t F q ε rc+ r r rc r rc t q rc rc r r e r rc t HG + KJ + ε + ε + ln ε + r q r C e teq C r ε + e j where eq + r q r The voltge cross the cpcitor is C e t eq ε C C r + e j F (c) the switch is then opened, r rε As t the cpcitor voltge pproches ε r+ r + the cpcitor dischrges through the voltmeter ts voltge decys exponentilly ccording rε to r e trc +
32 6 Direct Current Circuits ANSWES TO EEN POBLEMS P8 () 79 A; 4 P84 Ω P84 () 4 ; 9 65 P86 () 7 Ω ; 99 A in 4 Ω nd 9 Ω; 7 A in 7 Ω; 88 A in Ω P P8 () see the solution; no P8 see the solution P84 kω; kω; kω P844 () ma, 5 4 ; 67 ma, ; (c) ma ; P846 see the solution P848 () W; W P85 55 A P85 h P854 is 4 higher P86 47 Ω nd Ω P856 () see the solution; 8 ma; W; P88 () Ω ; nd (d) see the solution; see the solution; 47 A; kw (c) Ω; (e) Prllel P kω P8 74 ma ; 9 A; ε 6 P8 see the solution P86 Ps + Ps 4PP s p nd P P s s 4 P s P p P84 () 85 ma in ; 69 ma in ; 8 ma in ; c higher by 69 P86 () b + g ; + P86 A up in Ω ; 4 A up in 7 Ω ; see the solution A up in 8 Ω ; 8 A down in Ω; P864 see the solution P88 see the solution P866 b + gcln P8 8 W to the lethnd resistor; 5 W to ech 4 Ω; 45 W to the righthnd resistor P8 () 6 6 ma ; 5 µ C; (c) 96 A P84 98 s P868 () µ C ; increse by 444 µ C P87 see the solution P87 () 5 Ω; 4 A P874 () x 4 ; no; x 75 Ω P86 () 5 s; s; s (c) µ A + b µ Age t rε P876 () nd see the solution; (c) r e trc P88 () 9 s; 78 ms + P84 t C ln
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