27-1 CIRCUITS Pumping Charges WHAT IS PHYSICS? CHAPTER

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1 CICUITS 7- WHAT IS PHYSICS? You numer of electrcl devces you own nd mght even crry mentl lst of the devces you wsh you owned. Every one of those devces, s well s the electrcl grd tht powers your home, depends on modern electrcl engneerng. We cnnot esly estmte the current fnncl worth of electrcl engneerng nd ts products, ut we cn e certn tht the fnncl worth contnues to grow yerly s more nd more tsks re hndled electrclly. dos re now tuned electronclly nsted of mnully. Messges re now sent y eml nsted of through the postl system. eserch journls re now red on computer nsted of n lrry uldng, nd reserch ppers re now coped nd fled electronclly nsted of photocoped nd tucked nto flng cnet. The sc scence of electrcl engneerng s physcs. In ths chpter we cover the physcs of electrc crcuts tht re comntons of resstors nd tteres (nd, n Secton 7-9, cpctors). We restrct our dscusson to crcuts through whch chrge flows n one drecton, whch re clled ether drect-current crcuts or DC crcuts. We egn wth the queston: How cn you get chrges to flow? re surrounded y electrc crcuts. You mght tke prde n the CHAPTE 7 7- Pumpng Chrges If you wnt to mke chrge crrers flow through resstor, you must estlsh potentl dfference etween the ends of the devce. One wy to do ths s to connect ech end of the resstor to one plte of chrged cpctor.the troule wth ths scheme s tht the flow of chrge cts to dschrge the cpctor, quckly rngng the pltes to the sme potentl. When tht hppens, there s no longer n electrc feld n the resstor, nd thus the flow of chrge stops. To produce stedy flow of chrge, you need chrge pump, devce tht y dong work on the chrge crrers mntns potentl dfference etween pr of termnls.we cll such devce n emf devce, nd the devce s sd to provde n emf, whch mens tht t does work on chrge crrers. An emf devce s sometmes clled set of emf.the term emf comes from the outdted phrse electromotve force, whch ws dopted efore scentsts clerly understood the functon of n emf devce. In Chpter 6, we dscussed the moton of chrge crrers through crcut n terms of the electrc feld set up n the crcut the feld produces forces tht move the chrge crrers. In ths chpter we tke dfferent pproch: We dscuss the moton of the chrge crrers n terms of the requred energy n emf devce supples the energy for the moton v the work t does. A common emf devce s the ttery, used to power wde vrety of mchnes from wrstwtches to sumrnes. The emf devce tht most nfluences our dly lves, however, s the electrc genertor, whch, y mens of electrcl connectons (wres) from genertng plnt, cretes potentl dfference n our 705

2 706 CHAPTE 7 CICUITS homes nd workplces. The emf devces known s solr cells, long fmlr s the wng-lke pnels on spcecrft, lso dot the countrysde for domestc pplctons. Less fmlr emf devces re the fuel cells tht power the spce shuttles nd the thermoples tht provde onord electrcl power for some spcecrft nd for remote sttons n Antrctc nd elsewhere. An emf devce does not hve to e n nstrument lvng systems, rngng from electrc eels nd humn engs to plnts, hve physologcl emf devces. Although the devces we hve lsted dffer wdely n ther modes of operton, they ll perform the sme sc functon they do work on chrge crrers nd thus mntn potentl dfference etween ther termnls. The world s lrgest ttery energy storge plnt (dsmntled n 996) connected over 8000 lrge led-cd tteres n 8 strngs t 000 V ech wth cplty of 0 MW of power for 4 hours. Chrged up t nght, the tteres were then put to use durng pek power demnds on the electrcl system. (Courtesy Southern Clforn Edson Compny) Fg. 7- A smple electrc crcut, n whch devce of emf does work on the chrge crrers nd mntns stedy current n resstor of resstnce. ' 7-3 Work, Energy, nd Emf Fgure 7- shows n emf devce (consder t to e ttery) tht s prt of smple crcut contnng sngle resstnce (the symol for resstnce nd resstor s ). The emf devce keeps one of ts termnls (clled the postve termnl nd often leled ) t hgher electrc potentl thn the other termnl (clled the negtve termnl nd leled ).We cn represent the emf of the devce wth n rrow tht ponts from the negtve termnl towrd the postve termnl s n Fg. 7-. A smll crcle on the tl of the emf rrow dstngushes t from the rrows tht ndcte current drecton. When n emf devce s not connected to crcut, the nternl chemstry of the devce does not cuse ny net flow of chrge crrers wthn t. However, when t s connected to crcut s n Fg. 7-, ts nternl chemstry cuses net flow of postve chrge crrers from the negtve termnl to the postve termnl, n the drecton of the emf rrow. Ths flow s prt of the current tht s set up round the crcut n tht sme drecton (clockwse n Fg. 7-). Wthn the emf devce, postve chrge crrers move from regon of low electrc potentl nd thus low electrc potentl energy (t the negtve termnl) to regon of hgher electrc potentl nd hgher electrc potentl energy (t the postve termnl). Ths moton s just the opposte of wht the electrc feld etween the termnls (whch s drected from the postve termnl towrd the negtve termnl) would cuse the chrge crrers to do. Thus, there must e some source of energy wthn the devce, enlng t to do work on the chrges y forcng them to move s they do. The energy source my e chemcl, s n ttery or fuel cell. It my nvolve mechncl forces, s n n electrc genertor. Temperture dfferences my supply the energy, s n thermople; or the Sun my supply t, s n solr cell. Let us now nlyze the crcut of Fg. 7- from the pont of vew of work nd energy trnsfers. In ny tme ntervl dt, chrge dq psses through ny cross secton of ths crcut, such s.ths sme mount of chrge must enter the emf devce t ts low-potentl end nd leve t ts hgh-potentl end. The devce must do n mount of work dw on the chrge dq to force t to move n ths wy. We defne the emf of the emf devce n terms of ths work: dw dq (defnton of ). (7-) In words, the emf of n emf devce s the work per unt chrge tht the devce does n movng chrge from ts low-potentl termnl to ts hgh-potentl termnl. The SI unt for emf s the joule per coulom; n Chpter 4 we defned tht unt s the volt. An del emf devce s one tht lcks ny nternl resstnce to the nternl movement of chrge from termnl to termnl. The potentl dfference etween the termnls of n del emf devce s equl to the emf of the devce. For exm-

3 7-4 CALCULATING THE CUENT IN A SINGLE-LOOP CICUIT 707 PAT 3 ple, n del ttery wth n emf of.0 V lwys hs potentl dfference of.0 V etween ts termnls. A rel emf devce, such s ny rel ttery, hs nternl resstnce to the nternl movement of chrge. When rel emf devce s not connected to crcut, nd thus does not hve current through t, the potentl dfference etween ts termnls s equl to ts emf. However, when tht devce hs current through t, the potentl dfference etween ts termnls dffers from ts emf. We shll dscuss such rel tteres n Secton 7-5. When n emf devce s connected to crcut, the devce trnsfers energy to the chrge crrers pssng through t. Ths energy cn then e trnsferred from the chrge crrers to other devces n the crcut, for exmple, to lght ul. Fgure 7- shows crcut contnng two del rechrgele (storge) tteres A nd B, resstnce,nd n electrc motor M tht cn lft n oject y usng energy t otns from chrge crrers n the crcut. Note tht the tteres re connected so tht they tend to send chrges round the crcut n opposte drectons. The ctul drecton of the current n the crcut s determned y the ttery wth the lrger emf, whch hppens to e ttery B, so the chemcl energy wthn ttery B s decresng s energy s trnsferred to the chrge crrers pssng through t. However, the chemcl energy wthn ttery A s ncresng ecuse the current n t s drected from the postve termnl to the negtve termnl. Thus, ttery B s chrgng ttery A. Bttery B s lso provdng energy to motor M nd energy tht s eng dsspted y resstnce.fgure 7- shows ll three energy trnsfers from ttery B; ech decreses tht ttery s chemcl energy. m Chemcl energy lost y B M A () A B B Work done y motor on mss m Therml energy produced y resstnce Chemcl energy stored n A 7-4 Clcultng the Current n Sngle-Loop Crcut We dscuss here two equvlent wys to clculte the current n the smple sngleloop crcut of Fg. 7-3; one method s sed on energy conservton consdertons, nd the other on the concept of potentl. The crcut conssts of n del ttery B wth emf, resstor of resstnce,nd two connectng wres.(unless otherwse ndcted, we ssume tht wres n crcuts hve neglgle resstnce. Ther functon, then, s merely to provde pthwys long whch chrge crrers cn move.) Energy Method Equton 6-7 (P ) tells us tht n tme ntervl dt n mount of energy gven y dtwll pper n the resstor of Fg. 7-3 s therml energy. As noted n Secton 6-7, ths energy s sd to e dsspted. (Becuse we ssume the wres to hve neglgle resstnce, no therml energy wll pper n them.) Durng the sme ntervl, chrge dq dtwll hve moved through ttery B, nd the work tht the ttery wll hve done on ths chrge, ccordng to Eq. 7-, s dw dq dt. From the prncple of conservton of energy, the work done y the (del) ttery must equl the therml energy tht ppers n the resstor: Ths gves us dt dt.. The emf s the energy per unt chrge trnsferred to the movng chrges y the ttery. The quntty s the energy per unt chrge trnsferred from the movng chrges to therml energy wthn the resstor. Therefore, ths equton mens tht the energy per unt chrge trnsferred to the movng chrges s equl to the () Fg. 7- () In the crcut, B A ; so ttery B determnes the drecton of the current. () The energy trnsfers n the crcut. The ttery drves current through the resstor, from hgh potentl to low potentl. B Hgher potentl Fg. 7-3 A sngle-loop crcut n whch resstnce s connected cross n del ttery B wth emf. The resultng current s the sme throughout the crcut. Lower potentl

4 708 CHAPTE 7 CICUITS energy per unt chrge trnsferred from them. Solvng for, we fnd. (7-) Potentl Method Suppose we strt t ny pont n the crcut of Fg. 7-3 nd mentlly proceed round the crcut n ether drecton, ddng lgerclly the potentl dfferences tht we encounter. Then when we return to our strtng pont, we must lso hve returned to our strtng potentl. Before ctully dong so, we shll formlze ths de n sttement tht holds not only for sngle-loop crcuts such s tht of Fg. 7-3 ut lso for ny complete loop n multloop crcut, s we shll dscuss n Secton 7-7: LOOP ULE: The lgerc sum of the chnges n potentl encountered n complete trversl of ny loop of crcut must e zero. Ths s often referred to s Krchhoff s loop rule (or Krchhoff s voltge lw), fter Germn physcst Gustv oert Krchhoff. Ths rule s equvlent to syng tht ech pont on mountn hs only one elevton ove se level. If you strt from ny pont nd return to t fter wlkng round the mountn, the lgerc sum of the chnges n elevton tht you encounter must e zero. In Fg. 7-3, let us strt t pont, whose potentl s V,nd mentlly wlk clockwse round the crcut untl we re ck t, keepng trck of potentl chnges s we move. Our strtng pont s t the low-potentl termnl of the ttery. Becuse the ttery s del, the potentl dfference etween ts termnls s equl to.when we pss through the ttery to the hgh-potentl termnl,the chnge n potentl s. As we wlk long the top wre to the top end of the resstor, there s no potentl chnge ecuse the wre hs neglgle resstnce; t s t the sme potentl s the hgh-potentl termnl of the ttery. So too s the top end of the resstor. When we pss through the resstor, however, the potentl chnges ccordng to Eq. 6-8 (whch we cn rewrte s V ). Moreover, the potentl must decrese ecuse we re movng from the hgher potentl sde of the resstor.thus, the chnge n potentl s. We return to pont y movng long the ottom wre. Becuse ths wre lso hs neglgle resstnce, we gn fnd no potentl chnge. Bck t pont,the potentl s gn V.Becuse we trversed complete loop,our ntl potentl,s modfed for potentl chnges long the wy, must e equl to our fnl potentl; tht s, V V. The vlue of V cncels from ths equton, whch ecomes 0. Solvng ths equton for gves us the sme result, /,s the energy method (Eq. 7-). If we pply the loop rule to complete counterclockwse wlk round the crcut, the rule gves us 0 nd we gn fnd tht /. Thus,you my mentlly crcle loop n ether drecton to pply the loop rule. To prepre for crcuts more complex thn tht of Fg. 7-3, let us set down two rules for fndng potentl dfferences s we move round loop:

5 7-5 OTHE SINGLE-LOOP CICUITS 709 PAT 3 ESISTANCE ULE: For move through resstnce n the drecton of the current, the chnge n potentl s ;n the opposte drecton t s. EMF ULE: For move through n del emf devce n the drecton of the emf rrow, the chnge n potentl s ;n the opposte drecton t s. CHECKPOINT The fgure shows the current n sngle-loop crcut wth ttery B nd resstnce (nd wres of neglgle resstnce). () Should the emf rrow t B e drwn pontng leftwrd or rghtwrd? At ponts,,nd c,rnk () the mgntude of the current,(c) the electrc potentl, nd (d) the electrc potentl energy of the chrge crrers, gretest frst. c B 7-5 Other Sngle-Loop Crcuts In ths secton we extend the smple crcut of Fg. 7-3 n two wys. Internl esstnce Fgure 7-4 shows rel ttery, wth nternl resstnce r,wred to n externl resstor of resstnce. The nternl resstnce of the ttery s the electrcl resstnce of the conductng mterls of the ttery nd thus s n unremovle feture of the ttery. In Fg. 7-4,however,the ttery s drwn s f t could e seprted nto n del ttery wth emf nd resstor of resstnce r.the order n whch the symols for these seprted prts re drwn does not mtter. If we pply the loop rule clockwse egnnng t pont, the chnges n potentl gve us r 0. (7-3) Solvng for the current, we fnd. (7-4) r Note tht ths equton reduces to Eq. 7- f the ttery s del tht s, f r 0. Fgure 7-4 shows grphclly the chnges n electrc potentl round the crcut. (To etter lnk Fg. 7-4 wth the closed crcut n Fg. 7-4, mgne curlng the grph nto cylnder wth pont t the left overlppng pont t r el ttery () Potentl (V) r Emf devce Fg. 7-4 () A sngle-loop crcut contnng rel ttery hvng nternl resstnce r nd emf.() The sme crcut, now spred out n lne.the potentls encountered n trversng the crcut clockwse from re lso shown.the potentl V s rtrrly ssgned vlue of zero, nd other potentls n the crcut re grphed reltve to V. V r () V esstor V

6 70 CHAPTE 7 CICUITS the rght.) Note how trversng the crcut s lke wlkng round (potentl) mountn ck to your strtng pont you return to the strtng elevton. In ths ook, when ttery s not descred s rel or f no nternl resstnce s ndcted, you cn generlly ssume tht t s del ut, of course, n the rel world tteres re lwys rel nd hve nternl resstnce. () 3 Seres resstors nd ther equvlent hve the sme current ( ser- ). esstnces n Seres Fgure 7-5 shows three resstnces connected n seres to n del ttery wth emf. Ths descrpton hs lttle to do wth how the resstnces re drwn. ther, n seres mens tht the resstnces re wred one fter nother nd tht potentl dfference V s ppled cross the two ends of the seres. In Fg. 7-5, the resstnces re connected one fter nother etween nd, nd potentl dfference s mntned cross nd y the ttery. The potentl dfferences tht then exst cross the resstnces n the seres produce dentcl currents n them. In generl, eq When potentl dfference V s ppled cross resstnces connected n seres, the resstnces hve dentcl currents. The sum of the potentl dfferences cross the resstnces s equl to the ppled potentl dfference V. () Fg. 7-5 () Three resstors re connected n seres etween ponts nd. () An equvlent crcut, wth the three resstors replced wth ther equvlent resstnce eq. Note tht chrge movng through the seres resstnces cn move long only sngle route. If there re ddtonl routes, so tht the currents n dfferent resstnces re dfferent, the resstnces re not connected n seres. esstnces connected n seres cn e replced wth n equvlent resstnce eq tht hs the sme current nd the sme totl potentl dfference V s the ctul resstnces. CHECKPOINT In Fg. 7-5, f 3,rnk the three resstnces ccordng to () the current through them nd () the potentl dfference cross them, gretest frst. You mght rememer tht eq nd ll the ctul seres resstnces hve the sme current wth the nonsense word ser-. Fgure 7-5 shows the equvlent resstnce eq tht cn replce the three resstnces of Fg To derve n expresson for eq n Fg. 7-5, we pply the loop rule to oth crcuts. For Fg. 7-5,strtng t nd gong clockwse round the crcut, we fnd 3 0, or. (7-5) 3 For Fg. 7-5,wth the three resstnces replced wth sngle equvlent resstnce eq, we fnd eq 0, or. (7-6) eq Comprson of Eqs. 7-5 nd 7-6 shows tht eq 3. The extenson to n resstnces s strghtforwrd nd s eq n j j (n resstnces n seres). (7-7) Note tht when resstnces re n seres, ther equvlent resstnce s greter thn ny of the ndvdul resstnces.

7 7-6 POTENTIAL DIFFEENCE BETWEEN TWO POINTS 7 PAT Potentl Dfference Between Two Ponts We often wnt to fnd the potentl dfference etween two ponts n crcut. For exmple, n Fg. 7-6, wht s the potentl dfference V V etween ponts nd? To fnd out, let s strt t pont (t potentl V ) nd move through the ttery to pont (t potentl V ) whle keepng trck of the potentl chnges we encounter. When we pss through the ttery s emf, the potentl ncreses y.when we pss through the ttery s nternl resstnce r,we move n the drecton of the current nd thus the potentl decreses y r.we re then t the potentl of pont nd we hve V r V, or V V r. (7-8) To evlute ths expresson, we need the current. Note tht the crcut s the sme s n Fg. 7-4,for whch Eq.7-4 gves the current s Susttutng ths equton nto Eq. 7-8 gves us Now susttutng the dt gven n Fg. 7-6, we hve V V r. V V r r r. V V (7-9) (7-0) (7-) Suppose, nsted, we move from to counterclockwse, pssng through resstor rther thn through the ttery. Becuse we move opposte the current, the potentl ncreses y.thus, V V or V V. (7-) Susttutng for from Eq. 7-9, we gn fnd Eq Hence, susttuton of the dt n Fg. 7-6 yelds the sme result, V V 8.0 V. In generl, The nternl resstnce reduces the potentl dfference etween the termnls. r =.0 Ω = V = 4.0 Ω Fg. 7-6 Ponts nd,whch re t the termnls of rel ttery, dffer n potentl. To fnd the potentl etween ny two ponts n crcut, strt t one pont nd trverse the crcut to the other pont, followng ny pth, nd dd lgerclly the chnges n potentl you encounter. Potentl Dfference Across el Bttery In Fg. 7-6, ponts nd re locted t the termnls of the ttery. Thus, the potentl dfference V V s the termnl-to-termnl potentl dfference V cross the ttery. From Eq. 7-8, we see tht V r. (7-3) If the nternl resstnce r of the ttery n Fg. 7-6 were zero, Eq. 7-3 tells us tht V would e equl to the emf of the ttery nmely, V. However, ecuse r.0, Eq.7-3 tells us tht V s less thn. From Eq. 7-, we know tht V s only 8.0 V. Note tht the result depends on the vlue of the current through the ttery. If the sme ttery were n dfferent crcut nd hd dfferent current through t, V would hve some other vlue.

8 7 CHAPTE 7 CICUITS Fg. 7-7 () Pont s drectly connected to ground. () Pont s drectly connected to ground. r =.0 Ω = V () = 4.0 Ω Ground s tken to e zero potentl. r =.0 Ω = V () = 4.0 Ω Groundng Crcut Fgure 7-7 shows the sme crcut s Fg. 7-6 except tht here pont s drectly connected to ground, s ndcted y the common symol. Groundng crcut usully mens connectng the crcut to conductng pth to Erth s surfce (ctully to the electrclly conductng most drt nd rock elow ground). Here, such connecton mens only tht the potentl s defned to e zero t the groundng pont n the crcut. Thus n Fg. 7-7, the potentl t s defned to e V 0. Equton 7- then tells us tht the potentl t s V 8.0 V. Fgure 7-7 s the sme crcut except tht pont s now drectly connected to ground. Thus, the potentl there s defned to e V 0. Equton 7- now tells us tht the potentl t s V 8.0 V. Power, Potentl, nd Emf When ttery or some other type of emf devce does work on the chrge crrers to estlsh current,the devce trnsfers energy from ts source of energy (such s the chemcl source n ttery) to the chrge crrers. Becuse rel emf devce hs n nternl resstnce r,t lso trnsfers energy to nternl therml energy v resstve dsspton (Secton 6-7). Let us relte these trnsfers. The net rte P of energy trnsfer from the emf devce to the chrge crrers s gven y Eq. 6-6: P V, (7-4) where V s the potentl cross the termnls of the emf devce. From Eq. 7-3, we cn susttute V r nto Eq. 7-4 to fnd P ( r) r. (7-5) From Eq. 6-7, we recognze the term r n Eq. 7-5 s the rte P r of energy trnsfer to therml energy wthn the emf devce: P r r (nternl dsspton rte). (7-6) Then the term n Eq. 7-5 must e the rte P emf t whch the emf devce trnsfers energy oth to the chrge crrers nd to nternl therml energy. Thus, CHECKPOINT 3 A ttery hs n emf of V nd n nternl resstnce of. Is the termnlto-termnl potentl dfference greter thn, less thn, or equl to V f the current n the ttery s () from the negtve to the postve termnl, () from the postve to the negtve termnl, nd (c) zero? P emf (power of emf devce). (7-7) If ttery s eng rechrged, wth wrong wy current through t, the energy trnsfer s then from the chrge crrers to the ttery oth to the ttery s chemcl energy nd to the energy dsspted n the nternl resstnce r. The rte of chnge of the chemcl energy s gven y Eq. 7-7, the rte of dsspton s gven y Eq. 7-6, nd the rte t whch the crrers supply energy s gven y Eq. 7-4.

9 7-6 POTENTIAL DIFFEENCE BETWEEN TWO POINTS 73 PAT 3 Smple Prolem Sngle-loop crcut wth two rel tteres The emfs nd resstnces n the crcut of Fg. 7-8 hve the followng vlues: 4.4 V,. V, r.3, r.8, 5.5. () Wht s the current n the crcut? KEY IDEA Bttery r () c r Bttery We cn get n expresson nvolvng the current n ths sngleloop crcut y pplyng the loop rule. Clcultons: Although knowng the drecton of s not necessry, we cn esly determne t from the emfs of the two tteres. Becuse s greter thn,ttery controls the drecton of, so the drecton s clockwse. (These decsons out where to strt nd whch wy you go re rtrry ut, once mde, you must e consstent wth decsons out the plus nd mnus sgns.) Let us then pply the loop rule y gong counterclockwse gnst the current nd strtng t pont.we fnd r r 0. Check tht ths equton lso results f we pply the loop rule clockwse or strt t some pont other thn. Also, tke the tme to compre ths equton term y term wth Fg. 7-8,whch shows the potentl chnges grphclly (wth the potentl t pont rtrrly tken to e zero). Solvng the ove loop equton for the current, we otn 4.4 V. V r r A 40 ma. (Answer) () Wht s the potentl dfference etween the termnls of ttery n Fg. 7-8? KEY IDEA We need to sum the potentl dfferences etween ponts nd. Clcultons: Let us strt t pont (effectvely the negtve termnl of ttery ) nd trvel clockwse through ttery to pont (effectvely the postve termnl), keepng trck of potentl chnges. We fnd tht V r V, Potentl (V) r r c V = 4.4 V V r Fg. 7-8 () A sngle-loop crcut contnng two rel tteres nd resstor.the tteres oppose ech other; tht s, they tend to send current n opposte drectons through the resstor. () A grph of the potentls, counterclockwse from pont,wth thepotentl t rtrrly tken to e zero. (To etter lnk the crcut wth the grph, mentlly cut the crcut t nd then unfold the left sde of the crcut towrd the left nd the rght sde of the crcut towrd the rght.) whch gves us =. V Bttery esstor Bttery () V V r (0.396 A)(.3 ) 4.4 V V c 3.84 V 3.8 V, r V (Answer) whch s less thn the emf of the ttery. You cn verfy ths result y strtng t pont n Fg. 7-8 nd trversng the crcut counterclockwse to pont. We lern two ponts here. () The potentl dfference etween two ponts n crcut s ndependent of the pth we choose to go from one to the other. () When the current n the ttery s n the proper drecton, the termnl-to-termnl potentl dfference s low. Addtonl exmples, vdeo, nd prctce vlle t WleyPLUS

10 74 CHAPTE 7 CICUITS The current nto the juncton must equl the current out (chrge s conserved). c 3 3 d Fg. 7-9 A multloop crcut consstng of three rnches: left-hnd rnch d, rght-hnd rnch cd,nd centrl rnch d.the crcut lso conssts of three loops: left-hnd loop d,rght-hnd loop cd, nd g loop dc. 7-7 Multloop Crcuts Fgure 7-9 shows crcut contnng more thn one loop. For smplcty, we ssume the tteres re del. There re two junctons n ths crcut, t nd d, nd there re three rnches connectng these junctons. The rnches re the left rnch (d), the rght rnch (cd), nd the centrl rnch (d). Wht re the currents n the three rnches? We rtrrly lel the currents, usng dfferent suscrpt for ech rnch. Current hs the sme vlue everywhere n rnch d, hs the sme vlue everywhere n rnch cd,nd 3 s the current through rnch d.the drectons of the currents re ssumed rtrrly. Consder juncton d for moment: Chrge comes nto tht juncton v ncomng currents nd 3,nd t leves v outgong current.becuse there s no vrton n the chrge t the juncton, the totl ncomng current must equl the totl outgong current: 3. (7-8) You cn esly check tht pplyng ths condton to juncton leds to exctly the sme equton. Equton 7-8 thus suggests generl prncple: JUNCTION ULE: The sum of the currents enterng ny juncton must e equl to the sum of the currents levng tht juncton. Prllel resstors nd ther equvlent hve the sme potentl dfference ( pr-v ) () Fg. 7-0 () Three resstors connected n prllel cross ponts nd.() An equvlent crcut, wth the three resstors replced wth ther equvlent resstnce eq. () eq Ths rule s often clled Krchhoff s juncton rule (or Krchhoff s current lw). It s smply sttement of the conservton of chrge for stedy flow of chrge there s nether uldup nor depleton of chrge t juncton. Thus, our sc tools for solvng complex crcuts re the loop rule (sed on the conservton of energy) nd the juncton rule (sed on the conservton of chrge). Equton 7-8 s sngle equton nvolvng three unknowns. To solve the crcut completely (tht s, to fnd ll three currents), we need two more equtons nvolvng those sme unknowns.we otn them y pplyng the loop rule twce. In the crcut of Fg. 7-9, we hve three loops from whch to choose: the left-hnd loop (d), the rght-hnd loop (cd), nd the g loop (dc). Whch two loops we choose does not mtter let s choose the left-hnd loop nd the rght-hnd loop. If we trverse the left-hnd loop n counterclockwse drecton from pont,the loop rule gves us (7-9) If we trverse the rght-hnd loop n counterclockwse drecton from pont, the loop rule gves us (7-0) We now hve three equtons (Eqs. 7-8, 7-9, nd 7-0) n the three unknown currents, nd they cn e solved y vrety of technques. If we hd ppled the loop rule to the g loop, we would hve otned (movng counterclockwse from ) the equton 0. However, ths s merely the sum of Eqs. 7-9 nd 7-0. esstnces n Prllel Fgure 7-0 shows three resstnces connected n prllel to n del ttery of emf.the term n prllel mens tht the resstnces re drectly wred together on one sde nd drectly wred together on the other sde, nd tht potentl dfference V s ppled cross the pr of connected sdes. Thus, ll three resstnces hve the sme potentl dfference V cross them, producng current through ech. In generl,

11 7-7 MULTILOOP CICUITS 75 PAT 3 When potentl dfference V s ppled cross resstnces connected n prllel, the resstnces ll hve tht sme potentl dfference V. In Fg. 7-0,the ppled potentl dfference V s mntned y the ttery. In Fg. 7-0,the three prllel resstnces hve een replced wth n equvlent resstnce eq. esstnces connected n prllel cn e replced wth n equvlent resstnce eq tht hs the sme potentl dfference V nd the sme totl current s the ctul resstnces. You mght rememer tht eq nd ll the ctul prllel resstnces hve the sme potentl dfference V wth the nonsense word pr-v. To derve n expresson for eq n Fg. 7-0, we frst wrte the current n ech ctul resstnce n Fg. 7-0 s V, V, nd 3 V 3, where V s the potentl dfference etween nd.if we pply the juncton rule t pont n Fg. 7-0 nd then susttute these vlues, we fnd 3 V 3. (7-) If we replced the prllel comnton wth the equvlent resstnce eq (Fg. 7-0), we would hve V. (7-) eq Comprng Eqs. 7- nd 7- leds to eq 3. Extendng ths result to the cse of n resstnces, we hve (7-3) eq n j j (n resstnces n prllel). (7-4) For the cse of two resstnces, the equvlent resstnce s ther product dvded y ther sum; tht s, eq. (7-5) Note tht when two or more resstnces re connected n prllel, the equvlent resstnce s smller thn ny of the comnng resstnces.tle 7- summrzes the equvlence reltons for resstors nd cpctors n seres nd n prllel. CHECKPOINT 4 A ttery, wth potentl V cross t, s connected to comnton of two dentcl resstors nd then hs current through t. Wht re the potentl dfference cross nd the current through ether resstor f the resstors re () n seres nd () n prllel? Seres nd Prllel esstors nd Cpctors Tle 7- eq n Seres Prllel Seres Prllel j j esstors Cpctors Eq. 7-7 Eq. 7-4 Eq. 5-0 C eq n n n Eq. 5-9 eq j j Sme current through Sme potentl dfference Sme chrge on ll Sme potentl dfference ll resstors cross ll resstors cpctors cross ll cpctors C eq j C j C j j

12 76 CHAPTE 7 CICUITS Smple Prolem esstors n prllel nd n seres Fgure 7- shows multloop crcut contnng one del ttery nd four resstnces wth the followng vlues: 0, 0, V, 3 30, () Wht s the current through the ttery? Notng tht the current through the ttery must lso e the current through,we see tht we mght fnd the current y pplyng the loop rule to loop tht ncludes ecuse the current would e ncluded n the potentl dfference cross. Incorrect method: Ether the left-hnd loop or the g loop should do. Notng tht the emf rrow of the ttery ponts upwrd, so the current the ttery supples s clockwse, we mght pply the loop rule to the left-hnd loop, clockwse from pont.wth eng the current through the ttery, we would get 4 0 (ncorrect). However, ths equton s ncorrect ecuse t ssumes tht,,nd 4 ll hve the sme current. esstnces nd 4 do hve the sme current, ecuse the current pssng through 4 must pss through the ttery nd then through wth no chnge n vlue. However, tht current splts t juncton pont only prt psses through,the rest through 3. Ded-end method: To dstngush the severl currents n the crcut, we must lel them ndvdully s n Fg. 7-. Then, crclng clockwse from, we cn wrte the loop rule for the left-hnd loop s 4 0. Unfortuntely, ths equton contns two unknowns, nd ; we would need t lest one more equton to fnd them. Successful method: A much eser opton s to smplfy the crcut of Fg. 7- y fndng equvlent resstnces. Note crefully tht nd re not n seres nd thus cnnot e replced wth n equvlent resstnce. However, nd 3 re n prllel, so we cn use ether Eq. 7-4 or Eq. 7-5 to fnd ther equvlent resstnce 3.From the ltter, KEY IDEA (0 )(30 ) 50. We cn now redrw the crcut s n Fg. 7-c; note tht the current through 3 must e ecuse chrge tht moves through nd 4 must lso move through 3.For ths smple one-loop crcut, the loop rule (ppled clockwse from pont s n Fg. 7-d) yelds Susttutng the gven dt, we fnd whch gves us V (0 ) ( ) (8.0 ) 0, V 40 () Wht s the current through? (Answer) () We must now work ckwrd from the equvlent crcut of Fg. 7-d,where 3 hs replced nd 3.() Becuse nd 3 re n prllel, they oth hve the sme potentl dfference cross them s 3. Workng ckwrd: We know tht the current through 3 s 0.30 A. Thus, we cn use Eq. 6-8 ( V/) nd Fg. 7-e to fnd the potentl dfference V 3 cross 3.Settng 3 = from (), we wrte Eq. 6-8 s V 3 3 (0.30 A)( ) 3.6 V. The potentl dfference cross s thus lso 3.6 V (Fg. 7-f), so the current n must e, y Eq. 6-8 nd Fg. 7-g, V (c) Wht s the current 3 through 3? (Answer) We cn nswer y usng ether of two technques: () Apply Eq. 6-8 s we just dd. () Use the juncton rule, whch tells us tht t pont n Fg. 7-,the ncomng current nd the outgong currents nd 3 re relted y 3. Clculton: errngng ths juncton-rule result yelds the result dsplyed n Fg. 7-g: A 0.8 A 0. A. 3.6 V A. KEY IDEAS 0.8 A. KEY IDEAS (Answer) Addtonl exmples, vdeo, nd prctce vlle t WleyPLUS

13 7-7 MULTILOOP CICUITS 77 PAT 3 The equvlent of prllel resstors s smller. A = 0 Ω 3 = Ω = 8.0 Ω c c c () () (c) Applyng the loop rule yelds the current. Applyng V = yelds the potentl dfference. = 0.30 A = 0.30 A = 0 Ω = 0 Ω = V = 0.30 A 3 = Ω = V V 3 = 3.6 V = 0.30 A 3 = Ω 4 = 8.0 Ω 4 = 8.0 Ω = 0.30 A (d) c = 0.30 A (e) c Prllel resstors nd ther equvlent hve the sme V ( pr-v ). Applyng = V/ yelds the current. = 0.30 A 3 = 0.30 A 3 = 0. A = V = 0 Ω V = 3.6 V 3 = 30 Ω V 3 = 3.6 V = 0 Ω = V = 0 Ω V = 3.6 V 3 = 30 Ω V 3 = 3.6 V = 0.8 A = 0 Ω 4 = 8.0 Ω 4 = 8.0 Ω = 0.30 A c (f ) = 0.30 A c (g) Fg. 7- () A crcut wth n del ttery. ()Lel the currents.(c) eplcng the prllel resstors wth ther equvlent. (d) (g) Workng ckwrd to fnd the currents through the prllel resstors.

14 78 CHAPTE 7 CICUITS Smple Prolem Mny rel tteres n seres nd n prllel n n electrc fsh Electrc fsh re le to generte current wth ologcl cells clled electroplques, whch re physologcl emf devces. The electroplques n the type of electrc fsh known s South Amercn eel re rrnged n 40 rows, ech row stretchng horzontlly long the ody nd ech contnng 5000 electroplques. The rrngement s suggested n Fg. 7-; ech electroplque hs n emf of 0.5 V nd n nternl resstnce r of 0.5.The wter surroundng the eel completes crcut etween the two ends of the electroplque rry, one end t the nml s hed nd the other ner ts tl. () If the wter surroundng the eel hs resstnce w 800, how much current cn the eel produce n the wter? KEY IDEA We cn smplfy the crcut of Fg. 7- y replcng comntons of emfs nd nternl resstnces wth equvlent emfs nd resstnces. Clcultons: We frst consder sngle row. The totl emf row long row of 5000 electroplques s the sum of the emfs: row 5000 (5000)(0.5 V) 750 V. The totl resstnce row long row s the sum of the nternl resstnces of the 5000 electroplques: row 5000r (5000)(0.5 ) 50. We cn now represent ech of the 40 dentcl rows s hvng sngle emf row nd sngle resstnce row (Fg. 7-). In Fg. 7-, the emf etween pont nd pont on ny row s row 750 V. Becuse the rows re dentcl nd ecuse they re ll connected together t the left n Fg. 7-, ll ponts n tht fgure re t the sme electrc potentl. Thus, we cn consder them to e connected so tht there s only sngle pont.the emf etween pont nd ths sngle pont s row 750 V, so we cn drw the crcut s shown n Fg. 7-c. Electroplque Frst, reduce ech row to one emf nd one resstnce. r Ponts wth the sme potentl cn e tken s though connected electroplques per row 750 V 40 rows row row row row () w () row w row c Emfs n prllel ct s sngle emf. row = 750 V row row row c eplce the prllel resstnces wth ther equvlent. row c eq w w (c) (d) Fg. 7- () A model of the electrc crcut of n eel n wter. Ech electroplque of the eel hs n emf nd nternl resstnce r.along ech of 40 rows extendng from the hed to the tl of the eel,there re 5000 electroplques.the surroundng wter hs resstnce w.() The emf row nd resstnce row of ech row. (c) The emf etween ponts nd s row. Between ponts nd c re 40 prllel resstnces row.(d) The smplfed crcut, wth eq replcng the prllel comnton.

15 7-7 MULTILOOP CICUITS 79 PAT 3 Between ponts nd c n Fg. 7-c re 40 resstnces row 50,ll n prllel.the equvlent resstnce eq of ths comnton s gven y Eq. 7-4 s or eplcng the prllel comnton wth eq,we otn the smplfed crcut of Fg. 7-d.Applyng the loop rule to ths crcut counterclockwse from pont,we hve row w eq 0. Solvng for nd susttutng the known dt, we fnd eq 40 eq row 40 row j w eq j A 0.93 A. row, V (Answer) If the hed or tl of the eel s ner fsh, some of ths current could pss long nrrow pth through the fsh, stunnng or kllng t. () How much current row trvels through ech row of Fg. 7-? Becuse the rows re dentcl, the current nto nd out of the eel s evenly dvded mong them. Clculton: Thus, we wrte row 40 KEY IDEA 0.97 A A. (Answer) Thus, the current through ech row s smll, out two orders of mgntude smller thn the current through the wter.ths tends to spred the current through the eel s ody, so tht the eel need not stun or kll tself when t stuns or klls fsh. Smple Prolem Multloop crcut nd smultneous loop equtons Fgure 7-3 shows crcut whose elements hve the followng vlues: 3.0 V, 6.0 V,.0, 4.0. The three tteres re del tteres. Fnd the mgntude nd drecton of the current n ech of the three rnches. KEY IDEAS It s not worthwhle to try to smplfy ths crcut, ecuse no two resstors re n prllel, nd the resstors tht re n seres (those n the rght rnch or those n the left rnch) present no prolem. So, our pln s to pply the juncton nd loop rules. Juncton rule: Usng rtrrly chosen drectons for the currents s shown n Fg. 7-3, we pply the juncton rule t pont y wrtng 3. (7-6) An pplcton of the juncton rule t juncton gves only the sme equton, so we next pply the loop rule to ny two of the three loops of the crcut. Left-hnd loop: We frst rtrrly choose the left-hnd loop, rtrrly strt t pont,nd rtrrly trverse the loop n the clockwse drecton, otnng ( ) 0, where we hve used ( ) nsted of 3 n the mddle rnch. Susttutng the gven dt nd smplfyng yeld (8.0 ) (4.0 ) 3.0 V. (7-7) Fg. 7-3 A multloop crcut wth three del tteres nd fve resstnces. 3 ght-hnd loop: For our second pplcton of the loop rule, we rtrrly choose to trverse the rght-hnd loop counterclockwse from pont,fndng ( ) 0. Susttutng the gven dt nd smplfyng yeld (4.0 ) (8.0 ) 0. (7-8) Comnng equtons: We now hve system of two equtons (Eqs. 7-7 nd 7-8) n two unknowns ( nd ) to solve ether y hnd (whch s esy enough here) or wth mth pckge. (One soluton technque s Crmer s rule, gven n Appendx E.) We fnd 0.50 A. (7-9) (The mnus sgn sgnls tht our rtrry choce of drecton for n Fg. 7-3 s wrong, ut we must wt to correct t.) Susttutng 0.50 A nto Eq. 7-8 nd solvng for then gve us 0.5 A. (Answer)

16 70 CHAPTE 7 CICUITS Wth Eq. 7-6 we then fnd tht A 0.5 A 0.5 A. The postve nswer we otned for sgnls tht our choce of drecton for tht current s correct. However, the negtve nswers for nd 3 ndcte tht our choces for those currents re wrong.thus, s lst step here, we correct the nswers y reversng the rrows for nd 3 n Fg. 7-3 nd then wrtng = 0.50 A nd 3 = 0.5 A. (Answer) Cuton: Alwys mke ny such correcton s the lst step nd not efore clcultng ll the currents. Addtonl exmples, vdeo, nd prctce vlle t WleyPLUS r A Fg. 7-4 A sngle-loop crcut, showng how to connect n mmeter (A) nd voltmeter (V). c d V 7-8 The Ammeter nd the Voltmeter An nstrument used to mesure currents s clled n mmeter. To mesure the current n wre, you usully hve to rek or cut the wre nd nsert the mmeter so tht the current to e mesured psses through the meter. (In Fg. 7-4, mmeter A s set up to mesure current.) It s essentl tht the resstnce A of the mmeter e very much smller thn other resstnces n the crcut. Otherwse, the very presence of the meter wll chnge the current to e mesured. A meter used to mesure potentl dfferences s clled voltmeter. To fnd the potentl dfference etween ny two ponts n the crcut, the voltmeter termnls re connected etween those ponts wthout rekng or cuttng the wre. (In Fg. 7-4, voltmeter V s set up to mesure the voltge cross.) It s essentl tht the resstnce V of voltmeter e very much lrger thn the resstnce of ny crcut element cross whch the voltmeter s connected. Otherwse, the meter tself ecomes n mportnt crcut element nd lters the potentl dfference tht s to e mesured. Often sngle meter s pckged so tht, y mens of swtch, t cn e mde to serve s ether n mmeter or voltmeter nd usully lso s n ohmmeter, desgned to mesure the resstnce of ny element connected etween ts termnls. Such verstle unt s clled multmeter. 7-9 C Crcuts In precedng sectons we delt only wth crcuts n whch the currents dd not vry wth tme. Here we egn dscusson of tme-vryng currents. S Fg. 7-5 When swtch S s closed on,the cpctor s chrged through the resstor.when the swtch s fterwrd closed on,the cpctor dschrges through the resstor. C Chrgng Cpctor The cpctor of cpctnce C n Fg. 7-5 s ntlly unchrged. To chrge t, we close swtch S on pont.ths completes n C seres crcut consstng of the cpctor, n del ttery of emf,nd resstnce. From Secton 5-, we lredy know tht s soon s the crcut s complete, chrge egns to flow (current exsts) etween cpctor plte nd ttery termnl on ech sde of the cpctor. Ths current ncreses the chrge q on the pltes nd the potentl dfference V C ( q/c) cross the cpctor. When tht potentl dfference equls the potentl dfference cross the ttery (whch here s equl to the emf ), the current s zero. From Eq. 5- (q CV), the equlrum (fnl) chrge on the then fully chrged cpctor s equl to C. Here we wnt to exmne the chrgng process. In prtculr we wnt to know how the chrge q(t) on the cpctor pltes, the potentl dfference V C (t) cross the cpctor, nd the current (t) n the crcut vry wth tme durng the chrgng process. We egn y pplyng the loop rule to the crcut, trversng t

17 7-9 C CICUITS 7 PAT 3 clockwse from the negtve termnl of the ttery. We fnd q C 0. (7-30) The lst term on the left sde represents the potentl dfference cross the cpctor. The term s negtve ecuse the cpctor s top plte, whch s connected to the ttery s postve termnl, s t hgher potentl thn the lower plte. Thus, there s drop n potentl s we move down through the cpctor. We cnnot mmedtely solve Eq ecuse t contns two vrles, nd q.however,those vrles re not ndependent ut re relted y Susttutng ths for n Eq nd rerrngng, we fnd dq dt q C dq dt. (7-3) (chrgng equton). (7-3) Ths dfferentl equton descres the tme vrton of the chrge q on the cpctor n Fg To solve t, we need to fnd the functon q(t) tht stsfes ths equton nd lso stsfes the condton tht the cpctor e ntlly unchrged; tht s, q 0 t t 0. We shll soon show tht the soluton to Eq. 7-3 s q C ( e t/c ) (chrgng cpctor). (7-33) (Here e s the exponentl se,.78..., nd not the elementry chrge.) Note tht Eq does ndeed stsfy our requred ntl condton, ecuse t t 0 the term e t/c s unty; so the equton gves q 0. Note lso tht s t goes to nfnty (tht s, long tme lter), the term e t/c goes to zero; so the equton gves the proper vlue for the full (equlrum) chrge on the cpctor nmely, q C.A plot of q(t) for the chrgng process s gven n Fg The dervtve of q(t) s the current (t) chrgng the cpctor: dq dt e t/c (chrgng cpctor). (7-34) A plot of (t) for the chrgng process s gven n Fg. 7-6.Note tht the current hs the ntl vlue / nd tht t decreses to zero s the cpctor ecomes fully chrged. A cpctor tht s eng chrged ntlly cts lke ordnry connectng wre reltve to the chrgng current.a long tme lter, t cts lke roken wre. By comnng Eq. 5- (q CV ) nd Eq. 7-33, we fnd tht the potentl dfference V C (t) cross the cpctor durng the chrgng process s (chrgng cpctor). (7-35) Ths tells us tht V C 0 t t 0 nd tht V C when the cpctor ecomes fully chrged s t :. The Tme Constnt V C q C ( e t/c ) The product C tht ppers n Eqs. 7-33, 7-34, nd 7-35 hs the dmensons of tme (oth ecuse the rgument of n exponentl must e dmensonless nd q ( µ C) (ma) The cpctor s chrge grows s the resstor's current des out. C Tme (ms) / () Tme (ms) () 0 Fg. 7-6 () A plot of Eq. 7-33, whch shows the uldup of chrge on the cpctor of Fg () A plot of Eq. 7-34, whch shows the declne of the chrgng current n the crcut of Fg. 7-5.The curves re plotted for 000, C mf, nd 0 V; the smll trngles represent successve ntervls of one tme constnt t.

18 7 CHAPTE 7 CICUITS ecuse, n fct,.0.0 F.0 s). The product C s clled the cpctve tme constnt of the crcut nd s represented wth the symol t: t C (tme constnt). (7-36) From Eq. 7-33, we cn now see tht t tme t t ( C), the chrge on the ntlly unchrged cpctor of Fg. 7-5 hs ncresed from zero to q C ( e ) 0.63C. (7-37) In words, durng the frst tme constnt t the chrge hs ncresed from zero to 63% of ts fnl vlue C. In Fg.7-6,the smll trngles long the tme xes mrk successve ntervls of one tme constnt durng the chrgng of the cpctor.the chrgng tmes for C crcuts re often stted n terms of t. Dschrgng Cpctor Assume now tht the cpctor of Fg. 7-5 s fully chrged to potentl V 0 equl to the emf of the ttery.at new tme t 0, swtch S s thrown from to so tht the cpctor cn dschrge through resstnce. How do the chrge q(t) on the cpctor nd the current (t) through the dschrge loop of cpctor nd resstnce now vry wth tme? The dfferentl equton descrng q(t) s lke Eq. 7-3 except tht now, wth no ttery n the dschrge loop, 0.Thus, dq dt q C 0 The soluton to ths dfferentl equton s (dschrgng equton). (7-38) q q 0 e t/c (dschrgng cpctor), (7-39) where q 0 ( CV 0 ) s the ntl chrge on the cpctor.you cn verfy y susttuton tht Eq s ndeed soluton of Eq Equton 7-39 tells us tht q decreses exponentlly wth tme, t rte tht s set y the cpctve tme constnt t C. At tme t t, the cpctor s chrge hs een reduced to q 0 e,or out 37% of the ntl vlue.note tht greter t mens greter dschrge tme. Dfferenttng Eq gves us the current (t): dq dt q 0 C e t/c (dschrgng cpctor). (7-40) Ths tells us tht the current lso decreses exponentlly wth tme, t rte set y t.the ntl current 0 s equl to q 0 /C. Note tht you cn fnd 0 y smply pplyng the loop rule to the crcut t t 0; just then the cpctor s ntl potentl V 0 s connected cross the resstnce, so the current must e 0 V 0 / (q 0 /C)/ q 0 /C.The mnus sgn n Eq.7-40 cn e gnored;t merely mens tht the cpctor s chrge q s decresng. Dervton of Eq To solve Eq. 7-3, we frst rewrte t s The generl soluton to ths dfferentl equton s of the form dq dt q C. (7-4) q q p Ke t, (7-4)

19 7-9 C CICUITS 73 PAT 3 where q p s prtculr soluton of the dfferentl equton, K s constnt to e evluted from the ntl condtons, nd /C s the coeffcent of q n Eq To fnd q p,we set dq/dt 0 n Eq. 7-4 (correspondng to the fnl condton of no further chrgng), let q q p,nd solve,otnng To evlute K,we frst susttute ths nto Eq.7-4 to get q p C. (7-43) q C Ke t. Then susttutng the ntl condtons q 0 nd t 0 yelds 0 C K, or K C.Fnlly,wth the vlues of q p,,nd K nserted, Eq. 7-4 ecomes q C C e t/c, whch, wth slght modfcton, s Eq CHECKPOINT 5 The tle gves four sets of vlues for the crcut elements n Fg nk the sets ccordng to () the ntl current (s the swtch s closed on ) nd () the tme requred for the current to decrese to hlf ts ntl vlue, gretest frst. 3 4 (V) 0 0 ( ) C (mf) Smple Prolem Dschrgng n C crcut to vod fre n rce cr pt stop As cr rolls long pvement, electrons move from the pvement frst onto the tres nd then onto the cr ody.the cr stores ths excess chrge nd the ssocted electrc potentl energy s f the cr ody were one plte of cpctor nd the pvement were the other plte (Fg. 7-7).When the cr stops, t dschrges ts excess chrge nd energy through the tres, just s cpctor cn dschrge through resstor. If conductng oject comes wthn few centmeters of the cr efore the cr s dschrged, the remnng energy cn e suddenly trnsferred to sprk etween the cr nd the oject. Suppose the conductng oject s fuel dspenser. The sprk wll not gnte the fuel nd cuse fre f the sprk energy s less thn the crtcl vlue U fre 50 mj. When the cr of Fg. 7-7 stops t tme t 0, the cr ground potentl dfference s V 0 30 kv. The crground cpctnce s C 500 pf, nd the resstnce of ech tre s tre 00 G.How much tme does the cr tke to dschrge through the tres to drop elow the crtcl vlue U fre? KEY IDEAS () At ny tme t, cpctor s stored electrc potentl energy U s relted to ts stored chrge q ccordng to Eq. 5- (U q /C). () Whle cpctor s dschrgng, the chrge decreses wth tme ccordng to Eq (q q 0 e t/c ). Clcultons: We cn tret the tres s resstors tht re connected to one nother t ther tops v the cr ody nd t () () MDOG PST4 XP3I PM WNF 3N Bommn Schtuff True Vles Effectve cpctnce tre tre C tre tre (d) U fre U 0 GΩ GΩ 9.4 t (s) DIVE THU Tre resstnce Fg. 7-7 () A chrged cr nd the pvement cts lke cpctor tht cn dschrge through the tres. () The effectve crcut of the cr pvement cpctor, wth four tre resstnces tre connected n prllel. (c) The equvlent resstnce of the tres. (d) The electrc potentl energy U n the cr pvement cpctor decreses durng dschrge. ULTA 6 MOTEL (c) C

20 74 CHAPTE 7 CICUITS ther ottoms v the pvement. Fgure 7-7 shows how the four resstors re connected n prllel cross the cr s cpctnce, nd Fg. 7-7c shows ther equvlent resstnce. From Eq. 7-4, s gven y, tre tre tre tre or tre (7-44) 4 4 When the cr stops, t dschrges ts excess chrge nd energy through. We now use our two Key Ides to nlyze the dschrge. Susttutng Eq nto Eq. 5- gves U q C (q 0e t/c ) C q 0 C e t/c. (7-45) From Eq. 5- (q CV ), we cn relte the ntl chrge q 0 on the cr to the gven ntl potentl dfference V 0 : q 0 CV 0.Susttutng ths equton nto Eq.7-45 rngs us to U (CV 0) C e t/c CV 0 e t/c, or e t/c U (7-46) CV. 0 Tkng the nturl logrthms of oth sdes, we otn or t C (7-47) ln U CV 0. Susttutng the gven dt, we fnd tht the tme the cr tkes to dschrge to the energy level U fre 50 mj s t (5 0 9 )(500 0 F) ln ( J) (500 0 F)( V) 9.4 s. t C ln U CV 0, (Answer) Fre or no fre: Ths cr requres t lest 9.4 s efore fuel or fuel dspenser cn e rought sfely ner t. Durng rce, pt crew cnnot wt tht long. Insted, tres for rce crs nclude some type of conductng mterl (such s cron lck) to lower the tre resstnce nd thus ncrese the cr s dschrge rte. Fgure 7-7d shows the stored energy U versus tme t for tre resstnces of 00 G (the vlue we used n our clcultons here) nd 0 G. Note how much more rpdly cr dschrges to level U fre wth the lower vlue. Addtonl exmples, vdeo, nd prctce vlle t WleyPLUS Emf An emf devce does work on chrges to mntn potentl dfference etween ts output termnls. If dw s the work the devce does to force postve chrge dq from the negtve to the postve termnl, then the emf (work per unt chrge) of the devce s dw dq (defnton of ). (7-) The volt s the SI unt of emf s well s of potentl dfference. An del emf devce s one tht lcks ny nternl resstnce. The potentl dfference etween ts termnls s equl to the emf. A rel emf devce hs nternl resstnce. The potentl dfference etween ts termnls s equl to the emf only f there s no current through the devce. Anlyzng Crcuts The chnge n potentl n trversng resstnce n the drecton of the current s ;n the opposte drecton t s (resstnce rule). The chnge n potentl n trversng n del emf devce n the drecton of the emf rrow s ; n the opposte drecton t s (emf rule). Conservton of energy leds to the loop rule: Loop ule. The lgerc sum of the chnges n potentl encountered n complete trversl of ny loop of crcut must e zero. Conservton of chrge gves us the juncton rule: Juncton ule. The sum of the currents enterng ny juncton must e equl to the sum of the currents levng tht juncton. Sngle-Loop Crcuts The current n sngle-loop crcut contnng sngle resstnce nd n emf devce wth emf nd nternl resstnce r s (7-4) r, whch reduces to / for n del emf devce wth r 0. Power When rel ttery of emf nd nternl resstnce r does work on the chrge crrers n current through the ttery, the rte P of energy trnsfer to the chrge crrers s P V, (7-4) where V s the potentl cross the termnls of the ttery. The rte