Current and Resistance

Transcription

1 7 Currnt and Rsistanc CHPTER OUTLNE 71 Elctric Currnt 7 Rsistanc 7 Modl for Elctrical Conduction 74 Rsistanc and Tmpratur 75 Suprconductors 76 Elctric Powr NSWERS TO QUESTONS Q71 ndividual vhicls cars, trucks and motorcycls would corrspond to charg Th numbr of vhicls that pass a crtain point in a givn tim would corrspond to th currnt Q7 Voltag is a masur of potntial diffrnc, not of currnt Surg implis a flow and only charg, in coulombs, can flow through a systm t would also b corrct to say that th victim carrid a crtain currnt, in amprs Q7 Gomtry and rsistivity n turn, th rsistivity of th matrial dpnds on th tmpratur Q74 Rsistanc is a physical proprty of th conductor basd on th matrial of which it is mad and its siz and shap, including th locations whr currnt is put in and takn out Rsistivity is a physical proprty only of th matrial of which th rsistor is mad Q75 Th radius of wir B is tims th radius of wir, to mak its cross sctional ara tims largr Q76 Not all conductors oby Ohm s law at all tims or xampl, considr an xprimnt in which a variabl potntial diffrnc is applid across an incandscnt light bulb, and th currnt is masurd t vry low voltags, th filamnt follows Ohm s law nicly But thn long bfor th filamnt bgins to glow, th plot of bcoms non-linar, bcaus th rsistivity is tmpraturdpndnt Q77 conductor is not in lctrostatic quilibrium whn it is carrying a currnt, duh! f chargs ar placd on an isolatd conductor, th lctric filds stablishd in th conductor by th chargs will caus th chargs to mov until thy ar in positions such that thr is zro lctric fild throughout th conductor conductor carrying a stady currnt is not an isolatd conductor its nds must b connctd to a sourc of mf, such as a battry Th battry maintains a potntial diffrnc across th conductor and, thrfor, an lctric fild in th conductor Th stady currnt is du to th rspons of th lctrons in th conductor du to this constant lctric fild 15

2 16 Currnt and Rsistanc Q78 Th bottom of th rods on th Jacob s Laddr ar clos nough so that th supplid voltag is sufficint to produc dilctric brakdown of th air Th initial spark at th bottom includs a tub of ionizd air molculs Sinc this tub containing ions is warmr than th air around it, it is buoyd up by th surrounding air and bgins to ris Th ions thmslvs significantly dcras th rsistivity of th air Thy significantly lowr th dilctric strngth of th air, marking longr sparks possibl ntrnal rsistanc in th powr supply will typically mak its trminal voltag drop, so that it cannot produc a spark across th bottom nds of th rods singl continuous spark, thrfor will ris up, bcoming longr and longr, until th potntial diffrnc is not larg nough to sustain dilctric brakdown of th air Onc th initial spark stops, anothr on will form at th bottom, whr again, th supplid potntial diffrnc is sufficint to brak down th air Q79 Th conductor dos not follow Ohm s law, and must hav a rsistivity that is currnt-dpndnt, or mor likly tmpratur-dpndnt Q71 powr supply would corrspond to a watr pump; a rsistor corrsponds to a pip of a crtain diamtr, and thus rsistanc to flow; charg corrsponds to th watr itslf; potntial diffrnc corrsponds to diffrnc in hight btwn th nds of a pip or th ports of a watr pump Q711 Th amplitud of atomic vibrations incrass with tmpratur toms can thn scattr lctrons mor fficintly Q71 n a mtal, th conduction lctrons ar not strongly bound to individual ion cors Thy can mov in rspons to an applid lctric fild to constitut an lctric currnt Each mtal ion in th lattic of a microcrystal xrts Coulomb forcs on its nighbors Whn on ion is vibrating rapidly, it can st its nighbors into vibration This procss rprsnts nrgy moving though th matrial by hat Q71 Th rsistanc of coppr incrass with tmpratur, whil th rsistanc of silicon dcrass with incrasing tmpratur Th conduction lctrons ar scattrd mor by vibrating atoms whn coppr hats up Silicon s charg carrir dnsity incrass as tmpratur incrass and mor atomic lctrons ar promotd to bcom conduction lctrons Q714 currnt will continu to xist in a suprconductor without voltag bcaus thr is no rsistanc loss Q715 Suprconductors hav no rsistanc whn thy ar blow a crtain critical tmpratur or most suprconducting matrials, this critical tmpratur is clos to absolut zro t rquirs xpnsiv rfrigration, oftn using liquid hlium Liquid nitrogn at 77 K is much lss xpnsiv Rcnt discovris of matrials that hav highr critical tmpraturs suggst th possibility of dvloping suprconductors that do not rquir xpnsiv cooling systms Q716 n a normal mtal, suppos that w could procd to a limit of zro rsistanc by lngthning th avrag tim btwn collisions Th classical modl of conduction thn suggsts that a constant applid voltag would caus constant acclration of th fr lctrons, and a currnt stadily incrasing in tim On th othr hand, w can actually switch to zro rsistanc by substituting a suprconducting wir for th normal mtal n this cas, th drift vlocity of lctrons is stablishd by vibrations of atoms in th crystal lattic; th maximum currnt is limitd; and it bcoms impossibl to stablish a potntial diffrnc across th suprconductor Q717 Bcaus thr ar so many lctrons in a conductor (approximatly 1 8 lctrons m ) th avrag vlocity of chargs is vry slow Whn you connct a wir to a potntial diffrnc, you stablish an lctric fild vrywhr in th wir narly instantanously, to mak lctrons start drifting vrywhr all at onc

3 Chaptr 7 17 Q718 Currnt moving through a wir is analogous to a longitudinal wav moving through th lctrons of th atoms Th wav spd dpnds on th spd at which th disturbanc in th lctric fild can b communicatd btwn nighboring atoms, not on th drift vlocitis of th lctrons thmslvs f you lav a dirct-currnt light bulb on for a rasonably short tim, it is likly that no singl lctron will ntr on nd of th filamnt and lav at th othr nd Q719 Mor powr is dlivrd to th rsistor with th smallr rsistanc, sinc P R Q7 Th 5 W bulb has a highr rsistanc Th 1 W bulb carris mor currnt Q71 On ampr hour is 6 coulombs Th ampr hour rating is th quantity of charg that th battry can lift though its nominal potntial diffrnc Q7 Choos th voltag of th powr supply you will us to driv th hatr Nxt calculat th rquird rsistanc R as Knowing th rsistivity ρ of th matrial, choos a combination of wir lngth P R and cross sctional ara to mak K J H G K J You will hav to pay for lss matrial if you mak both ρ and smallr, but if you go too far th wir will hav too littl surfac ara to radiat away th nrgy; thn th rsistor will mlt SOLUTONS TO PROBLEMS Sction 71 P71 Q t Elctric Currnt a f Q t 1 4 s 1 1 C 6 N Q 1 1 C Clctron 15 lctrons P7 Th molar mass of silvr 17 9 g mol and th volum V is 4 6 a fa f 6 g ρv kg m m kg V ara thicknss 7 1 m 1 1 m m Th mass of silvr dpositd is m nd th numbr of silvr atoms dpositd is N kg 6 1 atoms 1 g 179 g 1 kg 1 V C s R 18 Ω Q N t 667 Cs 19 C KJ H G K J atoms 11 1 s 64 h

4 18 Currnt and Rsistanc af 1 t τ t P7 Qt dt τ z af a f 1 (a) Q τ τ 1 6 τ 1 a f b g (b) Q 1τ τ τ af (c) Q τ 1 τ P74 (a) Using k r mv k 6, w gt: v 19 1 m s r mr (b) Th tim for th lctron to rvolv around th proton onc is: 11 π r π m t 15 1 v ms 16 s Th total charg flow in this tim is C m s C, so th currnt is P75 Th priod of rvolution for th sphr is T π, and th avrag currnt rprsntd by this ω q qω rvolving charg is T π P76 q 4t + 5t+ 6 1 m H G cm K J 4 1 cm 1 m (a) a f dq 1 s 1t dt t 1 s t 1 s 17 (b) J 4 1 m 85 k m dq P77 dt q dq dt q z z z 1 14s a H G fsin K J 1 C L H G π K J O NM QP + 1 C 1π 1π 1π t dt s cos cos 65 C

5 Chaptr 7 19 P78 (a) J π m 99 5 k m 1 (b) J J1 ; so π 4 1 π r r m mm P79 (a) J π m 55 m J (b) rom J nv d, w hav n v d 55 m C 1 m s 1 m (c) Q rom t, w hav Q N t (This is about 8 yars!) 19 C s P71 (a) Th spd of ach dutron is givn by K 1 mv J kg v and v m s Th tim btwn dutrons passing a stationary point is t in q t Cs Ct or t s So th distanc btwn thm is vt ms s 1 1 m (b) On nuclus will put its narst nighbor at potntial V kq r N m C C V m This is vry small compard to th MV acclrating potntial, so rpulsion within th bam is a small ffct

6 11 Currnt and Rsistanc P711 W us nqv d n is th numbr of charg carrirs pr unit volum, and is idntical to th numbr of atoms pr unit volum W assum a contribution of 1 fr lctron pr atom in th rlationship abov or aluminum, which has a molar mass of 7, w know that vogadro s numbr of atoms, N, has a mass of 7 g Thus, th mass pr atom is 7 g 7 g N g atom dnsity of aluminum 7 g cm Thus, n mass pr atom gatom 8 n 6 1 atoms cm 6 1 atoms m Thrfor, vd nq or, v d 1 mm s m C 4 1 m 4 ms Sction 7 Rsistanc *P71 J E E 74 Vm σ 8 ρ 44 1 Ω m 1 Ω K J 1 V 1 7 m 1 V P71 5 5m R 4 Ω P714 (a) pplying its dfinition, w find th rsistanc of th rod, 15 V R 75Ω 75 k Ω 4 1 (b) Th lngth of th rod is dtrmind from th dfinition of rsistivity: R ρ Solving for and substituting numrical valus for R,, and th valu of ρ givn for carbon in Tabl 71, w obtain R ρ Ω5 1 m Ω m 56 m P715 R and R ρ 1 m : a6 mmf H G 1 mm ρ 7 a9 V f6 1 m : ρ Ω m a15 mf KJ 7 m

7 P716 J σe σ 1 NC π r π 1 m σ 55 Ω m b 1 1 a f ρ σ g b 181 Ω m g Chaptr P717 (a) Givn M ρ V ρ whr ρ d mass dnsity, d d M ρr ρr ρρ r d w obtain: Taking ρ r rsistivity, R ρ d M ρ d M MR 1 1 a5 f Thus, 18 m ρρ r d (b) V M, or π r M ρ d ρ d Thus, M r πρ d 1 1 a f π r m Th diamtr is twic this distanc: diamtr 8 µ m *P718 Th volum of th gram of gold is givn by ρ m V V m 1 kg m 4 1 m ρ 19 1 kg m m 8 ρ 44 1 Ω m 4 1 m R m P719 (a) Suppos th rubbr is 1 cm long and 1 mm in diamtr ρ 4ρ 41 Ω m 1 m R ~ ~ 1 π d π 1 m 1 1 4ρ Ω m 1 m (b) R ~ d π π 1 m 8 6 ~ 1 Ω 7 18 Ω Ω (c) 1 V ~ ~ 1 18 R 1 Ω 1 V ~ ~ Ω 9 16

8 11 Currnt and Rsistanc 9 g P7 Th distanc btwn opposit facs of th cub is H G K J 15 g cm ρ ρ ρ Ω m 7 (a) R Ω 777nΩ 5 1 m cm (b) V R Ω 1 5 g cm n 6 1 lctrons mol gmol 1 1 cm n KJ lctrons cm 1 m m 1 9 Cs nqv and v nq m C 5 m 6 b g 8 19 b g 8 µ ms P71 Originally, R ρ ρ ρ R inally, R f 9 9 P7 ρ π r r r ρ π r l b lg b Cug l Cu ρ l ρ Cu J 6 1 m P7 J σe so σ E 1 Vm Ω m a f ρ1 1 ρ ρ + ρ P74 R + d R Ω m a 5 m f+ 6 1 Ω m a 4 mf 1 m 78 Ω Sction 7 Modl for Elctrical Conduction P75 ρ m nq τ so m τ ρnq v d qe τ m E so Thrfor, E 181 V m 14 s

9 Chaptr 7 11 P76 (a) n is unaffctd (b) J so it doubls (c) J nv d so v d doubls (d) σ τ m is unchangd as long as σ dos not chang du to a tmpratur chang in th nq conductor P77 rom Equation 717, m τ nq ρ vτ ms 47 1 s 1 1 m 1 nm s Sction 74 Rsistanc and Tmpratur P78 t th low tmpratur T C w writ R C R + TC T C 1 αb g whr T t th high tmpratur T h, Rh R 1 + αb T h T g h 1 Thn afa1 f a8 f af C a18f and C 115 a H G 1 f K J 198 P79 R R 1 + α T 579 a f givs Ω a Ωf + C T Solving, T 14 1 C T C nd, th final tmpratur is T C C

10 114 Currnt and Rsistanc b g b g P7 R R + R R 1+ α T T + R 1+ α TT c n c c n n b g b g so Rc R R T T R T T cαc + nαn α n R Rn + R α α n α c Rn R 1 Rc R 1 α c KJ α n R n c n 1 k 1 L 4 1 CO Ω NM 5 1 CQP KJ R n 556 kω and R c 444 kω P71 (a) ρ ρ 1 + α T T 8 1 Ω m Ω m n α n α b g a f c 8 8 E Vm (b) J ρ 15 1 Ω m 6 m d (c) J J H G π K J π m 4 M 4 L M NM 4 m O P Q P 49 9 m (d) n 6 1 lctrons 6 98 g 7 1 g m J vd n 65 1 m lctrons m C b ga f () E Vm m 4 V P7 or aluminum, 8 lctrons m 659 µ ms α E 9 1 C 1 (Tabl 71) b ga f a f ρ ρ 1+ αe T 1+ α T R 1 + α T α C 1 (Tabl 191) R b g H G 1 + α E T 14Ω a1 + α Tf a f KJ 171 Ω

11 P7 R R 1 + α T R R Rα T R R α T R P74 ssuming linar chang of rsistanc with tmpratur, R R 1 + α T a f a f R 77 K Ω C P75 ρ ρ 1 + α T Ω a f or T W Rquir that ρw 4 ρ Cu so that T W 1 ρw 1 α ρ W W 1 KJ 45 1 C Thrfor, TW 47 6 C + T 67 6 C a KJ f 8 1 KJ 476 C Chaptr Sction 75 Suprconductors Problm 48 in Chaptr 4 can b assignd with this sction Sction 76 Elctric Powr P 6 W P76 1 V 5 1 V and R 5 4 Ω 6 *P77 P V 75 W b gb g P78 P hp 746 W hp W 5 P bg 448 P79 Th hat that must b addd to th watr is b gb ga f Q mc T 15 kg 4186 J kg C 4 C J Thus, th powr supplid by th hatr is 5 W Q 51 1 J P 419 W t t 6 s a f a f 11 V and th rsistanc is R P 419 W 8 9 Ω

12 116 Currnt and Rsistanc *P74 Th battry taks in nrgy by lctric transmission 6s a fa f H G JC 15 1 Cs 4 h K J P t t t puts out nrgy by lctric transmission 6s a fa tf 1 6 JC 18 1 Cs 4 h 1 h H G K J 1 h 49 J 469 J usful output 49 J (a) fficincy total input 469 J 5 (b) Th only plac for th missing nrgy to go is into intrnal nrgy: 469 J 49 J+ E E int 1 J int P741 (c) P P W imagin toasting th battry ovr a fir with 1 J of hat input: Q mc T Q 1 J kg C T 15 1 C mc 15 kg 975 J a f R 14 H G K J H G 1 K J 161 bg R P P P P KJ a f a f a f P % 1% 1 1% % 6 1% KJ a f a f a11 Vf a5 Wf P74 P 5 W R R 4 Ω (a) R ρ R so ρ 4 a4 Ωfπ 5 1 m Ω m b g (b) R R 1 + α T 4 Ω Ω a f a f 11 P R W 17 m

13 6 a fa Ωf ρ 15 1 Ω m 5 m P74 R 98 Ω π 1 m R V Chaptr V (a) E 597 Vm 5 m a f a fa f (b) P 149 V W αb g (c) R R 1 + T T 98 Ω C C 7 Ω a f a f a fa f 149 V 44 R 7 Ω P 149 V W a f a f a fa f K J 1 C 1 J 1 W s P744 (a) U q t 55 h 1 V K J 66 W h 66 kwh 1 s 1 V C 1 J KJ $ 6 (b) Cost H G 66 kwh K J 1 kwh 96 P745 P a f R af a1 f P R 1 8 W *P746 (a) Th rsistanc of 1 m of 1-gaug coppr wir is ρ ρ 4ρ Ω m 1m R π d π d π 5 1 m b g 8 Th rat of intrnal nrgy production is P R Ω 5 W l (b) P l R 4ρ π d P P l Cu l ρ ρ Cu P l Ω m 5 W 41 W Ω m Ω a f luminum of th sam diamtr will gt hottr than coppr

14 118 Currnt and Rsistanc *P747 Th nrgy takn in by lctric transmission for th fluorscnt lamp is 6s P t H G 11 Js a1 hf K J 1 h H G K J 6 $ 8 H G k K J cost 96 1 J kwh 1 or th incandscnt bulb, 96 1 W s J 6 J KJ H G h 6 s K J $ 88 6s P t H G 4 W a1 hf K J 1 h 7 $ 8 cost J KJ $ J saving $ $ 88 $ J P748 Th total clock powr is Js 6 H G 5 K J 4 1 H G s clocks K J 1 Jh clock 1 h W rom Q out in, th powr input to th gnrating plants must b: 1 in out Jh Q t W t Jh and th rat of coal consumption is Rat kg coal KJ 95 5 Jh 1 kg coal h 95 mtric ton h 6 1 J af a17 fa11 V f 187 W a 187 kw fa4 h f 4 49 kwh P749 P Enrgy usd in a 4-hour day H G K J cost 4 49 kwh $6 $ kwh a fa f b gb ga f P75 P 1 V 4 W E int 5 kg J kg C 77 C 161 kj Eint J t P 4 W 5 67 s

15 Chaptr P751 t oprating tmpratur, a fa f (a) P 15 1V 184W (b) Us th chang in rsistanc to find th final oprating tmpratur of th toastr a f 1 R R 1 + α T T T 441 C T C+ 441 C 461 C *P75 You pay th lctric company for nrgy transfrrd in th amount E P t 7 d (a) P t H G K J H G 86 4 s K J 1 J 4 W a wksf K J 1 wk 1 d 1 W s P t H G K J H G K J 7 d 4 h H G k 4 W a wksf K J 1 4 P t 1 wk 1 d 1 H G K J H G K J 7 d 4 h H G k K J H G 1 $ 4 W a wksf K J 1 wk 1 d 1 kwh 48 4 MJ kwh $1 61 (b) P t H G K J 1 h H G k K J H G 1 $ 97 W a minf K J 6 min 1 kwh $ (c) P t H G K J 1 h H G k K J H G 1 $ 5Wa4 minf K J 6 min 1 kwh $ 416 P75 Considr a 4-W blow dryr usd for tn minuts daily for a yar Th nrgy transfrrd to th dryr is b gb ga f KJ kwh P t 4 Js 6 sd 65 d J 1 kwh J W suppos that lctrically transmittd nrgy costs on th ordr of tn cnts pr kilowatt-hour Thn th cost of using th dryr for a yar is on th ordr of a fb g Cost kwh $ 1 kwh $ ~ $1

16 1 Currnt and Rsistanc dditional Problms a f V P754 (a) so P R R af a f 1 V a f a f 1 V R 576 Ω and R P 5 W P 1 W 144 Ω (b) P 5 W Q 1 C 8 1 V t t 1 C t 48 s 8 Th bulb taks in charg at high potntial and puts out th sam amount of charg at low potntial U 1 J 1 J (c) P 5 W t 4 s t t 5 W Th bulb taks in nrgy by lctrical transmission and puts out th sam amount of nrgy by hat and light b gb ga f (d) U P t 5 Js 86 4 sd d J Th lctric company slls nrgy H G K J H G K J Cost J $7 k W s h kwh 1 J KJ 6 H G s K J $1 6 $ 7 kwh 8 Cost pr oul $ J 6 kwh 6 1 J 1 1 Q *P755 Th original stord nrgy is Ui Q Vi C KJ (a) Whn th switch is closd, charg Q distributs itslf ovr th plats of C and C in paralll, Q prsnting quivalnt capacitanc 4C Thn th final potntial diffrnc is f for 4 C both (b) Q Th smallr capacitor thn carris charg C V C C Q f 4 4 charg C Q Q 4 C 4 Th largr capacitor carris d i H G K J Th largr (c) Th smallr capacitor stors final nrgy 1 1 C V C Q Q f 4C C capacitor posssss nrgy 1 C Q 4C Q K J C (d) Th total final nrgy is Q Q Q + Th loss of potntial nrgy is th nrgy C C 8C apparing as intrnal nrgy in th rsistor: Q Q Q + E int Eint C 8C 8C

17 Chaptr 7 11 P756 W find th drift vlocity from nqv nqv π r vd nq r 1 π m C π 1 m x x 1 m 8 v t s 7 yr 4 t v 4 1 m s d d ms P757 W bgin with th diffrntial quation α ρ ρ 1 d dt (a) Sparating variabls, ln ρ ρ KJ α TT z ρ dρ ρ T z ρ T αdt b g and ρ ρ α TT b g (b) rom th sris xpansion x 1 + x, ax << 1f, b g ρ ρ 1 + α TT a f 5 Ω P758 Th rsistanc of on wir is 1 mi 5 Ω mi K J Th whol wir is at nominal 7 kv away from ground potntial, but th potntial diffrnc btwn its two nds is R P759 ρ b ga f R 1 5 Ω 5 kv a f b g Thn it radiats as hat powr P 5 1 V 1 5 MW a f (m) R ( Ω) ρ ( Ω m) ρ Ω m (in agrmnt with tabulatd valu of Ω m in Tabl 71) P76 wirs 1 m a 18 Ω R 1 m 6 Ω m f a f a f a fb g (a) R hom lin V a f a fa f (b) P V 1 8 kw a f b g (c) P wirs R 11 6 Ω 46 W

18 1 Currnt and Rsistanc dv P761 (a) E 4 i 8 i dx a5 f V Vm m ρ 4 1 Ω m 5 m (b) R 4 π 1 1 m a 8 a f f 67 Ω (c) 4 V R 67 Ω 68 (d) J 68 8 i 1 i m i Mm m π 8 8 () ρj Ω m i m 8 i V m E a f dv x V P76 (a) E i i dx L ρ 4ρL (b) R π d (c) R Vπ d 4ρ L (d) J V i i ρ L V () ρ J i E L P76 R R + TT 1 αb g so T T n this cas, 1 6 P764 R L N O Q L NM O QP 1 R + M P T R α α 1, so T T a f 1 V Thrfor, R 6 α af C thus and 6 Ω C

19 P765 (a) P P 8 1 W so 1 V 667 Chaptr 7 1 U 1 J (b) t 5 1 P 8 1 W 7 b g and x v t ms 5 1 s 5 km ρ ρ 1+ α T T 1+ α T T P766 (a) W bgin with R 1+ α T T which rducs to R s b g b g b g b g αb g 1+ α bttg R 1+ α T T 1+ TT, 8 (b) or coppr: ρ 17 1 Ω m, α 9 1 C 1, and α C 1 Th simpl formula for R givs: R 8 a f ρ 17 1 π 1 1 a f a f 1 18 Ω R 1 8 Ω C 1 C C 1 4 Ω whil th mor complicatd formula givs: a18 Ωf C a8 C f C a8 Cf R C 8 C a f 1418 Ω P767 Lt α b th a tmpratur cofficint at C, and α b th tmpratur cofficint at C Thn ρ ρ 1 + α T Cf a, and ρ ρ 1+ α T C f must both giv th corrct rsistivity at any tmpratur T That is, w must hav: a f a f (1) ρ 1+ α T C ρ 1+ α T C a f, Stting T in quation (1) yilds: ρ ρ 1 α C a f and stting T C in quation (1) givs: ρ ρ 1+ α C Put ρ from th first of ths rsults into th scond to obtain: continud on nxt pag ρ ρ 1 α C 1 + α C a f a f

20 14 Currnt and Rsistanc a f a f 1 Thrfor 1+ α C 1 α C which simplifis to α α a f 1 α C rom this, th tmpratur cofficint, basd on a rfrnc tmpratur of C, may b computd for any matrial or xampl, using this, Tabl 71 bcoms at C : Matrial Tmp Cofficints at C Silvr 41 1 C Coppr 4 1 C Gold 6 1 C luminum 4 1 C Tungstn 49 1 C ron 56 1 C Platinum 45 1 C Lad 4 1 C Nichrom 4 1 C Carbon 5 1 C Grmanium 4 1 C Silicon 1 C P768 (a) thin cylindrical shll of radius r, thicknss dr, and lngth L contributs rsistanc dr ρd ρdr rl H G ρ bπ g π L K J dr r Th rsistanc of th whol annulus is th sris summation of th contributions of th thin shlls: R z r r a a H G KJ b ρ dr ρ rb ln π L r π L r (b) n this quation ρ r ln π L r w solv for H G b a KJ π ρ L V lnbr r b a g

21 Chaptr 7 15 P769 Each spakr rcivs 6 W of powr Using P R, w thn hav P 6 W 87 R 4 Ω Th systm is not adquatly protctd sinc th fus should b st to mlt at 87, or lss P77 E or dv E dx RE dq E dv E E σ dt R ρ ρ dx σ dv dx Currnt flows in th dirction of dcrasing voltag Enrgy flows as hat in th dirction of dcrasing tmpratur ρdx P771 R z z ρdx wy y whr y y + ρ dx ρl R w y + y y L x wy y R L z ρl wy y 1 y L 1 b g b g y ln g y1kj b 1 ln x L NM y 1 y + y L 1 x O QP L G P771 P77 rom th gomtry of th longitudinal sction of th rsistor shown in th figur, w s that ab rf ab af y h rom this, th radius at a distanc y from th bas is r aa bf y +b h dy or a disk-shapd lmnt of volum dr ρ π r : R ρ dy a b y h + b h z π a fb g G P77 Using th intgral formula du 1 z a au + b f, R aau a + bf ρ π h ab *P77 (a) ρ d Th rsistanc of th dilctric block is R σ κ Th capacitanc of th capacitor is C d d κ κ Thn RC σ d σ is a charactristic of th matrial only (b) a f 16 1 κ ρκ R 75 1 Ω m C 9 σ C C 14 1 N m Ω

22 16 Currnt and Rsistanc L M N H G V P774 xp M kt B KJ V 1 and R P O P Q 9 with 1 1, C, and k B 18 1 J K Th following includs a partial tabl of calculatd valus and a graph for ach of th spcifid tmpraturs (i) or T 8 K : af af af V R Ω (ii) or T K : af af af V R Ω G P774(i) (iii) or T K : af af af V R Ω G P774(ii) G P774(iii)

23 *P775 (a) Think of th dvic as two capacitors in paralll Th on on th lft has κ 1 1, 1 + x K J Th quivalnt capacitanc is κ 1 1 κ κ + + x x a x κ κxf d d d d d K J + K J + + Chaptr 7 17 (b) Th charg on th capacitor is Q C V Q a + x+ κ κxf d Th currnt is dq dq dx V Vv + + v a κf aκ 1f dt dx dt d d Th ngativ valu indicats that th currnt drains charg from th capacitor Positiv currnt is clockwis vaκ 1f d NSWERS TO EVEN PROBLEMS P7 64 h P7 171 Ω P74 (a) s th solution; (b) 15 m P74 15 Ω P76 (a) 17 ; (b) 85 k m P76 5, 4 Ω P78 (a) 99 5 k m ; (b) 8 mm P71 (a) 1 nm ; (b) no; s th solution P P74 (a) 5; (b) 1 J; (c) 151 C P74 (a) 17 m; (b) 4 W P71 M m P744 (a) 66 P714 (a) 75 kω ; (b) 56 m P Ω m P MΩ P7 (a) 777 nω ; (b) 8 µ m s P7 r r l Cu P74 78 Ω 19 P76 (a) nothing; (b) doubls; (c) doubls; (d) nothing P P7 carbon, 4 44 kω ; nichrom, 5 56 kω kwh ; (b) 96 P746 (a) 5 W; (b) 41 W; no P mtric ton h P75 67 s P75 (a) $161; (b) $5 8; (c) $416 P754 (a) 576 Ω and 144 Ω ; (b) 48 s; Th charg is th sam Th charg-fild systm is in a lowr-nrgy configuration (c) 4 s; Th nrgy ntrs by lctric transmission and xits by hat and lctromagntic radiation; (d) $16; nrgy; $J

24 18 Currnt and Rsistanc P756 7 yr P766 (a) s th solution; (b) 1418 Ω narly agrs with 14 Ω P758 5 MW rb P768 (a) R ρ π ln ; (b) ρ L V P76 (a) 116 V ; (b) 1 8 kw ; (c) 46 W π L r lnbr r i P76 (a) E V L ; (b) R 4 ρ V d ; (c) L π d π P77 s th solution 4 ρ L ; P77 s th solution i (d) J V ρ L ; () s th solution P774 s th solution P764 Ω a b a g