西尔斯现代物理答案30

30.1: a) V ,0.270/s)A 830(H)1025.3()/(412=⨯==-dt di M ε and is constant. b) If the second coil has the same changing current, then the induced voltage is the same and V .270.01=ε30.2: For a toroidal solenoid, .2/and ,/110B 1222r A i N i N M B πμ=ΦΦ=So, .2/210r N AN M πμ=30.3: a) H.1.96A)(6.52/Wb)0320.0()400(/122==Φ=i N M Bb) When Wb.107.11(700)/H)(1.96A)54.2(/A,54.2312B 21-⨯===Φ=N M i i30.4: a) H.106.82s)/A 0.242(/V 1065.1)/(/332--⨯=-⨯==dt di M ε b) A,20.1,2512==i NWb.103.2725/H)10(6.82A)20.1(/43212--⨯=⨯==Φ⇒N M i Bc) mV .2.45s)/A (0.360H)1082.6(/and s /A 360.0/3212=⨯===-dt Mdi dt di ε30.5: Ωs.1A)s /(V 1AC)s /(J 1A /J 1A /Nm 1A /Tm 1A /Wb 1H 1222=======30.6: For a toroidal solenoid, )./(//dt di i N L B ε=Φ= So solving for N we have:turns.238s)/A (0.0260Wb)(0.00285A)(1.40V)106.12()/(/3=⨯=Φ=-dt di i N B ε30.7: a) V.104.68s)/A (0.0180H)260.0()/(31-⨯===dt di L εb) Terminal a is at a higher potential since the coil pushes current through from b to a and if replaced by a battery it would have the + terminal at .a30.8: a) H.130.0m)120.0(2)m 1080.4()1800()500(2/252020m m =⨯==-πμπr A N μK K L b) Without the material, H.102.60H)130.0(500114m m -⨯===K L K L30.9: For a long, straight solenoid:.//and /200l A N μL l NiA μi N L B B =⇒=ΦΦ=30.10: a) Note that points a and b are reversed from that of figure 30.6. Thus, accordingto Equation 30.8, s./A 00.4H 0.260V 04.1-===--L V V dt dia b Thus, the current is decreasing.b) From above we have that .)s /A 00.4(dt di -= After integrating both sides of thisexpression with respect to ,t we obtainA.4.00s)(2.00A/s)(4.00A)0.12(s)/A 00.4(=-=⇒∆-=∆i t i30.11: a) H.0.250A/s)(0.0640/V )0160.0()/(/===dt di L εb) Wb.104.50(400)/H)(0.250A)720.0(/4-⨯===ΦN iL B30.12: a) J.540.02/A)(0.300H)0.12(2122===LI Ub) W.2.16)180(A)300.0(22=Ω==R I Pc) No. Magnetic energy and thermal energy are independent. As long as the current is constant, constant.=U30.13: πrAl N μLI U 4212202==turns.2850A)0.12()m 1000.5(J)(0.390 m)150.0(44224020=⨯==⇒-μπAI μπrU N30.14: a) J.101.73h)/s 3600h/day (24W)200(7⨯=⨯==Pt Ub) H.5406A)(80.0J)1073.1(22212722=⨯==⇒=I U L LI U30.15: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability μ is used in place of .0μ30.16: a) free space: J.3619)m 0290.0(2T)(0.560V 2V 30202====μμB u U b) material with J.04.8)m 0290.0()450(2T)(0.560V 2V 4503020m 2m ====⇒=μμK B u U K30.17: a) 32602002m 1.25T)(0.600J)1060.3(22Volume 2=⨯==⇒==μμμB U B Vol U u . b) T.9.11T 4.141m)(0.400J)1060.3(22236002=⇒=⨯==B Vol U B μμ30.18: a) .mT 35.4)m 0690.0(2)A 50.2()600(200===πμπμr NI Bb) From Eq. (30.10), .m /J 53.72)T 1035.4(2302302=⨯==-μμB u c) Volume .m 1052.1)m 1050.3()m 0690.0(22V 3626--⨯=⨯==ππrAd) .J 1014.1)m 1052.1()m /J 53.7(5363--⨯=⨯==uV Ue) .H 1065.3)m 0690.0(2)m 1050.3()600(26262020--⨯=⨯==πμπμr A N L J 1014.1)A 50.2()H 1065.3(21215262--⨯=⨯==LI U same as (d).30.19: a) .s /A 40.2H50.2V00.60When .==⇒=-=dt di i L iR dt di εb) When .A/s 800.0H 50.2)00.8()A 500.0(V 00.6A 00.1=Ω-=⇒=dt di i c) At A.413.0)1(8.00V00.6)1(s 200.0s)(0.250H)50.2/00.8()/(=-Ω=-=⇒=Ω--e e R εi t t L R d) As A.750.08.00V00.6=Ω=→⇒∞→R i t ε30.20: (a) mA,30A 030.01000V30max ===Ωi long after closing the switch.)bV4.0V 26V 30εV 26A)(0.0259Ω)(1000A0.0259e 1A 0.030)e(1R Battery L R )(max 1020=-=-=====⎪⎭⎫⎝⎛-=-=--V V Ri V i i μs μsR /L /t(or, could use s)20at μ==t L V dtdiL c)30.21: a) R L e R i t /),1(//=-=-τετ2121m ax m ax and ,)(1when 2/so /==-==--τ/t τ/t e e i i R εiμs L t τ/t 3.1750.0H)10(1.252)ln (R ln2and )(ln 321=Ω⨯===-- b) m ax 221m ax 221;Li U Li U ==2/when m ax m ax 21i i U U ==2929.02/11e so 211//=-==---τt τt e μs L t 30.7R /ln(0.2929)=-=30.22: a) A 13.2H0.115J)260.0(22212===⇒=L UI LI UV .256)(120A)13.2(=Ω==⇒IR εb) ⎪⎭⎫⎝⎛=====--20)/(222)/(2121212121and LI U e Li Li U Ie i t L R t L Rs.1032.321ln )2(120H 115.021ln 2214)/(2--⨯=⎪⎭⎫⎝⎛Ω-=⎪⎭⎫ ⎝⎛-=⇒=⇒R L t e t L R30.23: a) A.250.0240V600=Ω==RI εb) A.137.0A)250.0()s 10(4.00H)0.160/(240)/(04===-⨯Ω--e eI i tL R c) ,V 9.32)240()A 137.0(=Ω===iR V V ab cb and c is at the higher potential.d).s 1062.421ln )240()H 160.0(21ln 2142/1)/(02/1--⨯=⎪⎭⎫ ⎝⎛Ω-=⎪⎭⎫⎝⎛-=⇒==R L t e I i t L R30.24: a) .V 60and 00At ==⇒=bc ab v v t b) .0and V 60As →→⇒∞→bc ab v v tc) .V 0.24V 0.36V 0.60and V 0.36A 150.0When =-===⇒=bc ab v iR v i30.25: a) )1(00.8)V 00.6()1()1()H 50.2/00.8(2)/(2)/(0t t L R t L R e e R eI i P Ω----Ω=-=-==εεε ).1()W 50.4()s20.3(1te P ---=⇒b) 2)H 50.2/00.8(22)/(22)1(00.8)V 00.6()1(t t L R R e eR εR i P Ω---Ω=-== .)1()W 50.4(2)s20.3(1tR e P ---=⇒c) )()1()/(2)/(2)/()/(t L R tL R t L R t L R L e eRεe L εL e R εdt di iL P -----=⎪⎭⎫ ⎝⎛-== ).()W 50.4()s 40.6()s 20.3(11ttL eeP -----=⇒d) Note that if we expand the exponential in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor.30.26: When switch 1 is closed and switch 2 is open:.)/(ln 0)/(0000L R t i I te I i t LRI i t d L R i i d L R i dt di iR dt di L -=⇒-=⇒'-=''⇒-=⇒=+⎰⎰30.27: Units of ==ΩΩ=Ω=s /)s (/H /R L units of time.30.28: a) πf LCω21==.H 1037.2)F 1018.4()106.1(414131226222--⨯=⨯⨯==⇒ππC f L b) F.1067.3)H 1037.2()1040.5(4141113252m in 22m ax --⨯=⨯⨯==ππL f C30.29: a) )F 1000.6()H 50.1(2225-⨯===πLC πωπT s.rad 105,s 0596.0==ωb) .C 1020.7)V 0.12)(F 1000.6(45--⨯=⨯==CV Qc) .J 1032.4)V 0.12)(F 1000.6(212132520--⨯=⨯==CV Ud) .0)cos(,0At =⇒+===φφωt Q Q q t⎪⎪⎭⎫ ⎝⎛⨯⨯===--)F 1000.6)(H 50.1(s 0230.0cos)C 1020.7()cos(s,0230.054ωt Q q t C.1043.54-⨯-=Signs on plates are opposite to those at .0=te) )sin(,s 0230.0ωt ωQ dtdqi t -===A.0.0499H)10(6.00H)(1.50s 0.0230sinH)10H)(6.00(1.50C 107.20554-=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯-=⇒---iPositive charge flowing away from plate which had positive charge at .0=tf) Capacitor: .J 102.46F)102(6.00C)10(5.43235242---⨯=⨯⨯==C q U C Inductor: .J 101.87A)(0.0499H)50.1(2121322-⨯===Li U L30.30: (a) Energy conservation says (max ) = (max )C L U UA0.871H1012F1018V)22.5(CV 212136max 22max =⨯⨯===--L C V i LiThe charge on the capacitor is zero because all the energy is in the inductor. (b)LC T πωπ22==at 41 period: F)10(18H)10(122)2(414163--⨯⨯==ππLC Ts 1030.74-⨯=at 43 period: s 1019.2)s 1030.7(34334--⨯=⨯=T(c) C CV q μμ405)V (22.5F)18(0===30.31: F 0.30V1029.4C1015039μ=⨯⨯==--V Q C For an L-C circuit, LC ω1=and LC ωπT π22==mH 601.0)2(2==CT L π30.32: rad/s 1917)F 1020.3()H 0850.0(16=⨯=-ωa) C 1043.4srad 1917A 1050.874max maxmax max --⨯=⨯==⇒=ωi Q ωQ i b) From Eq. 31.26 2142722s 1917A 1000.5)C 1043.4(⎪⎪⎭⎫ ⎝⎛⨯-⨯=-=---LCi Q q.C 1058.37-⨯=30.33: a) )s A 80.2()F 1060.3()H 640.0(01622-⨯==⇒=+dt di LC q q LC dtq d C.1045.66-⨯=b) .V 36.2F1060.3C1050.866=⨯⨯==--C q ε30.34: a) .m ax m ax m ax m ax m ax LC i ωiQ ωQ i ==⇒=J 450.0)F 1050.2(2)C 1050.1(2.C 1050.1)F 1050.2()H 400.0()A 50.1(1025max2max 510max =⨯⨯==⇒⨯=⨯=⇒----C Q U Q b) 1410s 1018.3)F 1050.2()H 400.0(11222--⨯=⨯===πππωLC f (must double the frequency since it takes the required value twice per period).30.35: [].s s A A1s s C V ΩV C s ΩV C H F H ][222s LC LC =⇒=⋅⋅=⋅⋅=⋅⋅=⋅=⋅=30.36: Equation (30.20) is .0122=+q LC dtq d We will solve the equation using: .110)cos()cos(1).cos()sin()(cos 2222222LC ωLC ωt LC Q ωt Q ωq LC dt q d ωt Q ωdtqd ωt ωQ dt dq ωt Q q =⇒=⇒=+++-=+⇒+-=⇒+-=⇒+=ωφφφφφ30.37: a) .)(cos 2121222C ωt Q C q U C φ+==.1since ,)(sin 21)(sin 21212222222LCωCωt Q ωt Q L ωLi U L =+=+==φφ b) )(sin 21)(cos 2122222Total φφ+++=+=ωt Q L ωωt C Q U U U L C )(sin 121)(cos 212222φφ+⎪⎭⎫ ⎝⎛++=ωt Q LC L ωt C Q ))(sin )((cos 21222φφ+++=ωt ωt CQ Total 221U CQ ⇒=is a constant.30.38: a) )cos()2/(φ+'=-t ωAe q t L R)sin(22)cos(2).sin()cos(2)2/()2/(222)2/()2/(φφφφ+''++'⎪⎭⎫⎝⎛=⇒+''-+'-=⇒----t ωe L R A ωt ωe L R A dt q d t ωAe ωt ωe LR A dt dq t L R t L R t L R t L R ).cos()2/(2φ+''--t ωAe ωt L R222222222410122L RLC ωLC L R L R q LC q dt dq L R dt q d -='⇒=⎪⎪⎭⎫ ⎝⎛+-'-⎪⎭⎫ ⎝⎛=++⇒ωb) :0,,0====dtdqi Q q t At .4//122tan 2and cos 0sin cos 2and cos 22LR LC L RωL RQ ωL QR Q A A ωA LRdt dq Q A q --='-='--=⇒='--===⇒φφφφφ30.39: Subbing ,1,,,Ck R b L m q x →→→→we find: a) Eq. (13.41): .0:)27.30.(Eq 02222=++→=++LCqdt dq L R dt q d m kx dt dx m b dt x db) Eq. (13.43): .41:)28.30.(Eq 42222LR LC ωm b m k ω-='→-=' c) Eq. (13.42): ).cos(:)28.30.(Eq )cos()2/()2/(φφ+'=→+'=--t ωAe q t ωAe x t L R t m b30.40: .A V V C s F H 2Ω=⎥⎦⎤⎢⎣⎡⇒Ω=⋅Ω=⋅Ω==⎥⎦⎤⎢⎣⎡C L C L30.41: LC LC L R LC LC L R LC L R LC 61126114614122222-=⇒⎪⎭⎫ ⎝⎛-=⇒=-='ω. 4.45F)10(4.60H)6(0.2851F)10(4.60H)(0.2851)H 285.0(244Ω=⨯-⨯=⇒--R30.42: a) When s.rad 298F)10(2.50H)(0.45011,050=⨯===-LCωRb) We want 22220)95.0(411)41(95.0=-=-⇒=LC R LC L R LC ωω .8.83F)10(2.50(0.0975)H)4(0.450))95.0(1(452Ω=⨯=-=⇒-C LR30.43: a)b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph.30.44: a) ]s)240cos[()A 124.0((t πdtdL dt di Lε-=-= ).)s 240((sin )V 4.23())s 240sin(()240()A 124.0()H 250.0(t πt ε+=+=⇒ππb) ,0;V 4.23m ax ==i εsince the emf and current are ︒90out of phase. c) ,0;A 124.0m ax ==εi since the emf and current are ︒90 out of phase.30.45: a) ).(ln 22)(2)(000a b Nihμr dr Nih μhdr r πNi μhdr B bab a b a B ππ==⎪⎪⎭⎫ ⎝⎛==Φ⎰⎰⎰b) ).(ln 220a b πhN μi N L B =Φ= c) .2L 2)()/)(1(ln )/(ln 2022⎪⎭⎫ ⎝⎛-≈⇒⋅⋅⋅+-+-≈--=a a b πh N μa a b a a b a a b a b30.46: a) .122210122101110122122212l r πN N μl A N N μl IA N μIA A N A A I N I N M B B ===Φ=Φ= b) .1122210112102222dtdi l r πN N μdt di l A N μN dt d N εB ==Φ= c) .212221022121dtdi l r πN N μdt diM dt di M ε===30.47: a) .H 5.7)/A 00.4/()V 0.30()//(===⇒-=s dt di εL dtdiL ε b) .Wb 360)s 0.12)(V 0.30(==Φ⇒∆=Φ-Φ⇒Φ=f i f t εdt d ε c) W.1440/s)A 00.4)(A 0.48)(H 50.7(===dtdiLi P L.0104.0W 138240)0.60()A 0.48(22R =⇒=Ω==RL P PR i P30.48: a) ))s 0250.0/cos()A 680.0(()H 1050.3(3t dtddt di L πε-⨯==.V 299.0s0250.0)A 680.0)(H 1050.3(3max =⨯=⇒-πεb) .Wb 1095.5400)A 680.0)(H 1050.3(63m ax max--⨯=⨯==ΦN Li B c) .)s 0250.0/(sin )s 0250.0/)(A 680.0)(H 1050.3()(3t dtdiL t ππε-⨯-=-=V230.0)())s 0180.0)(s 6.125((sin )V 299.0(s)0180.0())s 6.125sin(()V 299.0()(11=⇒-=⇒-=⇒--t t t εεε30.49: a) Series: ,eq 2211dtdiL dt di L dt di L =+but i i i ==21for series components so eq 2121thus ,L L L dtdidt di dt di =+==b) Parallel: Now . where ,21eq 2211i i i dt diL dt di L dt di L +===.11 and But .So 121eq 2eq 1eq 2eq 22eq 121-⎪⎪⎭⎫⎝⎛+=⇒+=⇒==+=L L L dt di L L dt di L L dt di dt di L L dt di dt di L L dt di dt di dt di dt di30.50: a) ⎰=⇒=⇒=⋅.2200encl 0πriμB i μπr B I μd i Bb) .20ldr πriμBdA d B ==Φc) ⎰⎰==Φ=Φba b a B B ).a b (πilμr dr πil μd ln 2200 d) ).ln(20a b πμl i N L B=Φ=e) ).ln(4)ln(2212120202a b πli μi a b πμlLi U ===30.51: a) .2200encl 0πriμB i μr πB I μd =⇒=⇒=⋅⎰l Bb) .4)2(221)2(22020002dr r l i rdr l r i rdr l u udV dU B u πμππμμπμ=⎪⎪⎭⎫⎝⎛===⇒= c) )./(ln 442020a b πli μr dr πl i μdU U ba ba===⎰⎰ d) ),/(ln 2221022a b πμl i UL Li U ==⇒=which is the same as in Problem 30.50.30.52: a) ,22210110111111rπAN μr πi N μi A N i N L B =⎪⎪⎭⎫ ⎝⎛=Φ= r πAN μr πi N μi A N i N L B 22220220222222=⎪⎪⎭⎫ ⎝⎛=Φ=b) .2222122021022102L L r πA N μr πA N μr πA N N μM ==⎪⎪⎭⎫ ⎝⎛=30.53: E εμE μεB μB E εu u E B 00200022022==⇒=⇒= T.1017.2)V/m 650(600-⨯==με30.54: a) .1860A1045.60.123Ω=⨯==-V i V R f b) )/1(ln )/1(ln )1()/(f f t L R f i i RtL i i L Rt e i i --=⇒--=⇒-=- H.963.0))45.6/86.4(1(ln )s 1025.7)(1860(4=-⨯Ω-=⇒-L30.55: a) After one time constant has passed:.J 281.0)A 474.0)(H 50.2(2121A 474.0)1(00.8V 00.6)1(2211===⇒=-Ω=-=--Li U e e R i εOr, using Problem (30.25(c)):⎰⎰---==.)()50.4(7/30)40.6()20.3(dt e e W dt P U t t LJ 281.040.6)1(20.3)1()W 50.4(21=⎪⎪⎭⎫ ⎝⎛---=--e eb) ⎪⎭⎫⎝⎛-+=-=--⎰)1()W 50.4()1()W 50.4(1/0)/(tot e R L R L dt e U RL t L RJ 517.000.8H 50.2)W 50.4(1tot =Ω=⇒-e U c) dt e e U t L R R L t L R R )21()W 50.4()/(2/0)/(--⎰+-=⎪⎪⎭⎫ ⎝⎛---+=--)1(2)1(2)W 50.4(21e R L e R L R L J.236.0)168.0(00.8H50.2)W 50.4(=Ω=⇒R U The energy dissipated over the inductor (part (a)), plus the energy lost over the resistor (part (c)), sums to the total energy output (part (b)).30.56: a)J.1000.5240V 60)H 160.0(21212132220-⨯=⎪⎪⎭⎫ ⎝⎛Ω=⎪⎭⎫ ⎝⎛==R L Li U ε b) t L R L tL R e RRi dt di iL dt dU i L R dt di eRi )/(222)/(--=-==⇒-=⇒=εεW.52.4240)60()1000.4)(160.0/240(224-=Ω-=⇒-⨯-e V dt dU L c) In the resistor:.W 52.4240)V 60(.)1000.4)(160.0/240(22)/(2224=Ω===-⨯--e e R εR i dt dU t L R R d) tL R R e RεR i t P )/(222)(-==J,1000.5)240(2)H 160.0()V 60(232220)/(22-∞-⨯=Ω===⇒⎰R L R εe R εU t L R Rwhich is the same as part (a).30.57: Multiplying Eq. (30.27) by i , yields:.02121R 22222=++=⎪⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛+=++=-+C L R P P P Cq dt d Li dt d R i dt dq C q dt di Li R i i C q dt di Li i That is, the rate of energy dissipation throughout the circuit must balance over all of the circuit elements.30.58: a) If 243cos 832cos )cos(83QQ T T Q t Q q T t =⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛==⇒=ππω.2221221212)2(1)(1222222222B E U Cq C Q LC Q L Li U LCQ Q Q LCq Q LC i =====⇒=-=-=⇒b) The two energies are next equal when .85852T t πωt Q q =⇒=⇒=30.59: μF V U C CV U V C C C C C 222)V 0.12/()J 0160.0(2/2 so;V 0.1222221===== C πf L LCπf 2)2(1so 21==H 31.9gives Hz 3500μL f ==30.60: a) .V 0240.0F1050.2C 1000.646m ax=⨯⨯==--C Q V b) A 1055.1F)1050.2)(H 0600.0(1000.6221346max 22max ---⨯=⨯⨯==⇒=LCQi CQ Lic) .J 1021.7)A 1055.1)(H 0600.0(21218232m ax m ax --⨯=⨯==Li U d) If Cq C Q U U U U i i C L 224343J 1080.1412122max 8max max =⎪⎪⎭⎫⎝⎛==⇒⨯==⇒=-times.all for 2121.C 105.204322max6Cq Li U Q q +=⨯==⇒-30.61: The energy density in the sunspot is ./m J 10366.62/3402⨯==μB u B The total energy stored in the sunspot is .V u U B B = The mass of the material in the sunspot is .ρV m =;21so 2B B U mv U K == V u Vv ρB =221 The volume divides out, and /s m 102/24⨯==ρu v B30.62: (a) The voltage behaves the same as the current. Since , i V R ∝the scope must beacross the Ω150resistor.(b) From the graph, as ,V 25,→∞→R V t so there is no voltage drop across the inductor, so its internal resistance must be zero.)1(/m ax r t R e V V --=when .63.0)1(,m ax 1m ax V V V t e R ≈-==τFrom the graph, when τt V V =≈==ms 5.0 ,V 16 63.0m axH 075.0)150()ms 5.0(ms 5.0/=Ω=→=L R L(c) Scope across the inductor:30.63: a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid.b) At ∞→t the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the solenoid has some resistance .L R The final voltage across the solenoid is ,L IR where I is the final current in the circuit.c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit.d) As 0,=--∞→L IR IR εtV 50=ε and from the graph V 15=L R I (the final voltage across the inductor), soA 3.5/R V )35(and V 35===I R I e) Ω=== 4.3A)(3.5/V )15(so V ,15L R R I LL L V iR V εwhere ,0=-- includes the voltage across the resistance of the solenoid. V9.27,when so ,3.14,10V,50)]1(1[ so ),1(,tot /tot /tot ==Ω=Ω==--=-=-=--l τt L t L V τt R R εe R R εV e R εi iR εV τFrom the graph, L V has this value when t = 3.0 ms (read approximately from the graph), so .mH 43)3.14()ms 0.3(Then ms.0.3/tot =Ω===L R L τ30.64: (a) Initially the inductor blocks current through it, so the simplified equivalentircuit isA 333.0150V 50=Ω==R εi 0,A 333.0resistor)50with parallel in (inductor V 7.16it.through flows current no since 0V16.7A)333.0()50(V 33.3A)333.0()100(23142341===Ω====Ω==Ω=A A A V V V V V (b) Long after S is closed, steady state is reached, so the inductor has no potentialdrop across it. Simplified circuit becomesA 230.050V5.11A 153.075V5.11A,385.0V 11.5V 38.5V 500;V 5.38)A 385.0()100(A 385.0130V50/3214321=Ω==Ω===-====Ω==Ω==i i i V V V V R εi30.65: a) Just after the switch is closed the voltage 5V across the capacitor is zero and there s also no current through the inductor, so ,.054323V V V V V ==+= and since2435and ,0and 0V V V V == are also zero. 04=V means 3V reads zero.1V then must equal 40.0 V, and this means the current read by 1A is A.0.800)(50.0/V )0.40(=ΩA.800.0 so 0but ,14321432=====++A A A A A A A A A;800.041==A A all other ammeters read zero. V 0.401=V and all other voltmeters read zero.b) After a long time the capacitor is fully charged so .04=A The current through the inductor isn’t changing, so .02=V The currents can be calculated from the equivalent circuit that replaces the inductor by a short-circuit.:V0.24)0.50(A 480.0reads A;480.0)33.83(/V)0.40(11=Ω==Ω=I V A IThe voltage across each parallel branch is V 16.0V 4.02V 0.40=-V 0.16,05432====V V V V.that Note zero.reads A.320.0reads means V 0.16.A 160.0reads means V 0.1613243423A A A A A V A V =+==C 192)V (16.0F)0.12(so V 0.16)c 5μμ====CV Q Vd) At t = 0 and .0,2=∞→V t As the current in this branch increases from zero to 0.160 A the voltage 2V reflects the rate of change of current.30.66: (a) Initially the capacitor behaves like a short circuit and the inductor like an open circuit. The simplified circuit becomesA 500.0150V75=Ω==Ri εA,500.0V 0.50parallel)(in V50.0A)50.0()100(,0V 25.0A)50.0()50(2314431======Ω===Ω==A A A V V V V Ri V 2(b) Long after S is closed, capacitor stops all current. Circuit becomesV 0.753=V and all other meters read zero.(c) nC,5630V )(75nF)75(===CV q long after S is closed.30.67: a) Just after the switch is closed there is no current through either inductor and they act like breaks in the circuit. The current is the same through the ΩΩ15.0and 0.40 resistors and is equal to A;455.0A.455.0)0.150.40(V )0.25(41===Ω+ΩA A .032==A Ab) After a long time the currents are constant, there is no voltage across either inductor,and each inductor can be treated as a short-circuit . The circuit is equivalent to:A 585.0)73.42V )0.25(=Ω=IA.0.585reads 1A The voltage across each parallel branch is =Ω-)(40.0A)(0.585V 0.25 A.107.0)0.15(V)60.1(reads A.160.0)0.10V)60.1(reads A.320.0)0.5(V)60.1(reads V.1.60432=Ω=Ω=ΩA A A30.68: (a),s 50.0since ms 40.025m H10τR L τ>>===Ω steady state has been reached, for all practical purposes.A 00.225V 50=Ω==R i εThe upper limit of the energy that the capacitor can get is the energy stored in the inductor initially.C1090.0)F 1020()H 1010()A 00.2(212363max 0max 202max ---⨯=⨯⨯==→=→=Q LCi Q Li C Q U U L C (b) Eventually all the energy in the inductor is dissipated as heat in the resistor.J100.2)A 00.2()H 1010(212122320--⨯=⨯===Li U U L R30.69: a) At ,0=t all the current passes through the resistor ,1R so the voltage ab v is the total voltage of 60.0 V.b) Point a is at a higher potential than point .b c) V 0.60=cd v since there is no current through .2Rd) Point c is at a higher potential than point .be) After a long time, the switch is opened, and the inductor initially maintains thecurrent of .A 40.225.0V 0.6022=Ω==R i R ε Therefore the potential between a and b is .V 0.96)0.40()A 40.2(R 1-=Ω-=-=i v ab f) Point b is at a higher potential than point a .g) V 156)2540()A 40.2()(21-=Ω+Ω-=+-=R R i v cd h) Point d is at a higher potential than point c .30.70: a) Switch is closed, then at some later time:.V 0.15)A/s 0.50()H 300.0(A/s 0.50===⇒=dtdiL v dt di cdThe top circuit loop: 60.0 .A 50.10.40V0.60V 111=Ω=⇒=i R i The bottom loop: A.80.10.25V45.00V 0.15V 60222=Ω=⇒=--i R i b) After a long time: ,A 40.20.25V0.602=Ω=i and immediately when the switch is opened, the inductor maintains this current, so .A 40.221==i i30.71: a) Immediately after 1S is closed, ,V 0.36and ,0,00===cb ac v v i since the inductor stops the current flow.b) After a long time, =0i A 180.015050V 0.360=Ω+Ω=+R R ε,.V 0.27V 00.9V 0.36and ,V 00.9)50()A 18.0(00=-==Ω==cb ac v R i v c) ),1()A 180.0()()1()()s50()(total1total ttL R e t i e R t i ----=⇒-=ε()()().3)V 00.9(1)V 00.9(V 0.36)()(v and1)V 00.9()()()50()50(0)s50(0111ts ts cb tac e eR t i εt e R t i t v ------+=--=-=-==Below are the graphs of current and voltage found above.30.72: a) Immediately after 2S is closed, the inductor maintains the current A 180.0=i through .R The Kirchoff’s Rules around the outside of the circuit yield:.0and V 0.36)V 50()A 72.0(,A 072.0Ω50V360)50()150()18.0()150()18.0(V 0.360000=====⇒=--+=--+cb ac L v v i i R i iR εε b) After a long time, ,V 0.36=ac v and .0=cb v Thus,A 720.050V 0.3600=Ω==R εi A 720.0and ,02==s R i ic) and ,)A 180.0()()(,A 720.0)s 5.12()(total 01t R tL R R e t i e R εt i i ---=⇒== ()tt s e e A t i )s 5.12()s 5.12(1124)A 180.0()180.0()A 720.0()(-----=-=Below are the graphs of currents found above.30.73: a) Just after the switch is closed there is no current in the inductors. There is no current in the resistors so there is no voltage drop across either resistor. A reads zero and V reads 20.0 V.b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the inductors can be replaced by short-circuits. The circuit becomes equivalent toA 267.0)0.75()V 0.20(=Ω=a IThe voltage between points a and b is zero, so the voltmeter reads zero.c) Use the results of problem 30.49 to combine the inductor network into itsequivalent:Ω=0.75R is the equivalent resistance.V0.9V V 0.20so 0V 0.20V0.11)0.75)(A 147.0(A 147.0so ,ms 115.0,0.75,0.20ms 144.0)(75.0mH)8.10(/with ),1)((says Eq.(30.14)=-==--=Ω====Ω===Ω==-=-R L L R R τt V V V iR V i t R V εR L τe R εi30.74: (a) Steady state: A 600.0125V0.75=Ω==Ri ε(b) Equivalent circuit:F6.14F 351F 2511μμμ=+=s s C CEnergy conservation:202212Li C q =)F 106.14()H 1020()A 600.0(630--⨯⨯==LC i qC 41024.3-⨯=s1049.8)F 106.14()H 1020(22)2(4141463---⨯=⨯⨯====πππt LCLC T t30.75: a) Using Kirchhoff’s Rules: and ,01111R εi R i ε=⇒=-).1(0)(222222t L R e R i R i dt di L--=⇒=--εε b) After a long time, .and still,2211R εi R εi == c) After the switch is opened, ,))((21212tL R R e R εi i +-== and the current drops off. d) A 40-W light bulb implies .360W40V)(12022Ω===P V R If the switch is opened, and the current is to fall from 0.600A to 0.150 A in 0.0800 s,then:)s 0800.0)(H 0.22)360(())((2221)A 600.0(A 150.0)A 600.0(R t R R e e i +Ω-+-=⇒=.V 7.12)2.21)(A 600.0(0.21360)00.4ln(0800.0H 0.222222=Ω==⇒Ω=⇒+Ω=⇒R i εR R s e) Before the switch is opened, A 0354.0360V 7.1210=Ω==R i ε30.76: Series:.eq 2121212211dtdi L dt di M dt di M dt di L dt di L ≡+++ But .and 21122121M M M dt di dt didt di i i i ≡=+=⇒+=.and with ,and have We :Parallel .2or ,)2(So 21122112122eq 2121121eq eq 21M M M dtdi dt di dt di dt diL dt di M dt di L dtdi L dt di M dt di L M L L L dtdiL dt di M L L eq ≡==+=+=+++==++ To simplify the algebra let . and ,,21dtdiC dt di B dt di A ===So .,,eq 2eq 1C B A C L MA B L C L MB A L =+=+=+ Now solve for .of terms in and C B A.)2()()()2(0)()()(0)()()(.using 0)()(2111212112121C L L M L M B C L M B L L M B L M B M L C M L B L M B C M L B C A B L M A M L ---=⇒-=--⇒=-+---⇒=-+--⇒-==-+-⇒But ,)2()2()2()(21121211C L L M L M L L M L L M C L M C B C A --+---=----=-= or .2212C L L M L M A ---=Substitute A in B back into original equation. So C L C L L M L M M L L M C L M L eq 2112121)2()(2)(=---+--- .2eq 21212C L C L L M L L M =---⇒Finally, ML L M L L L 221221eq -+-=30.77: a) Using Kirchhoff’s Rules on the top and bottom branches of the circuit:).1()00).1(0)/1(0)/1(22022)1(22222222)(111112221t C R t t C R t t C t L R e εC Ce R R εt d i q e R εi C i R dt di C q R i εe R εi dt di L R i ε-'----=-='=⇒=⇒=--⇒=---=⇒=--⎰ b) .A 1060.95000V 0.48,0)1()0(3022011-⨯=Ω===-=e R εi e R εi c) As .0,A 92.10.25V 0.48)1()(:22111===Ω==-=∞∞→∞-∞-e R εi R εe R εi t A good definition of a “long time” is many time constants later. d) .)1()1()1(21)()1(2)(12121211t C R t L R t C R t R e R R e e R e R i i ----=-⇒=-⇒=εε Expanding the exponentials like :find we ,!32132 ++++=x x x e x⎪⎪⎭⎫ ⎝⎛-+-=+⎪⎭⎫ ⎝⎛- 2222122112121C R t RC t R R t L R t L R ,)(2122211R R t O C R R L R t =⋅⋅⋅++⎪⎪⎭⎫ ⎝⎛+⇒ if we have assumed that .1<<t Therefore: .s 106.1)F 100.2()5000(H 0.8)F 100.2)(5000)(H 0.8()1()1(113525222222---⨯=⎪⎪⎭⎫ ⎝⎛⨯Ω+⨯Ω=⇒⎪⎪⎭⎫ ⎝⎛+=⎪⎪⎭⎫ ⎝⎛+≈⇒t C R L C LR C R L R te) At .A 104.9)1(25V 48)1(:s 1057.13)25()(1131----⨯=-Ω=-=⨯=t t L R e e R i t ε f) We want to know when the current is half its final value. We note that the current 2i is very small to begin with, and just gets smaller, so we ignore it and find: ).1)(A 92.1()1(A 960.0)()(112111t L R t L R e e R i i ---=-===ε s 22.0)5.0ln(25H 0.8)5.0ln(500.01)(1=Ω==⇒=⇒-R L t e t L R30.78: a) Using Kirchoff’s Rules on the left and right branches:Left: .)(0)(121121εdtdi L i i R dt di L R i i ε=++⇒=-+- Right: .)(0)(221221εCq i i R C q R i i ε=++⇒=-+- b) Initially, with the switch just closed, .0and ,0221===q Rεi i c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat tedious exercise in bookkeeping that is left to the reader. We will show that the initial conditions are satisfied: .0)])0[cos(1()0()]cos()sin()2[(1()(0)0sin()sin(,0At 1112=-=⇒+-=====---Rεi ωt ωt ωRC e R εt i ωR εωt e ωR εq t βt βt d) When does 2i first equal zero? rad/s 625)2(112=-=RC LC ω .s 10256.1rad/s625785.00.7851.00)arctan( 1.00.F)1000.2)(400)(rad/s 625(22)tan(01)tan()2()]cos()sin()2([0)(36112-----⨯==⇒+=+=⇒+=⨯Ω+=+=⇒=+-⇒+-==t ωt ωRC ωt ωt ωRC ωt ωt ωRC e Rt i bt ε。