电路-电路原理-尼尔森Riedel(第九版)(2011)课后习题答案第6章.pdf

6 Inductance, Capacitance, and Mutual Inductance Assessment Problems AP 6.1 a ig= 8e300t 8e1200tA v = Ldig dt = 9.6e300t+ 38.4e1200tV,t 0+ v(0+) = 9.6 + 38.4 = 28.8V b v = 0when38.4e1200t= 9.6e300tort = (ln4)/900 = 1.54ms c p = vi = 384e1500t 76.8e600t 307.2e2400tW d dp dt = 0whene1800t 12.5e900t+ 16 = 0 Let x = e900tand solve the quadraticx2 12.5x + 16 = 0 x = 1.44766,t = ln1.45 900 = 411.05s x = 11.0523,t = ln11.05 900 = 2.67ms p is maximum at t = 411.05s e pmax= 384e1.5(0.41105) 76.8e0.6(0.41105) 307.2e2.4(0.41105)= 32.72W f W is max when i is max, i is max when di/dt is zero. When di/dt = 0, v = 0, therefore t = 1.54ms. g imax= 8e0.3(1.54) e1.2(1.54) = 3.78A wmax= (1/2)(4 103)(3.78)2= 28.6mJ 61 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 62CHAPTER 6. Inductance, Capacitance, and Mutual Inductance AP 6.2 a i = C dv dt = 24 106 d dte 15,000t sin30,000t = 0.72cos30,000t 0.36sin30,000te15,000tA,i(0+) = 0.72A b i ? 80 ms ? = 31.66mA,v ? 80 ms ? = 20.505V, p = vi = 649.23mW c w = ?1 2 ? Cv2= 126.13J AP 6.3 a v = ? 1 C ? Z t 0 idx + v(0) = 1 0.6 106 Z t 0 3cos50,000 xdx = 100sin50,000tV b p(t) = vi = 300cos50,000tsin50,000t = 150sin100,000tW,p(max)= 150W c w(max)= ?1 2 ? Cv2 max = 0.30(100)2= 3000J = 3mJ AP 6.4 a Leq= 60(240) 300 = 48mH b i(0+) = 3 + 5 = 2A c i = 125 6 Z t 0+(0.03e 5x)dx 2 = 0.125e5t 2.125A d i1= 50 3 Z t 0+(0.03e 5x)dx + 3 = 0.1e5t + 2.9A i2= 25 6 Z t 0+(0.03e 5x)dx 5 = 0.025e5t 5.025A i1+ i2= i AP 6.5 v1= 0.5 106 Z t 0+ 240 106e10 xdx 10 = 12e10t+ 2V v2= 0.125 106 Z t 0+ 240 106e10 xdx 5 = 3e10t 2V v1() = 2V,v2() = 2V W = ?1 2 (2)(4) + 1 2 (8)(4) ? 106= 20J 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems63 AP 6.6 a Summing the voltages around mesh 1 yields 4di1 dt + 8d(i2 + ig) dt + 20(i1 i2) + 5(i1+ ig) = 0 or 4di1 dt + 25i1+ 8di2 dt 20i2= 5ig+ 8dig dt ! Summing the voltages around mesh 2 yields 16d(i2 + ig) dt + 8di1 dt + 20(i2 i1) + 780i2= 0 or 8di1 dt 20i1+ 16di2 dt + 800i2= 16dig dt b From the solutions given in part (b) i1(0) = 0.4 11.6 + 12 = 0;i2(0) = 0.01 0.99 + 1 = 0 These values agree with zero initial energy in the circuit. At infi nity, i1() = 0.4A;i2() = 0.01A When t = the circuit reduces to . .i1() = ?7.8 20 + 7.8 780 ? = 0.4A;i2() = 7.8 780 = 0.01A From the solutions for i1and i2we have di1 dt = 46.40e4t 60e5t di2 dt = 3.96e4t 5e5t Also, dig dt = 7.84e4t Thus 4di1 dt = 185.60e4t 240e5t 25i1= 10 290e4t+ 300e5t 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 64CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 8di2 dt = 31.68e4t 40e5t 20i2= 0.20 19.80e4t+ 20e5t 5ig= 9.8 9.8e4t 8dig dt = 62.72e4t Test: 185.60e4t 240e5t 10 290e4t+ 300e5t+ 31.68e4t 40e5t +0.20 + 19.80e4t 20e5t ? = 9.8 9.8e4t+ 62.72e4t 9.8 + (300 240 40 20)e5t +(185.60 290 + 31.68 + 19.80)e4t ? = (9.8 + 52.92e4t) 9.8 + 0e5t+ (237.08 290)e4t ? = 9.8 52.92e4t 9.8 52.92e4t= 9.8 52.92e4t(OK) Also, 8di1 dt = 371.20e4t 480e5t 20i1= 8 232e4t+ 240e5t 16di2 dt = 63.36e4t 80e5t 800i2= 8 792e4t+ 800e5t 16dig dt = 125.44e4t Test: 371.20e4t 480e5t+ 8 + 232e4t 240e5t+ 63.36e4t 80e5t 8 792e4t+ 800e5t ? = 125.44e4t (8 8) + (800 480 240 80)e5t +(371.20 + 232 + 63.36 792)e4t ? = 125.44e4t (800 800)e5t+ (666.56 792)e4t ? = 125.44e4t 125.44e4t= 125.44e4t(OK) 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems65 Problems P 6.10 t 2s : iL= 103 2.5 Z t 0 3 103e4xdx + 1 = 1.2e 4x 4 ? ? ? ? t 0 + 1 = 0.3e4t+ 1.3A,0 t 2s iL(2) = 0.3e8+ 1.3 = 1.3A t 2s : iL= 103 2.5 Z t 2 3 103e4(x2)dx + 1.3 = 1.2e 4(x2) 4 ? ? ? ? t 2 + 1.3 = 0.3e4(t2)+ 1A,t 2s P 6.2a v = Ldi dt = (50 106)(18)e10t 10te10t = 900e10t(1 10t)V 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 66CHAPTER 6. Inductance, Capacitance, and Mutual Inductance b i(200ms) = 18(0.2)(e2) = 487.21mA v(200ms) = 900(e2)(1 2) = 121.8V p(200ms) = vi = (487.21 103)(121.8 106) = 59.34W c delivering 59.34W d i(200ms) = 487.21mA(from part b) w = 1 2Li 2 = 1 2(50 10 6)(0.48721)2 = 5.93J e The energy is a maximum where the current is a maximum: diL dt = 0when1 10t = 0ort = 0.1s imax= 18(0.1)e1= 662.18mA wmax= 1 2 (50 106)(0.66218)2= 10.96J P 6.3a 0 t 2ms : i = 1 L Z t 0 vsdx + i(0) = 106 200 Z t 0 5 103dx + 0 = 5000 200 x ? ? ? ? t 0= 25tA 2ms t :i = 106 200 Z t 2103(0)dx + 2 10 3 = 50mA b i = 25tmA,0 t 2ms;i = 50mA,t 2ms 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems67 P 6.4ai=0t 0 i=50tA0 t 5ms i=0.5 50tA5 t 10ms i=010ms t b v = Ldi dt = 20 103(50) = 1V0 t 5ms v = 20 103(50) = 1V5 t 10ms v=0t 0 v=1V0 t 5ms v=1V5 t 10ms v=010ms t p = vi p=0t 0 p=(50t)(1) = 50tW0 t 5ms p=(0.5 50t)(1) = 50t 0.5W5 t 10ms p=010ms t w=0t 0 w= Z t 0 (50 x)dx = 50 x 2 2 ? ? ? ? t 0= 25t 2 J0 t 5ms w= Z t 0.005(50 x 0.5)dx + 0.625 10 3 =25x2 0.5x ? ? ? ? t 0.005 +0.625 103 =25t2 0.5t + 2.5 103J5 t 10ms w=010ms 0for0 t 17.03sandi 0for17.03s t v = 0when12e10,000t= 80e40,000t . .e30,000t= 20/3sot = 63.24s Thus, v 0for63.24s t Therefore, p 0: w = Z 0 6e50,000tdt Z 0 8e80,000tdt Z 0 0.72e20,000tdt = 6e50,000t 50,000 ? ? ? ? 0 8e 80,000t 80,000 ? ? ? ? 0 0.72e 20,000t 20,000 ? ? ? ? 0 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems611 = (1.2 1 0.36) 104 = 16J Thus, the energy stored equals the energy extracted. P 6.8a v = Ldi dt v = 25 103 d dt10cos400t + 5sin400te 200t = 25 103(200e200t10cos400t + 5sin400t +e200t4000sin400t + 2000cos 400t) v = 25 103e200t(1000sin400t 4000sin400t) = 25 103e200t(5000sin400t) = 125e200tsin400tV dv dt = 125(e200t(400)cos400t 200e200tsin400t) = 25,000e200t(2cos400t sin400t) V/s dv dt = 0when2cos400t = sin400t . .tan400t = 2,400t = 1.11;t = 2.77ms b v(2.77 ms) = 125e0.55sin1.11 = 64.27V P 6.9ai= 1000 20 Z t 0 50sin250 xdx + 10 =2500cos250 x 250 ? ? ? ? t 0 +10 =10cos250tA bp=vi = (50sin250t)(10cos250t) =500sin250tcos250t p=250sin500tW w= 1 2 Li2 = 1 2 (20 103)(10cos250t)2 =1000cos2250tmJ w=(500 + 500cos 500t)mJ 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 612CHAPTER 6. Inductance, Capacitance, and Mutual Inductance cAbsorbing power:Delivering power: 2 t 4 ms0 t 2 ms 6 t 8 ms4 t 6 ms P 6.10i = (B1cos4t + B2sin4t)et/2 i(0) = B1= 10A di dt = (B1cos4t + B2sin4t)(0.5et/2) + et/2(4B1sin4t + 4B2cos4t) = (4B2 0.5B1)cos4t (4B1+ 0.5B2)sin4tet/2 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems613 v = 4di dt = (16B2 2B1)cos4t (16B1+ 2B2)sin4tet/2 v(0) = 60 = 16B2 2B1= 16B2 20 . .B2= 5A Thus, i = (10cos4t + 5sin4t)et/2,A,t 0 v = (60cos4t 170sin4t)et/2V,t 0 i(1) = 26,A;v(1) = 54.25V p(1) = (26)(54.25) = 339.57W delivering P 6.11p = vi = 40te10t 10te20t e20t W = Z 0 pdx = Z 0 40 xe10 x 10 xe20 x e20 xdx = 0.2J This is energy stored in the inductor at t = . P 6.12av(20s)=12.5 109(20 106)2= 5V (end of fi rst interval) v(20s)=106(20 106) (12.5)(400) 103 10 =5V(start of second interval) v(40s)=106(40 106) (12.5)(1600) 103 10 =10V(end of second interval) b p(10s) = 62.5 1012(105)3= 62.5mW,v(10s) = 1.25V, i(10s) = 50mA,p(10s) = vi = (1.25)(50m) = 62.5mW (checks) p(30s) = 437.50mW,v(30s) = 8.75V,i(30s) = 0.05A p(30s) = vi = (8.75)(0.05) = 62.5mW (checks) c w(10s) = 15.625 1012(10 106)4= 0.15625J w = 0.5Cv2= 0.5(0.2 106)(1.25)2= 0.15625J w(30s) = 7.65625J w(30s) = 0.5(0.2 106)(8.75)2= 7.65625J 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 614CHAPTER 6. Inductance, Capacitance, and Mutual Inductance P 6.13For 0 t 1.6s: iL= 1 5 Z t 03 10 3 dx + 0 = 0.6 103t iL(1.6s) = (0.6 103)(1.6) = 0.96mA Rm= (20)(1000) = 20k vm(1.6s) = (0.96 103)(20 103) = 19.2V P 6.14a i = 400 103 5 106 t = 80 103t0 t 5s i = 400 1035 t 20s i = 300 103 30 106 t 0.5 = 104t 0.520s t 50s q= Z 5106 0 8 104tdt + Z 15106 5106 0.4dt =8 104 t2 2 ? ? ? ? 5106 0 +0.4(10 106) =4 104(25 1012) + 4 106 =5C b v = 4 106 Z 5106 0 8 104xdx + 4 106 Z 20106 5106 0.4dx +4 106 Z 30106 20106(10 4x 0.5)dx = 4 106 8 104 x2 2 ? ? ? ? 5106 0 +0.4x ? ? ? ? 20106 5106 +104 x2 2 ? ? ? ? 30106 20106 0.5x ? ? ? ? 30106 20106 # = 4 1064 104(25 1012) + 0.4(15 106) +5000(900 1012400 1012) 0.5(10 106) = 18V v(30s) = 18V 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems615 c v(50s) = 4 106106+ 6 106+ 5000(2500 1012 400 1012) 0.5(30 106) = 10V w = 1 2Cv 2 = 1 2(0.25 10 6)(10)2 = 12.5J P 6.15av= 1 0.5 106 Z 500106 0 50 103e2000tdt 20 =100 103 e2000t 2000 ? ? ? ? 500106 0 20 =50(1 e1) 20 = 11.61V w= 1 2Cv 2 = 1 2(0.5)(10 6)(11.61)2 = 33.7J b v() = 50 20 = 30V w() = 1 2(0.5 10 6)(30)2 = 225J P 6.16a 0 t 10s C = 0.1F 1 C = 10 106 v = 10 106 Z t 0 0.05dx + 15 v = 50 104t + 15V0 t 10s v(10s) = 5 + 15 = 10V b 10s t 20s v = 10 106 Z t 10106 0.1dx + 10 = 106t 10 + 10 v = 106tV10 t 20s v(20s) = 106(20 106) = 20V c 20s t 40s v = 10 106 Z t 20106 1.6dx + 20 = 1.6 106t 32 + 20 v = 1.6 106t 12V,20s t 40s 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. 616CHAPTER 6. Inductance, Capacitance, and Mutual Inductance d 40s t v(40s) = 64 12 = 52V40s t P 6.17iC= C(dv/dt) 0 t 0.5 :iC= 20 106(60)t = 1.2tmA 0.5 t 1 :iC= 20 106(60)(t 1) = 1.2(t 1)mA P 6.18a w(0) = 1 2Cv(0) 2 = 1 2(0.20) 10 6(150)2 = 2.25mJ b v = (A1t + A2)e5000t v(0) = A2= 150V dv dt =5000e5000t(A1t + A2) + e5000t(A1) =(5000A1t 5000A2+ A1)e5000t dv dt (0) = A1 5000A2 i = C dv dt ,i(0) = C dv(0) dt 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems617 . . dv(0) dt = i(0) C = 250 103 0.2 106 = 1250 103 . .1.25 106= A1 5000(150) Thus, A1= 1.25 106+ 75 104= 2 106 V s c v = (2 106t + 150)e5000t i = C dv dt = 0.2 106 d dt(2 10 6t + 150)e5000t i= d dt (0.4t + 10 306)e5000t =(0.4t + 30 106)(5000)e5000t+ e5000t(0.4) =(2000t 150 103+ 0.4)e5000t =(0.25 2000t)e5000tA,t 0 P 6.19a i = C dv dt = 0,t 0 P 6.36a Rearrange by organizing the equations by di1/dt, i1, di2/dt, i2and transfer the igterms to the right hand side of the equations. We get 4di1 dt + 25i1 8di2 dt 20i2= 5ig 8dig dt 8di1 dt 20i1+ 16di2 dt + 80i2= 16dig dt b From the given solutions we have di1 dt = 320e5t+ 272e4t di2 dt = 260e5t 204e4t 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. Problems627 Thus, 4di1 dt = 1280e5t+ 1088e4t 25i1= 100 + 1600e5t 1700e4t 8di2 dt = 2080e5t 1632e4t 20i2= 20 1040e5t+ 1020e4t 5ig= 80 80e5t 8dig dt = 640e5t Thus, 1280e5t+ 1088e4t+ 100 + 1600e5t 1700e4t 2080e5t +1632e4t 20 + 1040e5t 1020e4t ? = 80 80e5t 640e5t 80 + (1088 1700 + 1632 1020)e4t +(1600 1280 2080 + 1040)e5t ? = 80 720e5t 80 + (2720 2720)e4t+ (2640 3360)e5t= 80 720e5t(OK) 8di1 dt = 2560e5t+ 2176e4t 20i1= 80 + 1280e5t 1360e4t 16di2 dt = 4160