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《电路》(英文版)Chpter 6 Problems

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I Let L= 1.25 H in Fig. 6-11, and determine v(t)if v(0) 1(02)=20A L 0.05F ig 6-11 For prob. I 2(a)What value of L in the circuit of Fig 6-11 will result in a transient response of the form, v(t)
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Basic circuit the Chpter6 Problems I Let L= 1.25 H in Fig. 6-11, and determine v(t)if v(0) 1(02)=20A L 0.05F ig 6-11 For prob. I 2(a)What value of L in the circuit of Fig 6-11 will result in a transient response of the form, v(t) Ae+Be? A 4 3 Element values in the circuit of Fig 6-12 are: vs=100V,Rs=10Q2, L=2mH, and C 20/3 u F. The switch closes at t=0, find V r(t) 4 A4 u F capacitor and a 10mH inductor are used in a critically damped parallel rlC circuit(a) Find R resistor. (b)At t=0 the capacitor voltage is 100V and it is absorbing 50w, write the response of the capacitor voltage versus t Vs 5 The switch in Fig. 6-13 has been closed for hours. It is opened at t=0, show that a 10V battery can be a lethal weapon by finding v at t=2.5T us 6 For the circuit shown in Fig. 6-14, find v c(t)and ic(t) 1200 20V 3 50(V Fig 6-14 For prob. 6 DaLian Maritime University

Basic Circuit Theory Chpter6 Problems 1 Let L = 1.25 H in Fig.6-11, and determine v(t) if v(0 ) = 100 V and: (a) i C (0 ) = 20A; (b) i (0 ) = 20A. + + L + - + 2Ω L L i C i R i 0.05F v Fig. 6-11 For prob. 1 . 2 (a) What value of L in the circuit of Fig.6-11 will result in a transient response of the form, v(t) = Ae + Be ?A 4 −4t −6t 3 Element values in the circuit of Fig.6-12 are : v = 100V, R = 10 S S Ω , L = 2mH, and C = 20/3 µ F. The switch closes at t = 0, find v (t) R 4 A 4 µ F capacitor and a 10mH inductor are used in a critically damped parallel RLC circuit. (a) Find R resistor. (b) At t = 0 the capacitor voltage is 100V and it is absorbing 50W, write the response of the capacitor voltage versus t. t=0 + - - + Rs C L R R v s v Fig. 6-12 For prob.4. 5 The switch in Fig.6-13 has been closed for hours. It is opened at t = 0, show that a 10V battery can be a lethal weapon by finding v at t = 2.5π µ s. - t=0 + v 10nF 2.5mH 10V 0.5Ω Fig. 6-13 For prob.5. 6 For the circuit shown in Fig.6-14, find v (t) and i (t). C C + - - + V 1200Ω C i C v u(t) F 3 10 µ 20V 50 20H Fig. 6-14 For prob.6. DaLian Maritime University 1

Basic Circuit Theory Chpter6 Problems 7 For the circuit shown in Fig. 6-15, find vc(t) and i,(t) 5u()A+ 5HFT C 6-15Fo 8 The strength of the current source in Fig. 6-16 is suddenly increased from 10A to 20A at t=0 finds(0 ),Vs(0),Vs(oo), and vs(t) 10A3g IH 80m Fig 6-16 For prob. 8. 9 A source-free parallel RLC circuit contains an inductor for which the product @ L is 10Q2 Ifs,=-6s and s2=-8s, find R, L, and C. 10. With reference to the circuit of Fig 6-17, let i(0)=40A and v(0)=40V. If L=12.5H,R= 0.1Q2, and C=0.2F, find v(t) Fig. 6-17 For prob. 10 Reference answers to Selected Problems 1:(a)200e--100eV(t>0);(b)-100e-+200eV(t>0) 2:(a)5/6H(b)v(t)=210e-2-190eV(t>0) 3:200(e 4:(a)2592;(b)(100+625e-0v(t>0) 5:-9992V 6:20+250e-100-150N,A 7:(a)-5 25--200800 (t>0)(b)-2+2(4e-200-00)(0) 8:(a)20V;(b)50V(c)40V;(d40e-23(0cos25t24sin2.5V(>0) 9:4959,1443H,1443mF 0:(a)-20e-+60eV(t>0) DaLian Maritime University

Basic Circuit Theory Chpter6 Problems 7 For the circuit shown in Fig.6-15, find v (t) and i (t). C L - + A 25 Ω L i 25µF C v 3 − 5u(t) 2.5mH 8 The strength of the current source in Fig.6-16 is suddenly increased from 10A to 20A at t = 0, find v (0 ), v (0 ), v ( ), and v (t). S − S + S ∞ S Fig. 6-15 For prob.7. - + 1H s v 3 Ω 2 Ω 80mF 10 A Fig. 6-16 For prob.8. 9 A source-free parallel RLC circuit contains an inductor for which the product ω 0 L is 10 . If s 1= - 6s and s 2 = - 8s , find R, L, and C. Ω −1 −1 10. With reference to the circuit of Fig.6-17, let i(0) = 40A and v(0) = 40V. If L = 12.5H, R = 0.1Ω , and C = 0.2F, find v(t) R L C C R i i i v − + Fig. 6-17 For prob.10. Reference Answers to Selected Problems 1: (a) 200e -100e V (t >0); (b) -100e +200e V ( t > 0). −2t −8t −2t −8t 2: (a) 5/6 H; (b) v(t)=210e -190e V (t > 0). −4t −6t 3: 200(e - e ). −5000t −15000t 4: (a) 25Ω ; (b) (100+6.25t) e V (t > 0). −5000t 5: -9992V. 6: 20+250(e -e )V, −100t −150t 24 1 (3e -2 e )A. −150t −100t 7: (a) –50+ 3 25 (16e -e )V (t >0); (b) –2+ −2000t −8000t 3 5 (4 e -e ) (t >0). −2000t −8000t 8: (a) 20V; (b) 50V;(c) 40V; (d)40+e (10 cos 2.5t+24 sin 2.5t)V (t>0). −2.5t 9: 4.95Ω , 1.443H, 14.43 mF. 10: (a) –20e +60e V (t >0). −10t −40t DaLian Maritime University 2

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