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H.W 1.1: In the circuit in figure 1.16, the switch has been closed for a long time before opening at t = 0. a) Find the value of I. so that v, (t) equals v, (0) when t = 1 ms b) Find the percentage of the stored energy that has been dissipated in the 10-2 resistor when t = 1 ms 9 k 30 mA( 10 2 3L Figure 1.16 Ans: (a) L= 14.43ml, (b) 75 %

H.W 1.1: In the circuit in figure 1.16, the switch has been closed for a long time before opening at t = 0. a) Find the value of I. so that v, (t) equals v, (0) when t = 1 ms b) Find the percentage of the stored energy that has been dissipated in the 10-2 resistor when t = 1 ms 9 k 30 mA( 10 2 3L Figure 1.16 Ans: (a) L= 14.43ml, (b) 75 %

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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H.W 1.1: In the circuit in figure 1.16, the switch has been closed for a long time before opening at t = 0. a) Find the value of L so that v. (t) equals v, (0) when t = 1 ms b) Find the percentage of the stored energy that has been dissipated in the 10-2 resistor when t = 1 ms
H.W 1.1: In the circuit in figure 1.16, the switch has been closed for a long time before opening at t = 0. a) Find the value of I. so that , (t) equals v, (0) when t = 1 ms b) Find the percentage of the stored energy that has been dissipated in the 10-2 resistor when t = 1 ms 9 k. + 30 mA (1 E1 kl 10 Ω υ,L Figure 1.16 Ans: (a) L- 14.43mH, (b) 75 % H.W 1.2: The switch in the circuit in figure 1.17 has been in the left position for a long time. At t = 0 it moves to the right position and stays there. a) Write the expression for the capacitor voltage, v(t) for t20 b) Write the expression for the current through the 40 k resistor, i(t) for t20, 5 kn 40 k. i 120 V 10 k 160 nF 325 k2 Ž10 kl Figure 1.17 Ans: v(t) = (80e-375t) V, t20, i(() = (1.6e-375t) ma, t2 0 H.W 1.3: The current and voltage at the terminals of the inductor in the circuit in figure 1.18 are: i(t) = (4 + 4e-40) A, t2 0 v(t) = (-80e-40t) V, t2 0* a) Specify the numerical values of Vs, R, lo and L. b) How many milliseconds after the switch has been closed does the energy stored in the inductor reach 9 J? R I= 0 L3v(t)
Transcribed Image Text:H.W 1.1: In the circuit in figure 1.16, the switch has been closed for a long time before opening at t = 0. a) Find the value of I. so that , (t) equals v, (0) when t = 1 ms b) Find the percentage of the stored energy that has been dissipated in the 10-2 resistor when t = 1 ms 9 k. + 30 mA (1 E1 kl 10 Ω υ,L Figure 1.16 Ans: (a) L- 14.43mH, (b) 75 % H.W 1.2: The switch in the circuit in figure 1.17 has been in the left position for a long time. At t = 0 it moves to the right position and stays there. a) Write the expression for the capacitor voltage, v(t) for t20 b) Write the expression for the current through the 40 k resistor, i(t) for t20, 5 kn 40 k. i 120 V 10 k 160 nF 325 k2 Ž10 kl Figure 1.17 Ans: v(t) = (80e-375t) V, t20, i(() = (1.6e-375t) ma, t2 0 H.W 1.3: The current and voltage at the terminals of the inductor in the circuit in figure 1.18 are: i(t) = (4 + 4e-40) A, t2 0 v(t) = (-80e-40t) V, t2 0* a) Specify the numerical values of Vs, R, lo and L. b) How many milliseconds after the switch has been closed does the energy stored in the inductor reach 9 J? R I= 0 L3v(t)
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