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3. In the circuit shown, R = 500 2, L = 0.64 H, C = 1 μF and I = -1 A. The initial voltage drop across the capacitor is 40 V and the initial inductor current is 0.5 A. Find the following a. The damped form of the solution b. iR(0*) c. ic (0*) dil(0*) d. dt e. iL(t) for t≥0 1=0 f. v(t) for t≥0 L R υ 3. a. underdamped b. c. d. iR (0+) = 80 mA = 0.080 A (0+)=-1.58 A diz (0+) dt = 62.5 A/s e. i(t)=1+e-1000t [1.5 cos(750t) + 2.0833sin (750t)] A f. v(t) = e 1000t [40 cos(750t) - 2053.33sin (750t)] V

3. In the circuit shown, R = 500 2, L = 0.64 H, C = 1 μF and I = -1 A. The initial voltage drop across the capacitor is 40 V and the initial inductor current is 0.5 A. Find the following a. The damped form of the solution b. iR(0*) c. ic (0*) dil(0*) d. dt e. iL(t) for t≥0 1=0 f. v(t) for t≥0 L R υ 3. a. underdamped b. c. d. iR (0+) = 80 mA = 0.080 A (0+)=-1.58 A diz (0+) dt = 62.5 A/s e. i(t)=1+e-1000t [1.5 cos(750t) + 2.0833sin (750t)] A f. v(t) = e 1000t [40 cos(750t) - 2053.33sin (750t)] V

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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3. In the circuit shown, R = 500 2, L = 0.64 H, C = 1 μF and I = -1 A. The initial voltage drop across the capacitor is 40 V and the initial inductor current is 0.5 A. Find the following a. The damped form of the solution b. iR(0*) c. ic (0*) dil(0*) d. dt e. iL(t) for t≥0 1=0 f. v(t) for t≥0 L R υ 3. a. underdamped b. c. d. iR (0+) = 80 mA = 0.080 A (0+)=-1.58 A diz (0+) dt = 62.5 A/s e. i(t)=1+e-1000t [1.5 cos(750t) + 2.0833sin (750t)] A f. v(t) = e 1000t [40 cos(750t) - 2053.33sin (750t)] V
Transcribed Image Text:3. In the circuit shown, R = 500 2, L = 0.64 H, C = 1 μF and I = -1 A. The initial voltage drop across the capacitor is 40 V and the initial inductor current is 0.5 A. Find the following a. The damped form of the solution b. iR(0*) c. ic (0*) dil(0*) d. dt e. iL(t) for t≥0 1=0 f. v(t) for t≥0 L R υ 3. a. underdamped b. c. d. iR (0+) = 80 mA = 0.080 A (0+)=-1.58 A diz (0+) dt = 62.5 A/s e. i(t)=1+e-1000t [1.5 cos(750t) + 2.0833sin (750t)] A f. v(t) = e 1000t [40 cos(750t) - 2053.33sin (750t)] V
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