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the characteristic equation are $12 = -a t Va – wi = -2, -50 and the corresponding response is v(1) = Aje + Ar We now apply the initial conditions to get A, and Az. 50 (8.5.1) v(0) = 5 = A, + Az (8.52) 5 + 0 1.923 x 10 x 10 dv(0) v(0) + Ri(0) -260 dt RC But differentiating Eq. (8.5.1). du 21 -24,e 50A2e dt

the characteristic equation are $12 = -a t Va – wi = -2, -50 and the corresponding response is v(1) = Aje + Ar We now apply the initial conditions to get A, and Az. 50 (8.5.1) v(0) = 5 = A, + Az (8.52) 5 + 0 1.923 x 10 x 10 dv(0) v(0) + Ri(0) -260 dt RC But differentiating Eq. (8.5.1). du 21 -24,e 50A2e dt

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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Can you explain how he made this step??! Example 8.5 In the parallel circuit of Fig. 8.13, find v(t) for t > 0, assuming v(0) = 5 V, i(0) = 0, L = I H, and C = 10 mF. Consider these cases: R = 1.923 N, R = 5N, and R = 6.25 N. Solution: I CASE 1 If R = 1.923 , = 26 2RC 2 x 1.923 X 10 x 103 1 = 10 VLC Vi x 10 x 10 Figure 8.13 A source-free parallel RLC circuit. Since a > wo in this case, the response is overdamped. The roots of the characteristic equation are $12 = -a ± Va - wi = -2, -50 and the corresponding response is U(1) = Aje1 + Ase S0 (8.5.1) We now apply the initial conditions to get A, and A2. (8.২) v(0) = 5 = A, + A2 v(0) + Ri(0) RC đu(0) 5 +0 = -260 dt 1.923 x 10 x 10 But differentiating Eq. (8.5.1). du = -2Aje2 - 50Ase 5or dt
Transcribed Image Text:Can you explain how he made this step??! Example 8.5 In the parallel circuit of Fig. 8.13, find v(t) for t > 0, assuming v(0) = 5 V, i(0) = 0, L = I H, and C = 10 mF. Consider these cases: R = 1.923 N, R = 5N, and R = 6.25 N. Solution: I CASE 1 If R = 1.923 , = 26 2RC 2 x 1.923 X 10 x 103 1 = 10 VLC Vi x 10 x 10 Figure 8.13 A source-free parallel RLC circuit. Since a > wo in this case, the response is overdamped. The roots of the characteristic equation are $12 = -a ± Va - wi = -2, -50 and the corresponding response is U(1) = Aje1 + Ase S0 (8.5.1) We now apply the initial conditions to get A, and A2. (8.২) v(0) = 5 = A, + A2 v(0) + Ri(0) RC đu(0) 5 +0 = -260 dt 1.923 x 10 x 10 But differentiating Eq. (8.5.1). du = -2Aje2 - 50Ase 5or dt
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