Question

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Part A Review Constants Find i(t) for t > 0. The switch in the circuit in (Figure 1) has been open for a long time before closing at t = 0. i(t) 30 30e-1000t mA 45,000te1000t i(t)= 15 - 15,000te-1000t _ 30e~1000t mA Figure 1 of 1 - 30e-1000t mA i(t)= 15+15,000te~1000t 30e-1000t mA i(t= 30 +15,000te-1000t O i(t) = 30 +45,000te 1000t - 30e-1000t mA 400 30e-1000t mA i(t= 15- 45,000te-1000t =0 + V3800 mH 125 uF 30e-1000t mA 12 V i(t)= 15+45,000te~1000t 30e-1000t mA i(t)= 30 - 15,000te-1000t Request Answer Submit

Answer 1

Find initial values v_c(0) & i_B (0) v_c(0) = 12 v i_c (0) = 0 A Now switch is closed c = 1.25 mu F L = 800 mH \r\n Apply Kcl 0 = v(s) - 12/s/400 + v (s) - 12/s/1/1.25 times 10^-6 s + v (s) - 0/800 times 10^- 3 s v (s) [1/400 + 1.25 times 10^-6 s + 1/800 times 10^-3 s] = 12/400 s + 12/s times 1.25 times 10^-6 s v (s) [2 times 10^-3 s + (1.25 times 10^-6 s times 800 times 10^-3 s) + 1/800 times 10^-3 s = 12/400 s + 12 times 1.25 times 10^-6 v (s) = [2 times 10^-3 s + 1 times 10^-6 s^2 + 1/800 times 10^-3 s] = 12 + 6 times 10^-3 s/400 s v (s) [1 + s^2 times