Question

Problem 1 (10 points): The switch in the circuit of Fig. 1 has been in position a for a long time. Att 0, the switch moves to position b 1. (4 points) Construct an s-domain circuit for t> 0 2. (4 points) Find Vo(s) and vo(t) for t> 0 3. (2 points) Find IL(s) and iL () fort > 0. t-0 Ri R2 R1 = 400 ohms, R2 = 1000 ohms, C = 6.25 nF L 16 mH, and vg-360 V Figure 1: Problem 1 Circuit

Answer 1

At t = 0 v - 360 dividedby400 + v dividedby400 + v dividedby1000 = 0 10 v - 360 + v + 4 v = 0 15 v = 3600 v = 240 i_l(o^-) = 0.24 A V_c(o^-) = 240 at t > 0 apply node analysis at V_0(s) V_0(s + 240 dividedbyS dividedby160 m dividedbyS) + V_0(s) dividedby1 k + V_0(s) + 3.84 m dividedby16 m s = 0 S V_0(s) + 240 dividedby160 m + V_0(s) dividedby1 k + V_0(s) dividedby16 m s + 3.84 m dividedby16 m s = 0 V_0(s) [s dividedby160 m + 1 dividedby1k + 1 dividedby16 m s] = -240 dividedby160 m - 3.84 m dividedby16 msV_0(s) [s dividedby160 m + 1 dividedby1k + 1 dividedby16 m s] = -1.5 10^- 6 - 0.24 per sec V_0(s) = -1.5 10^- 6 - 0.24 per sec dividedbyS dividedby160 m + 1 dividedby1k + 62.5 dividedbyS V_0(s) = - 240 - 38.4 10^6 dividedbyS dividedbyS + 160 k + 10^10 dividedbyS V_0(s) = s (- 240 - 38.4 10^6 dividedbyS dividedbyS^2 + 160 ks + 10^10) = s (- 240 - 38.4 10^6 dividedbyS dividedby (s + 80k)^2 + (60k)^2 = -240s - 39.4 10^6 dividedby (s + 80k)^2 + (60k)^2 = -240s dividedby (s + 80k)^2 + (60k)^2 - 38.4 10^6 dividedby (s + 80k)^2 + (60k)^2 = -240[s + 80k] dividedby (s + 80k)^2 + (60k)^2 + 80k 240 - 38.4 10^6 dividedby (s + 80k^2) + (60k)^2 V_0(t) = -240 e^- 80kt Cos 60 kt - 320 e^- 80kt Sin 60 kt V_0(t) = -240 e^- 80 kt [Cos 60 kt + 4 dividedby3 sin 60 kt] V = L di/dt V(s) = L [s I(s) - 0.24] V(s) dividedby s L + 0.24 dividedby s = i(s) i_l(s) = v_0(s) + 3.84m dividedby16 msi_l(s) = 62.5 V_0(s) dividedby s + 0.24 dividedby s = 62.5 dividedby s[-240 - 38.4 10^6 dividedby s dividedby s + 160 k + 10^10 dividedby s] + 0.24 dividedby s I_L(s) = 1 dividedby s[62.5 (- 240) - 62.5 38.4 10^6 dividedby s dividedby s + 160k + 10^10 dividedby s + 0.24] V(t) = L di(t)/dt integral v(t/L) dt = i(t) i(t) = 1 dividedby L integral - 240 e^- 80kt[cos 60kt + 4 dividedby 3 sin 60kt] dt = - 15 10^3 integral e^- 80kt[cos 60kt + 4 dividedby 3 sin 60kt] dt i(t) = -15 10^3 integral e^- 80kt[cos 60kt + 4 dividedby 3 sin 60kt] dt = 1 dividedby a^2 + b^2 e^ax[asinbx - bcosbx] 4 dividedby 3 e^- 80kt sin 60kt = 10^- 10 e^- 80kt[- 80ksin 60kt - 60kcos 60kt] = 4 dividedby 3[- 8 10^- 6 e^- 80kt sin 60kt - 6 10^6 e^- 80ktcos 60kt] integral e^ax cosbx dx = 1/a^2 + b^2 e^ax [acosbx + bsinbx] e^- 80kt cos 60kt = 1 dividedby 10^10 e^-80kt[- 80kcos 60kt + 60ksin 60kt] = 6 10^- 6e^- 80ktsin 60kt - 8 10^- 6 e^- 80kt cos 60kt i(t) = -15 10^3 [6 10^- 6 e^- 80kt sin 60kt - 32 dividedby 3 10^- 6 e^- 80kt sin 60kt - 8 10^-6 e^- 80kt cos 60kt - 8 10^- 6 e^- 80kt cos 60kt] i(t) = -15 10^3[- 10.67 e^- 80kt sin 60kt 10^- 6 - 16 10^- 6 e^- 30kt sin 60kt] i(t) = 160.05 10^- 3 6^- 80kt sin 60kt + 240 10^- 3 e^- 80kt sin 60kt