Question

16.20 - Solve this problem using time domain (ch.8 from the textbook), and then using Laplace - you should get the same answer. When solving using Laplace, you will need to find initial values of v_c and i_L and re-draw circuit with the "additional sources" that are representing initial values.

Find () for t>O in the circuit of Fig. 16.43. Figure 16.43: 10 S2 60Ω 1 mF 40 92 2.5 H Step-by-step solution Step 1 of 5 Refer to Figure 16.43 in the textbook. The switch is initially closed fort<0. So, the circuit remains in the steady state for 1<0 and the inductor and the capacitor are replaced by the short and open circuit respectively. Redraw the circuit for0 as shown in Figure 1 10Ω 60Ω 0 36 V+ Че) 40S2 Figure 1

Answer 1

The switch is initially closed for long time. So inductor and capacitor are short and open circuited respectively. i(0^-) = 0A & V_c(0^-) = 40/40 + 10 (36) V_c(0^-) = 28.2 V for t > 0: - This is source free RLC circuit. Alpha = R/2L = 100/5 = 20 & w_0 = 1/squareroot LC = 1/squareroot 2.5 times 10^-63 w_0 = 20 Here alpha = w_0 Leads to critical damping So, i(t) = [A + Bt]e^-alpha t = (A + Bt)e^-20 t at t = 0: i(0) = [A + B(0)] e^-20(0) implies A = i(0) = 0 L di(0)/dt + V(0^-) + Ri(0) = 0 from circuit, di(0)/dt = -V(0^-)/L = 28.8/2.5 di(t)/dt = B e^-20 t + Bt(-20)e^-20t [because A = 0] di(0)/dt = B = -28.8/2.5 = -11.52 implies B = -11.52 i(t) = -11.52 t middot e^-20 t A using Laplace Transform: - 1/SC = 1000/S, SL = 2.5S therefore V(0)/s = 28.8/S