Question

Q-7 (15 pts) For the RLC circuit shown below, the switch is closed at t-0s shorting the capacitor to gnd. Find the time domain expression for the transient current after the switching time instance. C-1uF R1-4000 i(t) L-1H 100V t-0s R2-4000

Answer 1

For t = 0^- V_1 (0^-) = V_c (0^+) = 100 volt i_(0^-) = i_(0^+) = 0 amp since inductor does not allow sudden charge in current and capacitor does not allow sudden charge in voltage for t > 0 Apply K V L at loop '2' Therefore 1/c integral i_2 dt + L di_2/dt + 400 I_= 0 rightarrow I(s)/10^- 6 s + 100/s + s I_2 (s) + 400 I_(s) = 0 \r\n I_2 (s) [10^6/s + s + 400] = -100/s rightarrow I_2 (s) [10^6 + s^2 + 400 s/s] = -100/s rightarrow I_2 (s) = -100/s^2 + 400 s + 10^6 = -100/(s + 200)^2 + (979.8)^2 = -100/979.8 times 979.8/(s + 200)^2 + (979.8)^2 Take Inverse Laplace I_+ 2 (t) = -100/979.8 e^- 200 t sin 979.8 t rightarrow i_2 (t) = -0.102 e^- 200 t sin (979.8 t).