Question

Problem # 3 (15 pts.) A) Given For the circuit element: Ic(t) AC Circuit Vc(tss 10 cos(1000t+30°) V B) Determine Step 1: The impedance of the capacitor, Step 2: The reactance of the capacitor, Step 3: The admittance of the capacitor, Step 4: The susceptance of the capacitor Step 5: The phasor current lcF in polar form, Step 6: The steady state current Ic(t)s.

Answer 1

V_c(t) = 10 cos(1000t + 30 degree)v v_c = 10 angle 30 degree v w = 1000 rad/sec impedance the capacitor z_c = 1/jwc = -j/1000 times 50 mu = -20 j z_c = -20j ohm reactance of the capacitor x_c = 1/wc = 1/1000 times 50 mu = 20 ohm x_c = 20 ohm admittance of the capacitor y_c = 1/zc = 1/- 20j ohm y_c = 0.05 j ohm substance of the capacitor BC = wc = 1000 times 5mu = 0.05 ohm B_c = 0.05 ohm phasor current I_c = v_c/Z_c = 10 angle 30 degree/- 20 j - 20 j implies 20 angle - 90 degree I_c = 10 angle 30 degree/20 angle - 90 degree = 0.5 angle 30 degree + 90 degree = 0.5 angle 120 degree A I_L = 0.5 angle 120 degree A steady state current I_c(t) = 0.5 (or) (1000t + 120 degree)a I_c(t)_ss = 0.5 cos(1000t + 120 degree)A.