Question

6. In the following circuit v,(t)-5sin(1x10 t) V and v,() -20cos(1x10+90 V Find 2o(t) using node voltage analysis (be careful with polarities). 395.5@18.4° mA 20HH 0.1H 2082 V(t) 20Ω zo(t) Vo(t)

Answer 1

V_a (t) = 5 sin(1 times 10^6 t) = 5 cos(90 - (1 times 10^6 t)) = 5 cos(-(10^6 t - 90 degrees)) = 5 cos(10^6 t - 90 degrees) V_a (t) = 5 angle -90 degrees V V_b (t) = 20 cos(1 times 10^6 t + 90 degrees)V V_b (t) = 20 angle 90 degrees V Here omega = 10^6 rad/s. Convert L & C values into phasor form 20 mu H implies j omega L implies j times 10^6 times 20 times 10^-6 implies j20 ohm 0.1 mu F implies -j/omega C implies -j/10^6 times 0.1 times 10^-6 implies -j10 ohm \r\n Apply Kcl at V V - 5 angle -90 degrees/-j10 + V/20 + V - 20 angle 90 degrees/20 = 0 V[1/-j10 + 1/20 + 1/20] = 5 angle -90 degrees/-j10 + 20 angle 90 degrees/20 V[0.1 + j0.1] = 0.5 + j V = 7.90 angle 18.43 degrees Now i_20 (t) = V/20