Problem 10.064 - Norton equivalent of a phasor circuit For the circuit shown in the figure below, find the Norton equivalent circuit at terminals a-b, where Is = 9 60° A. Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 180 degrees. Is ZN= IN= V V 60 Ω -o a j80Q2 bo ° Q A 40 92 -j30 Q

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Using mesh analysis answer the questions based on the circuit below 10 cos 10³t V ☐ 1₁=0.05-j0.75 2 ΚΩ www 1₂ = -0.035+j0.005 12 = -0.35+j0.5 H ell 1 μF HH lo 0.4 H H Phasor form of the inductor is j*400 ohms Phasor form of the inductor is j*100 ohms Phasor form of the capacitor is -j*1000 ohms Phasor form of the capacitor is -j*10 ohms Phasor form of 20* sin(1000t) is -j20 (20<-90) Phasor form of 20*sin(1000t) is-j10 (10<-90) ₁=0.0025-j0.0075 Iz 20 sin 10³t V