An AC source of RMS voltage 20 V with internal impedance ZS = (1+2j) Ω feeds a load of impedance ZL = (7+4j) Ω in the circuit given below. The reactive power is _________(a) 8 VAR(b) 16 VAR(c) 28 VAR(d) 32 VARThe question was asked in an online interview.Asked question is from Advanced Problems on Network Theory in portion Network Theory and Complex Circuit Diagram of Network Theory

An AC source of RMS voltage 20 V with internal impedance ZS = (1+2j) �

1.	An AC source of RMS voltage 20 V with internal impedance ZS = (1+2j) Ω feeds a load of impedance ZL = (7+4j) Ω in the circuit given below. The reactive power is _________(a) 8 VAR(b) 16 VAR(c) 28 VAR(d) 32 VARThe question was asked in an online interview.Asked question is from Advanced Problems on Network Theory in portion Network Theory and Complex Circuit Diagram of Network Theory
Answer» Right answer is (b) 16 VAR Easy explanation: CURRENT I = \(\frac{V}{Z_L+Z_S} = \frac{20∠0°}{8+6j}\) = \(\frac{20}{\sqrt{8^2+6^2}} = \frac{∠0°}{∠arc tan⁡(\frac{3}{4})}\) = \(\frac{20}{10}\) ∠-arc tan⁡(\(\frac{3}{4}\)) = 2∠-arc tan⁡(\(\frac{3}{4}\)) POWER consumed by load = \|I\|^2ZL = 4(7+4j) = 28 + j16 ∴ The REACTIVE power = 16 VAR.

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