What is the average power dissipation in an RLC series circuit with R = 10 $\Omega$, L = 0.1 H, C = 10 $\mu$F when driven at resonance by a 100 V-rms source?
A)100 W
B)500 W
C)1000 W
D)2 kW
E)700 W

Question

Quizplus · Accepted Answer

At resonance, the impedance of the circuit is purely resistive, and given by Z = R. Therefore, the current through the circuit is I = V/Z = 100/(10+0.1j10^6) = 9.95 A-rms. The power dissipated in the resistor is P = I^2R = 995 W. Therefore, the average power dissipation in the circuit is 995 W, which is closest to choice C, 1000 W.

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