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An LCR series circuit with R = 100 ohm is driven by an AC source of 200 V (rms) at angular frequency 300 rad/s. Removing only the capacitor makes the current lag the voltage by 60 deg; removing only the inductor makes the current lead the voltage by 60 deg. Find the current and the average power dissipated in the full LCR circuit.
- 2 A, 400 W
- 1 A, 100 W
- 2 A, 200 W
- 1 A, 200 W
Correct answer: 2 A, 400 W
Solution
Removing C: tan(60deg) = X_L/R => X_L = R*tan(60deg) = 100*sqrt(3). Removing L: tan(60deg) = X_C/R => X_C = 100*sqrt(3). Since X_L = X_C, the full LCR circuit is at resonance, so impedance Z = R = 100 ohm. Current I = V/Z = 200/100 = 2 A. Power P = I²*R = 2²*100 = 400 W (equivalently V*I since power factor = 1: 200*2 = 400 W).
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