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In a series LCR circuit, R = 200 ohm and the supply is 220 V, 50 Hz. When the capacitor is removed, the current lags the voltage by 30 deg; when the inductor is removed, the current leads the voltage by 30 deg. Find the power dissipated in the full LCR circuit.
- 242 W
- 305 W
- 210 W
- 0 W
Correct answer: 242 W
Solution
Removing C leaves R-L with tan(30) = X_L/R, so X_L = R*tan(30). Removing L leaves R-C with tan(30) = X_C/R, so X_C = R*tan(30). Thus X_L = X_C: the full circuit is at resonance, Z = R, power factor = 1, and P = V²/R = 220²/200 = 242 W.
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