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A series LCR circuit has a resistor of 200 Ω and is connected to a 220 V, 50 Hz AC supply. If the capacitor is removed first and then the inductor is removed, the voltage across the resistor in the circuit is found to lead the supply voltage by 30°. Also, when the inductor is removed, the current is observed to lead the voltage by 30°. The power dissipated in the circuit is
- 305 W
- 210 W
- Zero W
- 242 W
Correct answer: 242 W
Solution
The RL part and RC part give the same 30 degree phase, so X_L = X_C and the full circuit is at resonance with Z = R. Power = V^2/R = 220^2/200 = 242 W.
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