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A series LCR circuit having a resistance of 100 ohm is driven by an AC source of 400 V (rms) at an angular frequency of 300 rad/s. When the capacitor is disconnected from the circuit, the current lags behind the applied voltage by 60 degrees. When the inductor is disconnected instead, the current leads the applied voltage by 60 degrees. Find the rms current and the power dissipated in the complete LCR circuit.
- Current = 2 A, Power = 400 W
- Current = 4 A, Power = 800 W
- Current = 2 A, Power = 800 W
- Current = 4 A, Power = 1600 W
Correct answer: Current = 4 A, Power = 1600 W
Solution
From both conditions tan(60 deg) = sqrt(3), giving X_L = R*sqrt(3) = 100*sqrt(3) and X_C = R*sqrt(3) = 100*sqrt(3). Since X_L = X_C, the circuit is at resonance. At resonance Z = R = 100 ohm, so I = 400/100 = 4 A, and P = I² * R = 16 * 100 = 1600 W.
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