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An LCR series circuit with 10 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the power (in kW) dissipated in the LCR circuit.

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Detailed Solution

When C is removed, circuit is LR with $\tan ϕ = \frac{X_{L}}{R}$

When L is removed, circuit is CR with $\tan ϕ = \frac{X_{C}}{R}$

$X_{L} = X_{C}$ , resonance and hence

$X = X_{L} - X_{C} = 0, i.e., Z = R,$

So,

$I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{R} = \frac{200}{10}, P = V_{rms} {I_{rms}}_{} \cos ϕ \Rightarrow P = 200 \times 20 \times 1 \Rightarrow P = 4000 W = 4 kW$

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