JEE Main 2025 Apr 7 Shift 2, Physics Q22: LR, CR, LCR Circuits

JEE Main2025Apr 7, Shift 2Physics

An inductor of reactance 100 $Ω$ , a capacitor of reactance 50 $Ω$ , and a resistor of resistance 50 $Ω$ are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is _______ W.

Your answer

Solution

The impedance of the series LCR circuit is $Z = R^{2} + (X_{L} - X_{C})^{2} = 5 0^{2} + (100 - 50)^{2} = 2500 + 2500 = 502 Ω$ .

The average power dissipated (only across the resistor) is $P = V_{r m s}^{2} \frac{R}{Z ^{2}} = \frac{1 0 ^{2} \times 50}{( 50 2 ) ^{2}} = \frac{5000}{5000} = 1$ W.

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JEE Main 2025 Apr 7 Shift 2, Physics Q22: LR, CR, LCR Circuits

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