JEE Main 2025 Apr 4 Shift 2, Physics Q25: LR, CR, LCR Circuits

JEE Main2025Apr 4, Shift 2Physics

An inductor of self inductance 1 H connected in series with a resistor of $100 π$ ohm and an ac supply of $100 π$ volt, 50 Hz. Maximum current flowing in the circuit is ______ A.

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Solution

Inductive reactance at 50 Hz: $X_{L} = ω L = 2 π \cdot 50 \cdot 1 = 100 π Ω$ .

Resistance: $R = 100 π Ω$ .

Impedance: $Z = R^{2} + X_{L}^{2} = (100 π)^{2} + (100 π)^{2} = 100 π 2 Ω$ .

The supply voltage $V = 100 π$ V is the rms value. So

$I_{r m s} = \frac{V}{Z} = \frac{100 π}{100 π 2} = \frac{1}{2}$ A.

Maximum (peak) current: $I_{ma x} = 2 I_{r m s} = 2 \cdot \frac{1}{2} = 1$ A.

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JEE Main 2025 Apr 4 Shift 2, Physics Q25: LR, CR, LCR Circuits

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