An inductor of self inductance 1 H connected in series with a resistor of 100π Ω and an a.c. supply of 100π V, 50 Hz. Maximum current flowing in the circuit in ampere is (a) 0.5 (b) 1 (c) 1.5 (d) 2

Answer: (b). XL = 2π f L = 2π × 50 × 1 = 100π Ω Z = √(R² + XL²) = √((100π)² + (100π)²) = 100√(2) π Ω Irms = (100π)/(100√(2) π) = 1/√(2) A, so I₀ = √(2) Irms = 1 A. So the answer is (b) 1.

An inductor of self inductance 1 H connected in series with a… | MHT CET 2026 (12 May, Shift 2)

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RL and RC Series Circuits

An inductor of self inductance $1$ H connected in series with a resistor of $100\pi\ \Omega$ and an a.c. supply of $100\pi$ V, $50$ Hz. Maximum current flowing in the circuit in ampere is

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Answer: (b)

$X_L = 2\pi f L = 2\pi \times 50 \times 1 = 100\pi\ \Omega$
$Z = \sqrt{R^2 + X_L^2} = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\sqrt{2}\,\pi\ \Omega$
$I_{rms} = \frac{100\pi}{100\sqrt{2}\,\pi} = \frac{1}{\sqrt{2}}$ A, so $I_0 = \sqrt{2}\,I_{rms} = 1$ A.
So the answer is (b) $1$ .

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