An inductor of self inductance 1 H is 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.

Physics · Alternating Current

JEE Main 2025 — 4 April, Evening Shift — Question 61

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

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

$X_{L}=2 \pi \times 50 \times 1=100 \pi \Omega$ $Z=100 \pi \sqrt{2} \Omega$ $i_{\max }=\frac{100 \pi \sqrt{2}}{100 \pi \sqrt{2}}=1 \mathrm{~A}$

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Alternating Current
Topic: Series L-R, R-C, L-C Circuits with AC Source