Physics · Alternating Current

JEE Main 2026 — 5 April, Evening Shift — Question 22

A series LCR circuit with $R = 20\Omega$ , $L = 1.6\mathrm{H}$ and $C = 40\mu\mathrm{F}$ is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is $\Omega$ .

Answer: 200

Numerical answer — enter this value.

Step-by-step solution

Resonant frequency $\omega = 1/\sqrt{LC} = 1/\sqrt{1.6\times40\times10^{-6}} = 125$ rad/s. Inductive reactance $X_L = \omega L = 125\times1.6 = 200\Omega$ .

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Alternating Current
Topic: Series LCR Circuit and Power Factor