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JEE Main 20265 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 5 April
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

JEE Main 2026 · 5 April · Evening Shift · Physics · Alternating Current

A series LCR circuit with R = 20\ , L = 1.6 H and C = 40\ is...

A series LCR circuit with

R = 20\ \Omega

,

L = 1.6\text{ H}

and

C = 40\ \mu\text{F}

is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is _______

\Omega

.

Correct answer

0

Step-by-step solution

Given

L = 1.6\text{ H}

and

C = 40\ \mu\text{F} = 40 \times 10^{-6}\text{ F}

.

At resonance, the angular frequency is

\omega_r = \dfrac{1}{\sqrt{LC}}

.

The inductive reactance at resonance is

X_L = \omega_r L = \dfrac{L}{\sqrt{LC}} = \sqrt{\dfrac{L}{C}}

.

Substituting the given values:

X_L = \sqrt{\dfrac{1.6}{40 \times 10^{-6}}}

X_L = \sqrt{\dfrac{1.6}{4 \times 10^{-5}}}

X_L = \sqrt{4 \times 10^4}

X_L = 200\ \Omega

Answer:

200

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