JEE Main 2023PhysicsAlternating CurrentHardNumerical

JEE Main 2023Alternating Current Question with Solution

JEE Main 2023 (25 Jan Shift 1)

Question

An LCR series circuit of capacitance 62.5 nF and resistance of 50 Ω, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ____mH.(Take π2=10

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Show full solutionCorrect answer: 100
Correct answer
100

Step-by-step explanation

Given here, f=2 kHz=2000 Hz, C=62.5 nF=62.5×10-9 F and R=50 Ω

Resonance frequency (amplitude of the current will be maximum in this case) is given by f=12πLC, here, L is the inductance.

Putting the values, 

2000 Hz=12πL×62.5×10-9

L=14π2×20002×62.5×10-9=0.1 H=100 mH

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About this question

This is a previous-year question from JEE Main 2023, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.