JEE Main 2024PhysicsCapacitanceMediumMCQ

JEE Main 2024 — Capacitance Question with Solution

JEE Main 2024 (29 Jan Shift 1)

Question

A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be :

Choose an option

Show full solutionCorrect option: B
Correct answer
B1.5 A

Step-by-step explanation

By energy conservation, it follows that

12CV2=12LImax2   ...1

Equation (1) implies that

Imax=CLV=100×10-66.4×10-3×12=128=1.5 A

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About this question

This is a previous-year question from JEE Main 2024, covering the Capacitance 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.