JEE Main 2022PhysicsAlternating CurrentMediumNumerical

JEE Main 2022 — Alternating Current Question with Solution

JEE Main 2022 (24 Jun Shift 1)

Question

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 Vand frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is aπ A. The value of a is _____ .

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Show full solutionCorrect answer: 242
Correct answer
242

Step-by-step explanation

The reactance of the given inductor is,

XL=ωL=2πfL=2π×50×0.2=20π Ω.

The maximum current in the given circuit will be,

i0=2VrmsXL=2×22020π=242π A

Therefore, a=242

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About this question

This is a previous-year question from JEE Main 2022, 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.