JEE Main 2021 — Alternating Current Question with Solution
From: JEE Main 2021 (Online) 16th March Morning Shift
Question
A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8, L = 24 mH and C = 60 F. The value of power dissipated at resonant condition is 'x' kW. The value of x to the nearest integer is ____________.
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Show full solutionCorrect answer: 3906.25
Correct answer
3906.25
Step-by-step explanation
At resonance power (P)
P = 3906.25 w
P 4 Kw
P = 3906.25 w
P 4 Kw
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This is a previous-year question from JEE Main 2021, 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.