JEE Main 2021 — Alternating Current Question with Solution
From: JEE Main 2021 (Online) 26th August Morning Shift
Question
A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k, an inductor of inductive reactance XL = 250 and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take 2 = 10)
Choose an option
Show full solutionCorrect option: A
Correct answer
A4 F
Step-by-step explanation
From maximum average power
XL = XC
250 =
C = 4 106
XL = XC
250 =
C = 4 106
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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.