JEE Main 2021PhysicsCapacitanceEasyNumerical

JEE Main 2021 — Capacitance Question with Solution

JEE Main 2021 (31 Aug Shift 1)

Question

A capacitor of 50μF is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is _________ μC.

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Show full solutionCorrect answer: 100
Correct answer
100

Step-by-step explanation

At steady state current through branch of capacitor is zero. Potential difference across capacitor is 2 V. So, q=50×2=100μC

i=6(2+2+2)×103=1×10-3Amp
VA,B= potential across 50μF
VA,B=i×2×103=1×10-3×2×103=2 volt
Charge on 50μF
q=CV=50×10-6×2⇒100×10-6C=100μC

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About this question

This is a previous-year question from JEE Main 2021, 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.