JEE Main 2020PhysicsCapacitanceEasyMCQ

JEE Main 2020Capacitance Question with Solution

JEE Main 2020 (02 Sep Shift 2)

Question

10 μF capacitor is fully charged to a potential difference of 50 V After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is :

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Show full solutionCorrect option: A
Correct answer
A15 μF

Step-by-step explanation

V=C1V1+C2V2C1+C2
20=10×50+020+C
C=15μF

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About this question

This is a previous-year question from JEE Main 2020, 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.