JEE Main 2019PhysicsCapacitanceEasyMCQ

JEE Main 2019 — Capacitance Question with Solution

JEE Main 2019 (12 Apr Shift 2)

Question

In the given circuit, the charge on 4 μF capacitor will be:

Choose an option

Show full solutionCorrect option: C
Correct answer
C24 μC

Step-by-step explanation

The given circuit can be redrawn as:

Charge on 4 μF capacitor:
Q=4×64+6×10
Q=24 μC

in series combination charge remains same.

so,

Q4μF=Q6μF=24 μC

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About this question

This is a previous-year question from JEE Main 2019, 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.