JEE Main 2017PhysicsCapacitanceMediumMCQ

JEE Main 2017 — Capacitance Question with Solution

JEE Main 2017 (02 Apr)

Question

In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance C will be:

Choose an option

Show full solutionCorrect option: D
Correct answer
DCEr2r+r2

Step-by-step explanation

In steady state, current through capacitor will be zero, so current through r1 will be zero.

Hence, the net current flowing in the circuit will be,
I=Er+r2
The potential difference across the capacitor will be,
V2=I×r2=Er2r+r2

Therefore, charge on the capacitor plate will be,
q=CV2=CEr2r+r2

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About this question

This is a previous-year question from JEE Main 2017, 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.