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Capacitors

JEE Main level, numerical answer, medium

168 of 180

Question 168 of 180

A $2\mu F$ capacitor ${C}_{1}$ is first charged to a potential difference of $10V$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor ${C}_{2}$ of $8\mu F$ . The charge in ${C}_{2}$ on equilibrium condition is $\mu C$ . (Round off to the Nearest Integer)

Show answer and solution

Answer: 16

20=\left({C}_{1}+{C}_{2}\right)V\Rightarrow V=2volt.

\Rightarrow {Q}_{2}={C}_{2}V=16\mu C

=16

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Capacitors (Charging and Discarging of capacitors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.