Physics · Capacitors and R-C Circuits

JEE Main 2025 — 3 April, Evening Shift — Question 59

Using a battery, a 100 pF capacitor is charged to 60 V and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20 V , its capacitance is: (in pF )

Option A:
100
Option B:
400
Option C:
200
Correct
Option D:
600

Answer: C

Step-by-step solution

i.e. $Q=60 \times 100 \mathrm{pC}$

When new capacitance $C^{\prime}$ is

connected Then $Q=\left(C+C^{\prime}\right) V$

$60 \times 100=\left(100+C^{\prime}\right) \times 20$ $200=C^{\prime}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Evening Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Combination of Capacitors and Circuit Analysis