Physics · Capacitors and R-C Circuits

JEE Main 2024 — 30 January, Shift 1 — Question 52

A capacitor of capacitance $C$ and potential V has energy E. It is connected to another capacitor of capacitance 2 C and potential 2 V . Then the loss of energy is $\frac{x}{3} E$ , where x is

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Energy loss $=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}$ $=\frac{2}{3} . E$

$\therefore x=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Force Between Plates and Potential Energy Stored