Physics · Capacitors and R-C Circuits

JEE Main 2024 — 1 February, Shift 1 — Question 33

Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2 V . The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is

Option A:
$\frac{1}{4} \mathrm{CV}^{2}$
Correct
Option B:
$2 \mathrm{CV}^{2}$
Option C:
$\frac{1}{2} \mathrm{CV}^{2}$
Option D:
$\frac{3}{4} \mathrm{CV}^{2}$

Answer: A

Step-by-step solution

${{\text{V}}_{\text{C}}}=\frac{{{\text{q}}_{\text{net }\!\!~\!\!\text{ }}}}{{{\text{C}}_{\text{net }\!\!~\!\!\text{ }}}}=\frac{CV+2CV}{2\text{C}}$

${{\text{V}}_{\text{C}}}=\frac{3\text{ }\!\!~\!\!\text{ V}}{2}$

Loss of energy

$\begin{array}{*{35}{r}}{} & ~=\frac{1}{2}\text{C}{{\text{V}}^{2}}+\frac{1}{2}C{{(2V)}^{2}}-\frac{1}{2}2C{{\left( \frac{3V}{2} \right)}^{2}} \\{} & ~=\left( \frac{C{{V}^{2}}}{4} \right) \\\end{array}$

Answer key and solution verified before publishing.

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