JEE Main 2024 — Capacitance Question with Solution

Under the balanced state, there is no flow of charge through the capacitor. So, the path connecting the capacitor behaves as an open path.

The equivalent resistance of the entire circuit, under equilibrium condition, can be calculated as follows:

$\begin{array}{rcl}{R}_{eq}& =& 1+\frac{1}{{\displaystyle \frac{1}{4+4+4}}+{\displaystyle \frac{1}{4}}} \mathrm{\Omega }\\ & =& \left(1+3\right) \mathrm{\Omega }\\ & =& 4 \mathrm{\Omega }\end{array}$

Hence, the current through the entire circuit is given by

$\begin{array}{rcl}I& =& \frac{V}{R_{eq}}\\ & =& \frac{9}{4} \mathrm{A}\end{array}$

So, the current $\left(I_1\right)$ can be written as

$\begin{array}{rcl}{I}_1& =& \frac{9}{4}\times \frac{4}{16} \mathrm{A}\\ & =& \frac{9}{16} \mathrm{A}\end{array}$

The potential difference between the points $\mathrm{A}$ and $\mathrm{B}$ is,

$\begin{array}{rcl}{V}_A-V_B& =& I_{\mathit{1}}\times 8\\ & =& \frac{9}{16}\times 8\\ & =& \frac{9}{2} \mathrm{V}\end{array}$

Hence, the energy stored in the capacitor is given by

$\begin{array}{rcl}U& =& \frac{1}{2}C{\left(V_A-V_B\right)}^2\\ & =& \frac{1}{2}\times 4\times \frac{81}{4} \mathrm{\mu J}\\ & =& \frac{81}{2} \mathrm{\mu J}\end{array}$

$\therefore x=81$