QJEEM24Ph81a8: JEE Main 2024 Physics - Class 12 JEEM24Ph81a8 Solution: In the given figure, the charge stored in 6 μ F capacitor, when …

At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.

$$\begin{array}{rl}& {\mathrm{R}}_{\mathrm{eq}}=9\mathrm{\Omega }\\ & \mathrm{i}=\frac{9\mathrm{V}}{9\mathrm{\Omega }}=1\mathrm{A}\\ & \mathrm{\Delta }{\mathrm{V}}_{6\mathrm{\Omega }}=1\times 6=6\mathrm{V}\\ & {\mathrm{V}}_{\mathrm{A}}=3\mathrm{V}\end{array}$$

So, potential difference across 6$$\mu$$F is 6 V.

$$\begin{array}{rl}\text{ Hence }\mathrm{Q}& =\mathrm{C}\mathrm{\Delta }\mathrm{V}\\ & =6\times 6\times {10}^{-6}\mathrm{C}\\ & =36\mu \mathrm{C}\end{array}$$