A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference…
- $\frac{3 C_1}{n_1 n_2}$
- $\frac{9 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
- $\frac{3 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
- $\frac{9 C_1}{n_1 n_2}$
Solution
When connected in parallel, $\left(\mathrm{C}_{\mathrm{eq}}\right)_2=\mathrm{n}_2 \dot{\mathrm{C}}_2 ; \mathrm{V}_2=2 \mathrm{~V}$
Total energy is the same for both connections, $\mathrm{U}_1=\mathrm{U}_2$ $\begin{aligned} \therefore \quad & \frac{1}{2}\left(\mathrm{C}_{\mathrm{eq}}\right)_1 \mathrm{~V}_1^2=\frac{1}{2}\left(\mathrm{C}_{\mathrm{eq}}\right)_2 \mathrm{~V}_2^2 \quad \therefore\left(\because \mathrm{U}=\frac{1}{2} \mathrm{CV}^2\right) \\ & \frac{1}{2} \frac{\mathrm{C}_1}{\mathrm{n}_1} 36=\frac{1}{2} \mathrm{n}_2 \mathrm{C}_2 4 \\ & \mathrm{C}_2=\frac{9 \mathrm{C}_1}{\mathrm{n}_1 \mathrm{n}_2} \end{aligned}$
Asked in: MHT CET 2024 (16 May Shift 1)