A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference…

A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference 6 V . Another parallel combination of $\mathrm{n}_2$ capacitors, each of value $\mathrm{C}_2$ is charged by a source of potential' difference 2 V . Total energy of both the combinations is same. The value of $\mathrm{C}_2$ in terms of $\mathrm{C}_1$ is
  1. $\frac{3 C_1}{n_1 n_2}$
  2. $\frac{9 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
  3. $\frac{3 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
  4. $\frac{9 C_1}{n_1 n_2}$

Solution

When connected in series, $\left(\mathrm{C}_{\mathrm{eq}}\right)_1=\frac{\mathrm{C}_1}{\mathrm{n}_1} ; \mathrm{V}_1=6 \cdot \mathrm{~V}$
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)

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