The ratio of potential difference that must be applied across parallel and series combination of two…
The ratio of potential difference that must be applied across parallel and series combination of two capacitors $\mathrm{C}_1$ and $\mathrm{C}_2$ with their capacitance in the ratio $1: 2$ so that energy stored in these two cases becomes same is
$3:\sqrt {2}$
$\sqrt {2}:3$
$2:9$
$9:2$
Solution
Given: $\mathrm{C}_1: \mathrm{C}_2=1: 2$
$\begin{aligned}
\therefore \quad \mathrm{C}_2 & =2 \mathrm{C}_1 \\
\mathrm{C}_{\mathrm{P}} & =\mathrm{C}_1+\mathrm{C}_2=\mathrm{C}_1+2 \mathrm{C}_1=3 \mathrm{C}_1 \\
\mathrm{C}_{\mathrm{S}} & =\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{2 \mathrm{C}_1^2}{3 \mathrm{C}_1}=\frac{2}{3} \mathrm{C}_1
\end{aligned}$
Let $V_P$ and $V_S$ be the potentials applied across the parallel and series combinations respectively, then using $\mathrm{E}=\frac{1}{2} \mathrm{CV}^2$ we can write,
$\begin{aligned}
& \frac{1}{2} \mathrm{C}_{\mathrm{P}} \mathrm{V}_{\mathrm{P}}^2=\frac{1}{2} \mathrm{C}_{\mathrm{S}} \mathrm{V}_{\mathrm{s}}^2 \\
\therefore \quad & \frac{\mathrm{V}_{\mathrm{P}}^2}{\mathrm{~V}_{\mathrm{S}}^2}=\frac{\mathrm{C}_{\mathrm{S}}}{\mathrm{C}_{\mathrm{P}}} \\
& \frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{S}}}=\sqrt{\frac{\frac{2}{3} \mathrm{C}_1}{3 \mathrm{C}_1}}=\frac{\sqrt{2}}{3}
\end{aligned}$