Three capacitors of capacitances $25 \mu \mathrm{~F}, 30 \mu \mathrm{~F}$ and $45 \mu \mathrm{~F}$ are connected in parallel to a supply of 100 V . Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is $\frac{9}{\mathrm{x}} \mathrm{E}$. The value of $x$ is. $\_\_\_\_$
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Solution
In parallel combination: Potential difference is same across all
$$
\begin{aligned}
& \text { Energy }=\frac{1}{2}\left(\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3\right) \mathrm{V}^2 \\
& =\frac{1}{2}(25+30+45) \times(100)^2 \times 10^{-6}=0.5=\mathrm{E}
\end{aligned}
$$
In series combination: Charge is same on all.
$$
\begin{aligned}
& \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{1}{25}+\frac{1}{30}+\frac{1}{45} \\
& \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{(18+15+10)}{450}=\frac{43}{450} \Rightarrow \mathrm{C}_{\text {equ }}=\frac{450}{43} \\
& \text { Energy }=\frac{q^2}{2 \mathrm{C}_1}+\frac{q^2}{2 \mathrm{C}_2}+\frac{q^2}{2 \mathrm{C}_3} \\
& =\frac{\mathrm{Q}^2}{2}\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right] \\
& \frac{\left(\mathrm{V} \times \mathrm{C}_{\text {equ }}\right)^2}{2} \times \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{\mathrm{V}^2 \mathrm{C}_{\text {equ }}}{2} \\
& \frac{(100)^2}{2} \times \frac{450}{43} \times 10^{-6} \\
& \Rightarrow \frac{4.5}{86}=\frac{9}{x} E=\frac{9}{x} \times 0.5 \Rightarrow x=86
\end{aligned}
$$
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