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 \(\mathrm{E}\). 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 _______.

\(\text { Energy }=\frac{1}{2}\left(\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3\right) \mathrm{V}^2\)

\(=\frac{1}{2}(25+30+45) \times(100)^2 \times 10^{-6}=0.5=\mathrm{E}\)

In series combination: Charge is same on all.

\(\frac{1}{\mathrm{C}_{\text {equ }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{1}{25}+\frac{1}{30}+\frac{1}{45}\)

\(\frac{1}{\mathrm{C}_{\text {equ }}}=\frac{(18+15+10)}{450}=\frac{43}{450} \Rightarrow \mathrm{C}_{\text {equ }}=\frac{450}{43}\)

\(\text { Energy }=\frac{Q^2}{2 C_1}+\frac{Q^2}{2 C_2}+\frac{Q^2}{2 C_3}\)

\(=\frac{Q^2}{2}\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right]\)

\(\frac{\left(V \times C_{\text {equ }}\right.}{2} \times \frac{1}{C_{\text {equ }}}frac{V^2 C_{\text {equ }}}{2}\)

\(\frac{(100)^2}{2} \times \frac{450}{} \times 10^{-6}\)

\(\Rightarrow \frac{4.5}{86}=\frac{9}{x} E=\frac{9}{x} \times 0.5 \Rightarrow x=86\)