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___________.
JEE MAIN 2024, Diffcult
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In parallel combination : Potential difference is same across all
$\text { Energy }=\frac{1}{2}\left(\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3\right) \mathrm{V}^2$
$=\frac{1}{2}(25+30+45) \times(100)^2 \times 10^{-6}=0.5=\mathrm{E}$
In series combination: Charge is same on all.
$\frac{1}{\mathrm{C}_{\text {equ }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{1}{25}+\frac{1}{30}+\frac{1}{45}$
$\frac{1}{\mathrm{C}_{\text {equ }}}=\frac{(18+15+10)}{450}=\frac{43}{450} \Rightarrow \mathrm{C}_{\text {equ }}=\frac{450}{43}$
$\text { Energy }=\frac{Q^2}{2 C_1}+\frac{Q^2}{2 C_2}+\frac{Q^2}{2 C_3}$
$=\frac{Q^2}{2}\left[\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right]$
$\frac{\left(V \times C_{\text {equ }}\right.}{2} \times \frac{1}{C_{\text {equ }}}frac{V^2 C_{\text {equ }}}{2}$
$\frac{(100)^2}{2} \times \frac{450}{} \times 10^{-6}$
$\Rightarrow \frac{4.5}{86}=\frac{9}{x} E=\frac{9}{x} \times 0.5 \Rightarrow x=86$
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