In parallel combination : Potential difference is same across all $\begin{aligned}
& \text { Energy }=\frac{1}{2}\left(\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}3\right) \mathrm{V}^2 \
& =\frac{1}{2}(25+30+45) \times(100)^2 \times 10^{-6}=0.5=\mathrm{E}
\end{aligned}Inseriescombination:Chargeissameonall.\begin{aligned}
& \frac{1}{\mathrm{C}{\text {equ }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}2}+\frac{1}{\mathrm{C}3}=\frac{1}{25}+\frac{1}{30}+\frac{1}{45} \
& \frac{1}{\mathrm{C}{\text {equ }}}=\frac{(18+15+10)}{450}=\frac{43}{450} \Rightarrow \mathrm{C}{\text {equ }}=\frac{450}{43} \
& \text { Energy }=\frac{\mathrm{Q}^2}{2 \mathrm{C}_1}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_2}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_3} \
& =\frac{\mathrm{Q}^2}{2}\left[\frac{1}{\mathrm{C}1}+\frac{1}{\mathrm{C}2}+\frac{1}{\mathrm{C}3}\right] \
& \frac{\left(\mathrm{V} \times \mathrm{C}{\text {equ }}\right)^2}{2} \times \frac{1}{\mathrm{C}{\text {equ }}}=\frac{\mathrm{V}^2 \mathrm{C}{\text {equ }}}{2} \
& \frac{(100)^2}{2} \times \frac{450}{43} \times 10^{-6} \
& \Rightarrow \frac{4.5}{86}=\frac{9}{\mathrm{x}} \mathrm{E}=\frac{9}{\mathrm{x}} \times 0.5 \Rightarrow \mathrm{x}=86
\end{aligned}$
Three capacitors of capacitances 25 μ F, 30 μ F and 45 μ F are connected in parallel to a supply of 100 ~V. Energy stored in the above combination is…
JEE Main 2024 — Physics Electromagnetism
Three capacitors of capacitances 25μF,30μF and 45μF 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 x9E. The value of x is _____.
Held on 5 Apr 2024 · Verified 6 Jul 2026.
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