Question 127 of 180
Three capacitors of capacitances and are connected in parallel to a supply of . Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is . The value of is _____.
Show answer and solution
Answer: 86
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.
$\begin{aligned}
& \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{1}{25}+\frac{1}{30}+\frac{1}{45} \\
& \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{(18+15+10)}{450}=\frac{43}{450} \Rightarrow \mathrm{C}_{\text {equ }}=\frac{450}{43} \\
& \text { Energy }=\frac{\mathrm{Q}^2}{2 \mathrm{C}_1}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_2}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_3} \\
& =\frac{\mathrm{Q}^2}{2}\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right] \\
& \frac{\left(\mathrm{V} \times \mathrm{C}_{\text {equ }}\right)^2}{2} \times \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{\mathrm{V}^2 \mathrm{C}_{\text {equ }}}{2} \\
& \frac{(100)^2}{2} \times \frac{450}{43} \times 10^{-6} \\
& \Rightarrow \frac{4.5}{86}=\frac{9}{\mathrm{x}} \mathrm{E}=\frac{9}{\mathrm{x}} \times 0.5 \Rightarrow \mathrm{x}=86
\end{aligned}$
$\begin{aligned}
& \text { Energy }=\frac{1}{2}\left(\mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3\right) \mathrm{V}^2 \\
& =\frac{1}{2}(25+30+45) \times(100)^2 \times 10^{-6}=0.5=\mathrm{E}
\end{aligned}$
In series combination: Charge is same on all.
$\begin{aligned}
& \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}=\frac{1}{25}+\frac{1}{30}+\frac{1}{45} \\
& \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{(18+15+10)}{450}=\frac{43}{450} \Rightarrow \mathrm{C}_{\text {equ }}=\frac{450}{43} \\
& \text { Energy }=\frac{\mathrm{Q}^2}{2 \mathrm{C}_1}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_2}+\frac{\mathrm{Q}^2}{2 \mathrm{C}_3} \\
& =\frac{\mathrm{Q}^2}{2}\left[\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\frac{1}{\mathrm{C}_3}\right] \\
& \frac{\left(\mathrm{V} \times \mathrm{C}_{\text {equ }}\right)^2}{2} \times \frac{1}{\mathrm{C}_{\text {equ }}}=\frac{\mathrm{V}^2 \mathrm{C}_{\text {equ }}}{2} \\
& \frac{(100)^2}{2} \times \frac{450}{43} \times 10^{-6} \\
& \Rightarrow \frac{4.5}{86}=\frac{9}{\mathrm{x}} \mathrm{E}=\frac{9}{\mathrm{x}} \times 0.5 \Rightarrow \mathrm{x}=86
\end{aligned}$
