Physics - Capacitor Question with Solution | TestHub

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