A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

Initial energy stored in capacitor 2μF

${\mathrm{U}}_{\mathrm{i}}=\frac{1}{2}2{\left(\mathrm{V}\right)}^2={\mathrm{V}}^2$

Final voltage after switch 2 is ON

${\mathrm{V}}_{\mathrm{f}}=\frac{{\mathrm{C}}_1{\mathrm{V}}_1}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{2\mathrm{V}}{10}=0.2\mathrm{V}$

Final energy in both the capacitors

$$\begin{gathered} {\mathrm{U}}_{\mathrm{f}}=\frac{1}{2}\left({\mathrm{C}}_1+{\mathrm{C}}_2\right){{\mathrm{V}}_{\mathrm{f}}}^2 \\ =\frac{1}{2}10{\left(\frac{2\mathrm{V}}{10}\right)}^2=0.2{\mathrm{V}}^2 \end{gathered}$$

So, energy dissipated = $\frac{{\mathrm{V}}^2-0.2{\mathrm{V}}^2}{{\mathrm{V}}^2} \mathrm{x} 100=80\%$