Question

A $2\mu \text{F}$ capacitor is changed as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to the position $2$ is,

• A. $75\%$
• B. $0\%$
• C. $20\%$
• D. $80\%$

Answer 1

The correct option is D $80\%$

When switch will move to position $2$, both the capacitors will be in parallel connection.
$\text{Fractional loss in energy}=\frac{\Delta U}{U}$,
where $\Delta U=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}(V_1-V_2{)}^2$
$=\frac{1}{2}\frac{2\times 8}{10}(V-0{)}^2=\frac{8}{10}{V}^2$
$\text{Initial energy}=U=\frac{1}{2}{C}_1{V}_1^2=\frac{1}{2}\left(2\right)V^2=V^2$
$\therefore \text{percent dissipation}=\frac{\Delta U}{U}\times 100\%$
$=\frac{8}{10}\times 100\%$
$=80\%$
Hence, option $\left(D\right)$ is correct.