Question

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

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

Answer 1

The correct option is D $80\%$

Energy stored in capacitor is given by,
$U_i=\frac{1}{2}C{V}^2=\frac{1}{2}\times 2(V{)}^2$
After connecting $S$ to 2,
$V_{common}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}=\frac{2V+8\times 0}{10}$
$V_{common}=\frac{V}{5}$
Energy stored in capacitor after changing position of the key is given by
$\therefore U_f=\frac{1}{2}(2+8){\left(\frac{V}{5}\right)}^2=\frac{V^2}{5}$

Percentage of energy dissipated, $=\frac{(U_i-U_f)}{U_i}\times 100=\frac{V^2-\frac{V^2}{5}}{V^2}\times 100$
$=80\%$