A 2 μ F capacitor is charged as shown in the figure. The percentage...

q_i=C_i V=2 V=q$(say) This charge will remain constant after switch is shifted from position 1 to position 2.$$\begin{aligned}U_i & =\frac{1}{2} \frac{q^2}{C_i}=\frac{q^2}{2 \times 2}=\frac{q^2}{4}\\ U_f & =\frac{1}{2} \frac{q^2}{C_f}=\frac{q^2}{2 \times 10}=\frac{q^2}{20} \end{aligned}$$\therefore$ Energy dissipated $=U_i-U_f=\frac{q^2}{5}$ This energy dissipated $\left(=\frac{q^2}{5}\right)$ is $80 \%$ of the initial stored energy \left(=\frac{q^2}{4}\right)$. Analysis of Question (i) This question is moderately tough. (ii) In a capacitor circuit, redistribution of charge takes place under following three conditions. (a) A switch is closed. (b) A closed switch is opened. (c) A switch is shifted from one position to another position. In the redistribution of charge, energy is dissipated.