Physics - Capacitor Question with Solution | TestHub

$q_i=C_{i}V=2V=q$ (say) This charge will remain constant after switch is shifted from position 1 to position 2. $$\begin{array}{rl}{U}_i& =\frac{1}{2}\frac{q^2}{C_i}=\frac{q^2}{2\times 2}=\frac{q^2}{4}\\ U_f& =\frac{1}{2}\frac{q^2}{C_f}=\frac{q^2}{2\times 10}=\frac{q^2}{20}\end{array}$$ $\therefore$ 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.