A $2 \mu \mathrm{F}$ capacitor is charged as shown in the figure. The percentage of its stored energy…
A $2 \mu \mathrm{F}$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position 2 , is
$0 \%$
$20 \%$
$75 \%$
$80 \%$
Solution
$q_i=C_i V=2 V=q$ (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
