Question

A $2 \mu 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

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

Answer 1

Answer: (D)

$q_i=C_iV=2V=q$ (say)
This charge will remain constant after switch is shifted from position $1$ to position $2$.
$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} $
$\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 $\left(=\frac {q^2}{4}\right).$