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)

$q_i=C_{i}V=2V=q$ (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 $\left(=\frac{q^2}{4}\right).$