Question

A $2\mu F$ capacitor is charged as shown in 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

The correct option is D $80\%$

We have,

$E=\frac{1}{2}\left(2\mu F\right)\cdot v^2$..........(1)

$E=\frac{1}{2}\left(\frac{C_1{C}_2}{C_1+C_2}\right)\cdot v^2$

$E=\frac{1}{2}\left(\frac{\left(2\mu F\right)\left(8\mu F\right)}{10\mu F}\right)\cdot v^2$

$\therefore E=\frac{1}{2}\left(1.6\mu F\right)\cdot v^2$.......(2)

comparing (1) and (2), we have,

Percentage loss of energy $=\frac{1.6}{2}\times 100=80$%