A `2muF` capacitor is charged as shown in the figure. The percentage of its stored energy disispated after the switch S is turned to poistion 2 is

Options

A. 0
B. 0.2
C. 0.75
D. 0.8

Answer

D

Explanation

`q_1=c_1v=2v=q` (say)
the charge will remain constant after switch is shifted from possition 1 to possition 2
`Ui=1/2q^(2)/c_1=q^(2)/(2xx2)=q^(2)/4`
`U _f=1/2 q^(2)/c _f=q^(2)/(2xx10)=q^(2)/20 therefore "Energy dissiplated"=U_i-U_f=q^(2)/5`
This energy dissiplated `(q^(2)/5)` is 80% of the initial stored energy `(-q^(2)/4)`