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. `20%`
C. `75%`
D. `80%`

Answer

D

Explanation

`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)=(1)/(2)(q^(2))/(C_(i))=(q^(2))/(2xx2)=(q^(2))/(4)`
`U_(f)=(1)/(2)(q^(2))/(C_(i))=(q^(2))/(2xx10)=(q^(2))/(20)`
`therefore` Energy dissipated `(=(q^(2))/(5))` is `80%` of the intial stored energy `(=(q^(2))/(4)).`