Question

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

A. `0%`
B. `20%`
C. `75%`
D. `80%`

Answer 1

`U_(i)=(1)/(2)(2)V^(2)=V^(2)`
After connecting S to 2.
`V_("common")=(C_(1)V_(1)+C_(2)V_(2))/(C_(1)+C_(2))=(2V+8xx0)/(2+8)=(V)/(5)`
`U_(f)=(1)/(2)(2+8)((V)/(5))^(2)=(V^(2))/(5)`
Percent energy dissipated `=(U_(i)-U_(f))/(U_(i))xx100`
`=(V^(2)-(V^(2))/(5))/(V^(2))xx100=80%`