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

Initially, the energy stored in `2 muF` capacitor is
`U_(i) = (1)/(2) CV^(2) = (1)/(2) (2xx10^(-6)) V^(2) = V^(2) xx10^(-6) J`
Initially, the charge stored in `2 muF` capacitors is `Q_(1) = CV = (2xx10^(_6)) V = 2V xx10^(-6)` coulomb, when swich `S` is turned to position 2, the charge flows and both the capacitors share charges till a common potential `V_(C)` is reached, FIg. Therefore

Finally, the energy stored in both the capacitors
`U_(f) = (1)/(2) [(2+8)xx10^(-6)] ((V)/(5))^(2) = (V^(2))/(5) xx10^(-6) J`
`%` Loss of energy, `Delta U = (U_(i) - U_(f))/(U_(i)) xx100`
`= ((V^(2) - V^(2)//5)xx10^(-6))/(V^(2) xx10^(-6)) xx100 = 80%`