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

Answer

A

Explanation

Initial energy stored in capacitor `2 muF`
`U_(i) = (1)/(2)2(V)^(2) = V^(2)`
Final voltage after switch `2` is `ON`
`V_(f) = (C_(1)V_(1))/(C_(1)+C_(2)) = (2V)/(10) = 0.2 V`
Final energy in both the capacitors
`U_(f) = (1)/(2) (C_(1) + C_(2)) V_(f)^(2) = (1)/(2) 10 ((2V)/(10))^(2) = 0.2 V^(2)`
So energy dissipated `= (V^(2) - 0.2 V^(2))/(V^(2)) xx 100 = 80%`