A capacitor of `2 mu F` is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is:

Options

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

Answer

D

Explanation

Initial energy stored in capacitor `2muF`
`U_i=1/(2)2(V)^2=V^2`
Final voltage after switch 2 is ON
`V_f= (C_1V_1)/(C_1+C_2)=(2V)/10 = 0.2` V
Final energy in both the capacitors
`U_i=1/(2)(C_1+C_2)V_f^2=1/(2)10((2V)/10)^2=0.2V^2`
So energy dissipated `=(V^2-0.2V^2)(/V^2) xx 100 = 80%`