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. (a) 0%
B. (b) 20%
C. (c) 75%
D. (d) 80%

Answer

D

Explanation

When S and I are connected
The `2muF` capacitor gets charged. The potential difference across its plates will be V.
The potential energy stored in `2muF` capacitor
`U_i=1/2CV^2=1/2xx2xxV^2=V^2`
When S and 2 are connected
The `8muF` capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is disispated as heat. The energy loss is
`DeltaU=1/2(C_1+C_2)/(C_1+C_2)(V_1-V_2)^2`
Here, `C_1=2muF`, `C_28muF`, `V_1=V`, `V_2=0`
`:.` `DeltaU=1/2xx(2xx8)/(2+8)(V-0)^2=4/5V^2`
The percentage of the energy disispated `=(DeltaU)/(U_i)xx100`
`=(4/5V^2)/(V^2)xx100=80%`