In given circuit initially switch `S_(1)` is closed for long time and then opened and now `S_(2)` is closed. The total energy dissipated in `4 Omega` resistor is equal to :-

Options

A. 0.5 mJ
B. 0.05 mJ
C. 0.1 mJ
D. 10 mJ

Answer

C

Explanation

`H_(1)+H_(2)=(1)/(2)CV^(2)`
`(H_(1))/(H_(2))=(R_(1))/(R_(2))=(2)/(4)=(1)/(2)`
`H_(1)=(H_(2))/(2)`
`(H_(2))/(2)+H_(2)=(1)/(2)CV^(2)`
`H_(2)=(1)/(3)CV^(2)`
`=(1)/(3)xx3xx10^(-6)xx(10)^(2)=10^(_4)J`
`= 0.1 mJ`