Initially `K_(1)` is closed, now if `K_(2)` is also closed, find heat dissipated in the resistances of connecting wires

Options

A. `(1)/(2)CV^(2)`
B. `(2)/(3)CV^(2)`
C. `(1)/(3)CV^(2)`
D. `(1)/(4)CV^(2)`

Answer

C

Explanation

When `K_(1)` is closed
Q=CV
energy `U_(1)=1/2CV^(2)`
When `K_(2)` is also closed

Equivalebnt circuit

`C_(eq)=C+(2C)/(3)=(5C)/(6)CV^(2)`
Charges supplied by battery after closing `K_(2)`
`=5/2CV-CV=2/3CV`
Energy supplied by bettery `=U_(r)-U_(i)+DeltaH`
`2/3CV^(2)=5/6CV^(2)-1/2CV^(2)+DeltaH`
`thereforeDeltaH=1/3CV^(2)`