Two capacitors A and B of capacities C and 2C are connected in parallel and the combination is connected to a battery of V volts. After the charging is over , the battery is removed. Now a dielectric slab of `K=2` is inserted between the plates of A so as fill the completely. The energy lost by the system during the sharing of charges is
(i) With battery of parallel combination :

Explanation

`C_1=C,C_2=2C,V=V`
`C_P=C_1+C_2=3C,q=3CV`

Initial Energy stored,
`U_(i)=(1)/(2)C_(p)V^(2)=(3)/(2)CV^2`.