Two capacitors `C_(1)=3muF` and `C_(2)=6muF` in series, are connected in parallel to a third capacitor `C_(3)=4muF`. This arrangement is then connected to a battery of e.m.f., =`30V`, as shown. The energy lost by the battery in charging the capacitors

Options

A. `900muJ`
B. `1800muJ`
C. `2700muJ`
D. `3600muJ`

Answer

C

Explanation

`V_(1)=((6)/(3+6))xx30=20V`
`V_(2)30-V_(1)=10V`
Heat produced
`H=[(1)/(2)xx3(20-0)^(2)+(1)/(2)xx6(10-0)^(2)+(1)/(2)xx4(30-0)^(2)]xx10^(-6)`
`=2700muJ`