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

Options

A. `22 xx 10^(-6) J`
B. `11 xx 10^(-6) J`
C. `((32)/(2)) xx 10^(-6) J`
D. `((16)/(3)) xx10^(-6) J`

Answer

B

Explanation

`E_(lost)=(1)/(2)C_(eff)V^(2)`.