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

Answer

B

Explanation

`Q=C_(eq)V=((2xx6)/8+4)=2=11muC`
`U_i = QV = 11 xx 2 = 22muJ`
`U_f =1/2 QV = 11muJ`
Energy lost `= 22 -11 = 11 mu J = 11 xx10^(-6) J`