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)/(3)) xx 10^(-6) J`
D. `((16)/(3)) xx 10^(-6) J`

Answer

B

Explanation

`C_(eq) = (C_(1)C_(2))/(C_(1) + C_(2)) + C_(3) = (2 xx 6)/(2 + 6) + 4 = 5.5 muF`
Energy supplied `(E) = QV = CV^(2) = 22 xx 10^(-6) J`
`P.E.` stored
`(U) = (1)/(2) C_(eq)V^(2) = (1)/(2) xx 5.5 xx (2)^(2) = 11 xx 10^(-6) J`
`rArr` Energy lost `= E - U = 11 xx 10^(-6) J`