Two capcitors, 3 `muF` and `4 muF,` are individually charged across a 6 V battery. After being disconnected from the battery, they are connected together with the negative plate of one attached to the positive plate of the other. What is the final total energy stored ?

Options

A. `1.26 xx10^(-4) J`
B. `2.57 xx10^(-4) J`
C. `1.26 xx10^(-6) J`
D. `2.57 xx10^(-6) J`

Answer

D

Explanation

`Q_1= C_1 V= 18muC , Q_2 =C_2 V = 4xx 6 = 24muC`

Total charge = `Q_1+Q_2 = 6muC`
`C_(eq) = C_1 +C_2=3+4=7muC`
`U_f =Q^2/(2C_(eq))=((6)^2xx10^(-12))/(2xx7 xx10^(-6)) = 36/14xx10^(-6) = 2.57 xx10^(-6) J`
`= 2.57 xx10^(-6) J`