A capacitor of capacitance value `1 muF` is charged to 30 V and the battery is then disconnected. If it is connected across a `2 muF` capacotor, then the energy lost by the system is

Options

A. `300muF`
B. `450muF`
C. `225muF`
D. `150muF`

Answer

A

Explanation

`E_(1)=(1)/(2)C_(1)V_(1)^(2)=450xx10^(-6)J`
common potential `V=(q_(1)+q_(2))/(sigma_(1)+sigma_(2))=10`volt, `E_(2)=(1)/(2)(C_(1)+C_(2))V^(2)=150xx10^(-6)J`
Loss=`(450-150)muJ=300muJ`