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

Options

A. A)`300muJ`
B. B)`450muJ`
C. C)`225muJ`
D. D)`150muJ`

Answer

A

Explanation

Energy
`E_(1)=(1)/(2)C_(1)V_(1)^(2)=(1)/(2)xx1xx10^(-6)xx(30)^(2)`
`=450xx10^(-6)J`
Common potential
`V=(q_(1)+q_(2))/(C_(1)+C_(2))`
`=(1xx30+0)/(1+2)=10"volt"`
`E_(2)=(1)/(2)(C_(1)+C_(2))V^(2)`
`=(1)/(2)(1+2)xx10^(-6)xx(10)^(2)`
`=1.5xx100xx10^(-6)`
`=150xx10^(-6)J`
Loss of energy `=E_(2)-E_(1)=300muJ`