The enrgy of a charged capacitor is U. Another identical capacitor is connected parallel to the first capacitor, after disconnecting the battery. The total energy of the system of these capacitors will be

Options

A. `(U)/(4)`
B. `(U)/(2)`
C. `(3U)/(2)`
D. `(2U)/(4)`

Answer

B

Explanation

Common potential `(C_(1)V_(0)+C_(2)xx0)/(C_(1)+C_(2))=(C_(2)V_(0))/(C_(1)+C_(2))`
`U_("before")=(1)/(2)C_(1)V_(0)^(2)`
`U_("after")=(1)/(2)C_(1)[(C_(1)V_(0))/(C_(1)+C_(2))]^(2)+(1)/(2)C_(2)[(C_(1)V_(0))/(C_(1)+C_(2))]^(2)`
`=(1)/(2)[(C_(1)V_(0))/(C_(1)+C_(2))]^(2)(C_(1)+C_(2))`
`rArr" "(U_("before"))/(U_("after"))=(C_(1)+C_(2))/(C_(1))`
`"Here, "C_(1)=C_(2)=C`
`therefore" "(U_("before"))/(U_("after"))=(2C)/(C)" "rArr=(U)/(2)`