1.

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

Answer» Correct Answer - B
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)`


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