Two identical capacitors, have the same capacitance C. One of them is charged to potential `V_1` and the other `V_2`. The negative ends of the capacitors are connected together. When the poistive ends are also connected, the decrease in energy of the combined system is

Options

A. `C(V_(1)^(2) - V_(2)^(2))//4)`
B. `C(V_(1)^(2) + V_(2)^(2)//4)`
C. `C(V_(1)-V_(2))^2//4)`
D. `C(V_(1)+V_(2))^2//4)`

Explanation

Initially,

Initial energy `=(1)/(2)C(V_(1)^(2)+V_(2)^(2))q_(1)'=q_(2)'=CV_(1)+CV_(2)`
`(q_(q))/(C)=(q_(2))/(C)` or `q_(1)` (charge conservation)
`q_(1)=(C(V_(1)+V_(2))/(2))`
Final energy `=(C(V_(1)+V_(2))^(2))/(4)`
Therefore, change in energy `=` Initial energy `-` Final energy
`=(1)/(2)C(V_(1)^(2)+V_(2)^(2)+2V_(1)V_(2))`
`=(C)/(4)[2V_(1)^(2)+2V_(2)^(2)-V_(1)^(2)-V_(2)^(2)]=((C)/(4))(V_(1)-V_(2))^(2)`