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

Answer

C

Explanation

Initial energy of the combined system `U_(1)=(1)/(2)CV_(1)^(2) +(1)/(2)CV_(2)^(2)=(C)/(2)(V_(1)^(2) +V_(2)^(2))`
On joining the two condensers in parallel, common potential,
`V = (CV_(1) + CV_(2))/(C+C) = (V_(1)+V_(2))/(2)`
`therefore` Final energy of the combined system
`U_(2) = (1)/(2) (C+C) ((V_(1)+V_(2))/(2))^(2)`
Decrease in energy
`DeltaU = U_(1) - U_(2) = (1)/(2)C(V_(1)^(2) + V_(2)^(2)) -(1)/(2) (2C) ((V_(1)+V_(2))/(2))^(2)`
` = (C)/(2)[2(V_(1)^(2)+V_(2)^(2))-(V_(1) +V_(2))^(2)]`
`=(C)/(4)[2V_(1)^(2)0 + 2V_(2)^(2)-V_(1)^(2) - V_(2)^(2) - 2V_(1)V_(2)]=(C)/(4) (V_(1) -V_(2))^(2)`