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

Answer

C

Explanation

In the initial energy of the system is given by
`U_(i)=1/2CV_(1)^(2)+1/2CV_(2)^(2)=1/2C(V_(1)^(2)+V_(2)^(2))`
When the capacitors are joined the potential difference is
`V=(q_(1)+q_(2))/(C_(1)+C_(2))=(CV_(1)+CV_(2))/(C+C)=(V_(1)+V_(2))/2`
The final energy of the system is given by
`U_(f)=1/2(C+C)V^(2)=1/2(2C)V^(2)`
Decrease in energy
`U_(i)-U_(f)=1/2C(V_(1)^(2)+V_(2)^(2))`
`-1/2(2C)((V_(1)+V_(2))/2)^(2)=1/4C(V_(1)-V_(2))^(2)`