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 positive ends are also connected, the decrease in energy of the combined system is

Options

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

Answer

C

Explanation

Initial energy of the system
`U_(i) = (1)/(2) CV_(1)^(2) + (1)/(2)CV_(2)^(2)`
When the capacitors are joined, common potetnial
`V = (CV_(1) + CV_(2))/(2C) = (V_(1) + V_(2))/(2)`
Final energy of the system
`U_(f) = (1)/(2)(2C)V^(2) = (1)/(2)2C ((V_(1) + V_(2))/(2))^(2) = (1)/(4) C (V_(1) + V_(2))^(2)`
Decrease in energy `= U_(i) - U_(f) = (1)/(4) C(V_(1) - V_(2))^(2)`