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. `1/4 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

`U_i =1/2 C(C_1^2+V_2^2)`
Common potential 
`V=(C_1V_1+C_2V_2)/(2C) =(V_1+V_2)/(2)`
`U_f =1/2 (2C)V^2=(C(V_1+V_2)^2)/(4)`
Loss of energy `=U_i -U_f`
`=1/2C[V_1^2+V_2^2]-C/4 [V_1^2+V_2^2+2V_1V_2]`
`=1/4C[V_1^2+V_2^2-2V_1V_2]=1/4 C(V_1-V_2)^2`
Aliter :
Use formula
Loss of energy `=1/2 ((C_1C_2)/(C_1+C_2))(V_1-V_2)^2`