The capacities of two conductors are `C_(1)` and `C_(2)` and their respectively potentials are `V_(1)` and `V_(2)`. If they are connected by a thin wire, then the loss of energy will be given by

Options

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

Answer

C

Explanation

Initial energy `U_(i) = (1)/(2)C_(1)V_(1)^(2) + (1)/(2)C_(2)V_(2)^(2)`. Final energy
`U_(f) = (1)/(2)(C_(1) + C_(2))V^(2)` (where `V = (C_(1)V_(1) + C_(2)V_(2))/(C_(1)C_(2)))`
Hence energy loss
`DeltaU = U_(i) - U_(f) = (C_(1)C_(2))/(2(C_(1) + C_(2))) (V_(1) - V_(2))^(2)`