A condenser of capacity `C_(1)` is charged to a potential `V_(0)`. The electrostatic energy stored in it is `U_(0)`. It is connected to another uncharged condenser of capacity `C_(2)` in parallel. The energy dissipated in the process is

Options

A. `(C_(2))/(C_(1) + C_(2))U_(0)`
B. `(C_(1))/(C_(1) + C_(2)) U_(0)`
C. `((C_(1) - C_(2))/(C_(1) + C_(2)))U_(0)`
D. `(C_(1)C_(2))/(2(C_(1) + C_(2)))U_(0)`

Answer

A

Explanation

Loss of energy during sharing `= (C_(1)C_(2)(V_(1)-V_(2)^(2)))/(2(C_(1) + C_(2)))`
In the equation, put `V_(2) = 0, V_(1) = V_(0)`
`:.` Loss of energy `= (C_(1)C_(2)V_(0)^(2))/(2(C_(1) + C_(2)))`
`= (C_(2)U_(0))/(C_(1) + C_(2)) [ :' U_(0) = (1)/(2) C_(1)V_(0)^(2)]`