The capacitor `C_(1)` in the figure initially carries a charge `q_(0)`. When the switch `S_(1)` and `S_(2)` are shut, capacitor `C_(1)` is connected in series to a resistor `R` and a second capacitor `C_(2)`, which initially does not carry any charge.
If the heat lost in the resistor after a long time of closing the switch is `(q_(0)^(2)C_(2))/(kC_(1)(C_(1)+C_(2)))`, then Find the value of `k`.

Explanation

Electrostatic energy at `t=0` is
`U(0)=(q_(0)^(2))/(2C_(1))`
Final energy `=U(oo)=(q_(0)^(2))/(2(C_(1)+C_(2)))`
`DeltaU=U(0)-U(oo)=(q_(0)^(2)C_(2))/(2C_(1)(C_(1)+C_(2)))`