The capacitor `C_1` in the figure shown initially carries a charge `q_0`. When the switches `S_1` and `S_2` are closed, capacitor `C_1` is connected in series to a resistor `R` and a second capacitor `C_2` which is initially uncharged.

the total head dissipated in the circuit during the discharging process of `C_1` is

Options

A. `q_0^2/(2C_1^2)xxC`
B. `q_0^2/(2C)`
C. `(q_0^2C_2)/(2C_1^2)`
D. `q_0^2/(2C_1C_2)`

Answer

A

Explanation

Common potential in steady state when they finnaly come in parallel is
`V=("Total charge")/("Total capacity")=q_0/(C_1+C_2)`
total heat dissipated `=U_i-U_f`
`=q_0^2/(2C_1)-1/2(C_1+C_2) (q_0/(C_1+C_3))^2`
`=(q_0^2/(2C_1))((C_1C_2)/(C_1+C_2))`