Question

The capactor `C_1` in the figure initially carries a charge `q_0`. When the i switch `S_1` and `S_2` are closed, capacitor `C_1` is connected to a resistor `R` and a second capacitor `C_2`, which initially does not carry any charge.
(a) Find the charges deposited on the capacitors in steady state and the current through R as a function of time.
(b) What is heat lost in the resistor after a long time of closing the switch?

Answer 1

Correct Answer - A::B::C
In steady state, capacitors will be in parallel. Charge will distribute in direct ratio of their capacity.
`:. q_1=(C_1/(C_1+C_2))q_0`
and `q_2=(C_2/(C_1+C_2))q_0`
Initial emf in the circuit is potential difference across capacitor `C_1` or `q_0/C_1`.
Therefore, initial current would be
`i_0=(q_0//C_1)/(R)=(q_0)/(C_1R)`
Current as function of time will be `i=i_0e^(-t//tau_C)`
Here, `tau_C=((C_1C_2)/(C_1+C_2))R`
b. `U_i=1/2q_0^2/C_1` and `U_f=1/2q_0^2/(C_1+C_2)`
Heat lost in the resistor
`=U_i-U_f=q_0^2/2[C_2/(C_1(C_1+C_3))]`