A capacitor of capacitance `C_(1)` is charged to a potential difference V and then connected with an uncharged capacitor of capacitance `C_(2)` a resistance R. The switch is closed at t = 0. Choose the correct option(s):

Options

A. The initial current through the resistance is `V/R`
B. The current through the resistance as a function of time is `i(t)=(V)/(R)e^(-t/tau)` where `tau=R((C_(1)C_(2))/(C_(1)+C_(2)))`
C. The charge on capacitor `C_(1)` as a function of time is `q_(1)(t)= ((C_(1)V)/(C_(1)+C_(2)))(C_(1)+C_(2)e^(-t/tau))`
D. After a long time, the potential difference across the capacitor `C_(1) is ((C_(1))/(C_(1)+C_(2)))V`

Answer

A, B, C, D

Explanation

After a long time, when the current through the circuit has become zero, the potential difference across the two capacitors is equal (since KVL must be satisfied)
So, `(q_(1))/(C_(1))=(q_(2))/(C_(2))` Also, conserving charge `q_(1)+q_(2)=C_(1)V`.
Solving, we get `q_(1)=((C_(1)^(2))/(C_(1)+C-(2)))V`.