Calculation of `tau_C` Equivalent resistance across capacitor after short circuiting the battery is
`R_(n et)=(R+(6R)(3R))/(6R+3R)=3R`
`tauC=(C)R_(net)=3RC`
Calculation of steady state charge `q_0` at `t=oo` capacitor is fully charged and no current flows through it.
`=(V/(9R))(3R)`
`=V/3`
`q_0=(CV)/3`
Now, let charge on the capacitor at any time `t` be `q` and current through it is `i_1`. Then, `q=q_0(1-e^(-t/(tauC)))`...........i
Applying Kirchoff's second law in loop ACDFA,we have
`-6ir-3i_2R+V=0`
or `2i+i_2=V/(3R)` ............ii
Apply Kirchoff's junction law at B,we have
`i=i_1+i_2` ..............iii
Solving eqn i , ii and iii we have
`i_2=V/(9R)-2/3i_1=V/(9R)-(2q_0)/(3tau_C)e^(-t/(tauC))` and `i=V/(9R)+q_0/(3tau_C)e^(-t/(tau_C))`
