In the circuit shownin figure switch `S` is closed at time t=0. Find the current through different wire and charge stored on the capacitor at any time `t`.

Explanation

Calculation of equivalent time constant

In the circuit shown in figure, after short circuiting the battery 3R and 6R are parallel, so their combined resistance is
`((6R)(3R))/(6R+3R)=2R`
Now this 2R is in serics with the remaining R.
Hence, `R_("net") = 2R+R=3R , tau_(c) =(R_("net"))C = 3RC`
Calculation of steady state charge
At `t = oo`, capacitor is fully charged and no current flows through it.
P.D across capacitor = P.D across 3R
`=((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// tau_(C)))` i.e., `q=q_(0)(1-e^(-(t)/(3RC)))` and `i_(1)=(dq)/(dt)=(q_(0))/(t_(c))e^(-t // t_(c))=(q_(0))/(3RC)e^(-(t)/(3RC))` ...........(i)

Applying Kirchoff.s second law in loop ACDFA, we have `-6iR-3i_(2)R+V =0`
`2i+i_(2)=(V)/(3R)` ....... (ii)
Applying Kirchoff.s junction law at B, we have `i=i_(1)+i_(2)` ...... (iii)
solving Eqs. (i), (ii) and (iii), we have `i_(2)=(V)/(9R)-(2)/(3)i_(1)=(V)/(9R)-(2q_(0))/(3t_(c))e^(-t//t_(c))` where `q_(0) =(CV)/(3)` i.e., `i_(2)=(V)/(9R)-(2q_(0))/(3RC)e^(-(t)/(3RC))`
`i=(V)/(9R)+(q_(0))/(3t_(c))e^(-t//t_(c))=(V)/(9R)-(q_(0))/(3RC)e^(-(t)/(3RC))`