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 series with the remaing `R`.
Hence, `R_("net") = 2R+R = 3R, tau_(c ) = (R_("net")) C = 3RC`
Calculation of steady state charge `q_(0)`:
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))/(tau_(c ))e^(-t//tau_(c )) = (q_(0))/(3RC)e^((1)/(3RC))` .....(i)

Applying kirchhoff'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))`