Analyze the given circuit in the steady state condition. Charge on the capacitor is`q_0=16muC`

(a) Find the current in each branch
(b) Find the emf of the battery.
(c) If now the battery is removed and the points `A` and `C` are shorted. Find the time during which charge on the capacitor becomes
`8muC`

Explanation

`V_(B)=V_(D)=(q_0)/(C )=16/4= 4V`

`B` to `D` path `B Y X D`
`V_B+i_(2)xx2+i_(2)xx1-i_(1)xx4=V_(D)`
`4i_(1)-3i_(2)=V_(B)-V_(D) = 4`..(i)
`B` to `D` path `BUVD`
`V_(B) -i_2xx3-i_2 xx 3 + i_1 xx 4 = V_(D)`
`4i_(1)-6i_(2)=V_(D)-V_(B) =-4` ..(ii)
Solving (i) and (ii)
`i_(2) = 8/3 A, i_(1) = 3A`
`X` to `X` path `XDVX`
`V_X - i_1 xx 4-i_(1)xx4+E=V_(X)`
1E = 8i_(1)=24V`.