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

For parts (a) and (b) we have

For BAD, `V_(B) +2(I - i_(1)) + 1(I - i_(1)) - 4i_(1) = V_(D)`
or `3i - 7i_(1) = -4`
For BCD, `6i - 10i_(1) = 4 or i_(1) = 3A, I = 17//3A`
Current in `AC = i_(1) = 3A. In ABC, i=i_(1) = 8//3A`
`E = 8i_(1) = 24V `
`i = 2i_(3) = i_(1)+i_(2)` (i)

`(q)/(C) = 2i_(1) +i_(1) +4i_(3) = 3i_(1) + 4i_(3)` (ii)
Also `(q)/(C) = 3i_(2)+3i_(2) +4i_(3) = 6i_(2) +4i_(3)` (iii)
From Eqs. (ii) and (iii), we have `6i_(2) = 3i_(2) or i_(1) = 2i_(2)`
Solve to get `i = (q)/(4C) or (-dt)/(dt) = (q)/(4C) or q = q_(0) r^(-t//4C)`
or `5.92 = 16e^(t)/(4xx4xx10^(-6) or t =16mus`