In the circuit shown in Figure, the battery is an ideal one, with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0 .
(a) Find the charge Q on the capacitor at time t.
(b) Find the current in AB at time t. What is its liniting value as `t rarr oo`:
.

Answer

A, B, C

Explanation

Let at any time t charge on capacitor C be Q. Let currents are
as shown in fig. Since charge Q will increase with time 't'
therefore `i_1 = (dQ)/(dt)`

(a) Applying Kirchoff's second law in the loop MNABM.
`V = (i-i_1)R + iR`
or `V = 2iR -i_1R ..........(i)`
Similarly, applying Kirchoff's second law in loop MNSTM,
we have
`V = i_1R + Q/C + iR` ......(ii)
Eliminating i from equation (1) and (2), we get
`V = 3i_1R + (2Q)/C or 3i_1R = V - (2Q/C)`
or `i_1 = 1/(3R) (V-(2Q)/C) or (dQ)/(dt) = 1/(3R)(V-(2Q)/C)`
or `(dQ)/(V-(2Q)/C) = (dt)/(3R) or int_(0)^(Q) (dQ)/(V-(2Q)/C) = int_(0)^t (dt)/(3R)`
This equation gives Q `= (CV)/2 (1-e^(-2t//3RC))`
(b) `i_1 = (dQ)/(dt)`
`i_1 = d/(dt) [ (CV)/2 (1-e^(-2t//3RC))]`
`=(CV)/2 xx 2/(3RC) xx e^(-2t/3RC) = V/(3R) e^(-2t//3RC)`
From equation(i)
`i = (V+i_1R)/(2R) = (V+V/3e^(-2t//3RC))/(2R)`
`:. Current through AB`
`i_2 = i - i_1 = (V + V/3 e^(-2t//3RC))/(2R)` - V/R e^(-2t//3RC)`
`i_2 = V/(2R) - V/(6R) e^(-2t//3RC)`
`i_2 = V/(2R) as t rarr oo` .