Question

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 limiting value
as $t\to \infty ?$

Answer 1

Let at any time I charge on capacitor C be Q and currents are
as shown. Since, charge Q will increase with time t.
Therefore,
$i_1=\frac{dQ}{dt}$
(a) Applying Kirchhoff's second law in loop MNABM
$V=(i-i_1)R+iR$
or $V=2iR-i_{1}R...(i)$
Similarly, applying Kirchhoff's second law in loop
MNSTM, we have
$V=i_{1}R+\frac{Q}{C}+iR...(ii)$
Eliminating i from Eqs. (i) and (ii), we get
$V=3i_{1}R+\frac{2Q}{C}$
or $3i_{1}R=V-\frac{2Q}{C}$
or $i_1=\frac{1}{3R}(V-\frac{2Q}{C})$
or $\frac{dQ}{dt}=\frac{1}{3R}(V-\frac{2Q}{C})$
or $\frac{dQ}{V-\frac{2Q}{C}}=\frac{dt}{3R}$
or ${\int }_0^Q\frac{dQ}{V-\frac{2Q}{C}}={\int }_0^t\frac{dt}{3R}$
This equation gives
$Q=\frac{CV}{2}(1-e^{-2t/3RC})$
(b) $i_1=\frac{dQ}{dt}=\frac{V}{3R}{e}^{-2t/3RC}$
From Eq. (i)
$i=\frac{V+i_{1}R}{2R}=\frac{V+\frac{V}{3}{e}^{-2t/3RC}}{2R}$
$\therefore$ Current through AB
$i_2=i-i_1=\frac{V+\frac{V}{3}{e}^{-2t/3RC}}{2R}-\frac{V}{3R}{e}^{-2t/3RC}$
$i_2=\frac{V}{2R}-\frac{V}{6R}{e}^{-2t/3RC}$
$i_2=\frac{V}{2R}ast\to \infty$