Question

The switches in figure (a) and (b) are closed at t = 0 and reopened after a long time at t = t₀.


(a) The charge on C just after t = 0 is εC.
(b) The charge on C long after t = 0 is εC.
(c) The current in L just before t = t₀ is ε/R.
(d) The current in L long after t = t₀ is ε/R.

Answer 1

(b) The charge on C long after t = 0 is εC.
(d) The current in L long after t = t₀ is ε/R.

The charge on the capacitor at time ''t'' after connecting it with a battery is given by,
$Q=C\epsilon \left[1-e^{-t/RC}\right]$
Just after t = 0, the charge on the capacitor will be
$Q=C\epsilon \left[1-e^0\right]=0$
For a long after time, $t\to \infty$
Thus, the charge on the capacitor will be
$$\begin{gathered} Q=C\epsilon \left[1-e^{-\infty }\right] \\ \Rightarrow Q=C\epsilon \left[1-0\right]=C\epsilon \end{gathered}$$

The current in the inductor at time ''t'' after closing the switch is given by
$I=\frac{V_{\mathrm{b}}}{R}\left(1-e^{-tR/L}\right)$
Just before the time t₀, current through the inductor is given by
$I=\frac{V_{\mathrm{b}}}{R}\left(1-e^{-t_{0}R/L}\right)$
It is given that the time t₀ is very long.
∴ $t_0\to \infty$
$I=\frac{\epsilon }{R}\left(1-e^{-\infty }\right)=\frac{\epsilon }{R}$
When the switch is opened, the current through the inductor after a long time will become zero.