Question

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

Answers are Both options (b) & (c)

Explanation:

(b) The charge on C long after t = 0 is $\varepsilon C.$

(d) The current in L long after t = $t_0$ is \frac {\varepsilon }{R}.

The charge on the capacitor at time ''t'' after connecting it with a battery is given by,

$Q = C\varepsilon [1 - e^-^t^/^R^C]$

Just after t = 0, the charge on the capacitor will be

$Q = C\varepsilon [1 - e^0] = 0$

For a long after the time, $t \rightarrow  \infty$

$t \rightarrow \infty$  

Thus, the charge on the capacitor will be

$Q = C\varepsilon \times [1 - e^{-\infty }]$

⇒ $Q = C\varepsilon \times [1 - 0] = C\varepsilon$

The current in the inductor at time ''t'' after closing the switch is given by

$I = \frac {V_b}{R}\times (1 - e^{-tR/L})$

Just before the time $t_0$, the current through the inductor is given by

$I = \frac {V_b}{R}\times (1 - e^{-t_0R/L})$

It is given that the time $t_0$ is very long.

∴ $t_0 \rightarrow \infty$

I = $\frac {\varepsilon}{R} \times (1 - e^{-\infty }) = \frac {\varepsilon }{R}$

When the switch is opened, the current through the inductor after a long time will become zero.