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[1-e^(-t"/RC")]`

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

`Q=C epsilon[1-e^0]=0`

For a long after time, t → ∞

Thus, the charge on the capacitor will be

`Q = Cepsilon[1-e^(-infty)]`

`rArr Q=Cepsilon[1-0]=Cepsilon`

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

`I=V_b/R(1-e^(-tR"/L"))`

Just before the time t₀, current through the inductor is given by

`I=V_b/R(1-e^(-t_0R"/L"))`

It is given that the time t₀ is very long.

∴ t₀ → ∞

`I=epsilon/R(1-e^-infty)=epsilon/R`

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