In the circuit shown, the switch $$S_1$$ is closed at time $$t = 0$$ and the switch $$S_2$$ is kept open. At some later time $$(t_0)$$, the switch $$S_1$$ is opened and $$S_2$$ is closed. The behaviour of the current I as a function of time t is given by:

The current in a series LR circuit grows exponentially when connected to a DC source (charging phase) and decays exponentially when the source is disconnected and short-circuited (decaying phase).

Charging phase ($$0 \le t \le t_0$$):

With $$S_1$$ closed and $$S_2$$ open, the inductor stores magnetic energy. The current increases non-linearly with a concave-down exponential trajectory:  $$I(t) = \frac{\varepsilon}{R}\left(1 - e^{-t/\tau}\right)$$

Decaying phase ($$t > t_0$$):

With $$S_1$$ opened and $$S_2$$ closed, the battery is isolated, and the stored inductor current flows through the closed loop formed by $$S_2$$. The current decreases with a concave-up exponential decay trajectory:  $$I(t) = I(t_0)e^{-(t-t_0)/\tau}$$