RL circuit is when a resistor is placed in series with an inductor. When the RL circuit is connected with a dc supply $V_s$, the energizing begins when the switch is closed. The formulation w.r.t. time is:

$V_s=IR + L \frac{dI}{dt}$.

By solving the equation, we can derive:

$I = \frac{V_s}{R}(1 - e^{-Rt/L})$, $V_R=IR=V_s(1-e^{-Rt/L})$, $V_L=L \frac{dI}{dt}=V_s e^{-Rt/L}$.

There is also a time constant $L/R$ in the unit of seconds. For example, if an 10mH inductor and a series 10 Ohm resistor in a RL circuit, the time constant is $10^{-3}$ seconds. Similar to RC time constant, it will take five time constant (which is 5ms in this case) to consider the current in the circuit reaches the full value.

De-energizing RL circuit is a bit tricky. When the switch is opened, a rapid collapse of the magnetic field induces a high voltage, which may burn or melt the circuit. (To be more specific, large amount of energy are released in a very short time.) A common approach to mitigate this is to connect a suitable capacitor and resistor in series across the contacts, or called snubber network.

If the excitation is removed without breaking the circuit, we can formulate this: $V_s=IR+L \frac{dI}{dt}=0$.

The solution is:

$I=\frac{V_s}{R}e^{-Rt/L}$, $V_R=IR=V_s e^{-Rt/L}$, $V_L=L\frac{dI}{dt}=-V_s e^{-Rt/L}$.