Let's put an inductor (i.e., a coil with an inductance L) in series with a battery of emf ε and a resistor of resistance R. This is known as an RL circuit. There are some similarities between the RL circuit and the RC circuit, and some important differences.

First consider what happens with the resistor and the battery. When the switch is closed we have a current; when the switch is opened again we have no current. Now add an inductor to the circuit. When we close the switch now the current tries to jump up to the same value we had with the resistor but the inductor opposes this because a change in current means a change in flux for the coil. If the inductor adds negligible resistance to the circuit the current eventually reaches the same value it had with the resistor but the current follows an exponential curve to get there.

To see why, apply Kirchoff's loop rule. With the switch closed we have:

ε - IR - L dI/dt = 0.

Solving this for current gives I(t) = I_{o} [1 - e^{-t/τ} ]

where I_{o} = ε/R is the maximum current

and the time constant τ = L/R

The potential difference across the resistor has a similar form:

$Delta;V_{R} = ε [1 - e^{-t/τ} ]

The potential difference across the inductor is:

$Delta;V_{L} = ε e^{-t/τ}

The graph of current as a function of time in the RL circuit has the same form as the graph of the capacitor voltage as a function of time in the RC circuit, while the graph of the inductor voltage as a function of time in the RL circuit has the same form as the graph of current vs. time in the RC circuit.

If we take our previous circuit, wait for a while for the current to level off, and then open the switch so the battery is no longer in the circuit, what happens. If the circuit had just a resistor the current would drop immediately to zero. With the inductor the change in current means a change in magnetic flux so the inductor opposes the change. The current does eventually reach zero but it takes some time to get there.

To see why, apply Kirchoff's loop rule. With the switch opened we have:

IR + L dI/dt = 0

or dI/dt = -IR/L

Solving this for current gives I(t) = I_{o} e^{-t/τ}

The potential difference across the resistor has a similar form:

$Delta;V_{R} = ε e^{-t/τ}

The potential difference across the inductor is negative because it's acting like a battery hooked up in the opposite direction:

$Delta;V_{L} = -ε e^{-t/τ}

Once again, the graph of current as a function of time in the RL circuit has the same form as the graph of the capacitor voltage as a function of time in the discharging RC circuit, while the graph of the inductor voltage as a function of time in the RL circuit has the same form as the graph of current vs. time in the discharging RC circuit.