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.

IR

+

L

dI dt

=

0

or

dI dt

=

–IR L

Solving this for current gives I(t) = I_o e^-t/t

The potential difference across the resistor has a similar form:

DV_R = e e^-t/t

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

DV_L = -e e^-t/t

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.