Consider the circuit below of a cell, switch, resistor and inductor:

Let's assume that the circuit initially starts with the switch open, and that at a certain time t_{0} the switch is abruptly closed.

What current will initially flow? What current will flow after a long time has passed?

This is a very interesting circuit, and one that can take the reader quite some time to understand if not familiar with the properties of inductors. It turns out that at time t_{0} when the switch is first closed that the current flowing will be 0 Amps. This seems very strange at first, but it is worth some explanation. The switch is assumed to be "perfect", in the sense that it can instantly close. The initial current will be 0 Amps at t_{0} but will start to rise as time progresses. In other words, the current is only zero at t_{0}, and starts to rise from that point on. For the very briefest of moments (infinitessimally small), the current is zero. From the "Rate of Current Increase in an Ideal Inductor for a Fixed Applied Voltage" page, it is understood that the rate of current rise in the inductor is proportional to the voltage applied across the inductor terminals, and is inversely proportional to the inductance.

The first interesting effect to note is that at time t_{0} there is no current flowing through the resistor, and so from the formula V = IR there can be no voltage drop across the resistor.

This leads to the interesting conclusion that at time t_{0} the rate of rise of current in the inductor is E/L amps per second. After a very small time period there will be a current flowing through the resistor and inductor, say I_{t}. At this time, the resistor will "drop" a certain voltage I_{t}R, and so the voltage applied across the inductor terminals will be somewhat lower at (E - I_{t}) volts. So the rate of current rise in the inductor slows down as time progresses. In practice, it is oserved that the current rise slows down, and eventually (after an infinite time!) the inductor behaves as if it wasn't there - the current is then E/R amps. A graph of current versus time looks like this:

MathML formula | Formula Image | Straight text formula |
---|---|---|

$$i\left(t\right)=I\left(1-{e}^{-\left(\frac{R}{L}\right)t}\right)$$ | i(t) = I(1-e^(-(R/L)t)) |

Where

i(t) = Current at time t seconds after switch has been closed, Amps

I = Current that would flow if the inductor was not present and the switch was closed, Amps

L = inductance, Henries

R = resistance, Ohms

# "Real World" RL Circuits

Consider a "real world" example. The cell, interconnecting wires, switch and inductor will all have some resistance. So long as these resistances are small in comparison to that of the resistor, however, then the formulas already given will give results that are accurate enough for all practical circuits.