The RL Circuit

RL circuits are very similar to RC circuits however the time constant \(\tau\) is given by: \(\tau = \frac{L}{R}\) Where \(L\) is the equivalent inductance of the circuit and \(R\) is the equivalent resistance of the circuit. The unit of the time constant is seconds.

Find the time constant of the following circuit:

The inductance of the circuit is given by the single inductor so \(L = 12mH\). The resistance is the parallel combination of \(R_1\) and \(R_2\) so \(R = \frac{R_1 \times R_2}{R_1 + R_2} = 250\Omega\). Plugging these into our trusty formula and we find that: \(\tau = \frac{12 \times 10^{-3}}{250} = 48\mu\)s

In this course we claim that an inductor has no resistance and that after the current in the circuit settles to a constant value there is no voltage drop across any inductor. Remember though that while an inductor has a changing current through it there will be some voltage across it.

The "final" current through an inductor is the current that would pass through that point on the circuit if the inductor were replaced with a piece of resistorless wire.

Find the current through the inductor in the circuit below after a long time.

"After a long time" just means assume everything has gotten to its final, constant value. So we replace the inductor with a piece of wire and calculate the current that would run through that wire. We see that the current \(I_L = \frac{10V}{100\Omega} = 100\)mA So after a long time (or as we sometimes say, after all the transients have died out) the current through the inductor will be 100mA.

The current through an inductor is given by: \(\Large I_L = I_f(1-e^{\frac{-t}{\tau}})\) Where \(I_f\) is the "final" steady-state current that will flow through the inductor and \(t\) is the time in seconds since the circuit was closed.

When voltage is applied across an inductor the current through it will increase to 63.2% of its final value in one time constant, 86.5% in two time constants, 95% in three time constants, 98.2% in four time constants and 99.3% in five time constants.

Find the voltage across the inductor after the switch has been closed for 1 time constant.

For this one we'll first find the current running through the inductor after 1 time constant. Then we can find the voltage drop across the resistor and use KVL to find the voltage that must be across the inductor. Easy peasy. First we need \(I_f\) so replacing the inductor with a piece of wire we see that \(I_f = \frac{25}{1000} = 25\)mA Now we don't need it but the time constant \(\tau = \frac{L}{R} = \frac{0.1}{1000} = 100\mu\)s Since we're waiting 1 time constant \(t = \tau\) so \(\frac{-t}{\tau} = -1\) Now we can use our formula: \(I_L = I_f(1-e^{\frac{-t}{\tau}} = 25\times 10^{-3} (1 - e^{-1}) = 15.8\)mA Next we find the voltage drop across our resistor which, using Ohm's Law, gives \(V_R = I\times R = 0.0158 \times 1000 = 15.8\)V And finally using KVL we see that \(V_L = 25 - 15.8 = 9.2\)V We could have skipped some of the calculation by noting that the current through the inductor after 1 time constant was going to be \(0.025 \times 0.632 = 0.0158\) (by our rules of thumb above) however I wanted to show the full, general solution method.

If the circuit you're working on isn't a simple, series inductor and resistor combination then we'll need to use Thevenin's theorem to turn it into a simple series inductor and resistor combination. By finding the Thevenin resistance between the inductor's terminals we can use that as the resistance in our time constant formula.

Find the time constant of the following circuit:

Since this isn't a series inductor resistor combination we'll need to find the Thevenin equivalent resistance between the inductor's terminals. So we'll find the Thevenin equivalent resistance of the following circuit: Remember that when finding the Thevenin equivalent resistance we replace voltage sources by short circuits and current sources with open circuits. Now if you stare at it a bit you might see that, beginningg at node A and working around to node B we have the parallel combination of the two resistors \(= 50\Omega\). So now we can use our trusty formula: \(\tau = \frac{L}{R} = \frac{0.6}{50} = 12\)ms