Inductor equations

Now we're going to talk about the two forms of the inductor equation and get familiar with these things. I'm going to do some examples to show you how the inductor equations work.

So we know that the inductor equation is the voltage across an inductor is a factor called L, the inductance, times dI/dt. So the voltage is proportional to the slope or the rate of change of current. Let me do a quick review of the two letters that are used as variables for inductors.

When we have a resistor, we have we have a we call it an R, and it's in units of ohms, and that's named after Ohm himself. And when we have a capacitor, the units are C for capacitance and F for farad, that's named after Michael Faraday. And for the inductor, we have two other letters we use: L and H. L is the name of the component, and that's named after Heinrich Lens, and he did some research in induction, and the unit is named after Henry. So that's the corresponding R, C, and L are the names of the components, and ohms, farads, and henries are the units.

Okay, let's get back to our inductor. There's another form of the inductor equation where we write it in terms of just I in terms of V. To get to that point, we need to get rid of this derivative here. So what I'm going to do is take the integral of both sides of this equation.

We take the integral of V with respect to time, and that equals L times the integral of dI/dt with respect to time. These two terms go away, and we have the integral of dI. The antiderivative of dI is what? What function has a derivative of dI? That would be just I.

So let me flip this all around here. Now I'm going to write I on the left side equals 1 over L times the integral of V dt. Now we always put limits on this integral, so the limits in this case, our time goes from minus infinity till some time now. This tells us that the current through an inductor depends on the voltage across that inductor for the entire life, all the way back to time equals minus infinity.

And that's not so convenient; we don't want to track this inductor's voltage back that far in time. So what we'll do instead is we'll say, let's say at t equals zero that we know the current, some I of zero. Then we can change the limits on our integral here to be zero on the bottom, one over L, integral from time is zero until now, time T of V of T times dt.

In order to account for everything that's happened before time equals zero, we'll just say, what was the current at time equals zero? So that's I of zero. And I'll do one more little change; I want this variable up here to be small t, and that means I have to replace this T inside here with some other symbol as a dummy variable.

So now I'm going to get I of t, little t, equals 1 over L integral from 0 to little t of V dτ plus I of zero. So this is the integral form; there's the integral form of the inductor equation, and here is the derivative form. We'll use both these.

What I'm going to do now is we're going to do an exercise where we use this one down here. So in our circuit, we have a voltage source with a time-pattern voltage in it, and we have a current going through this thing, and we want to find out what's this current.

So we're connected to an inductor; it's a 10 millihenry inductor, and the voltage waveform starts at zero at time equals zero. The voltage goes up to two volts, then after two milliseconds it goes to minus two volts, after two more milliseconds it goes to plus two volts, and on and on. So we know V of T, and what we're going to figure out is what's I through the inductor.

To do that, we're going to use our inductor equation here, the integral form of the inductor equation. So I actually need some room on the screen here, so let me take a second and clean out some stuff.

Okay, we're ready to go. So let's look at what's going on here before time equals zero. So then the before state, we're going to make an assumption; we're going to assume that I is zero a long time ago. And if we look at our chart here, it says that V is zero as well.

If I take this fact and I look at this equation right here, if V is zero and L is some positive number, we can say that then we know that dI/dt equals zero. So this circuit is pretty much doing nothing before this waveform starts.

I'm going to make a little time chart of what we discover. That's time, and this will be current here. So before zero, before time is zero, we can say that the current is zero. Now we plotted in part of our curve.

Okay, now something happens. Now the voltage goes up to 2 volts, and we're going to use this equation, the integral form of the inductor equation, to figure out what happens to I of t now that the voltage has changed.

So we can say that I equals 1 over L times the integral from 0 to t of V of t. Now V in this little region of time here is 2 volts.

I is equal to 1 over 10 millihenries times 2 times the integral from 0 to t of dτ plus I zero, and we decided I zero was the current that it started at right there; I zero is zero.

Let's keep working on this. We can say that I equals 1 over 10 millihenries times 2, which comes out of the integral, and now it's the integral from 0 to t of dτ. The 0 goes away. That equals 2 over 10 millihenries times what? What's this integral evaluate to?

It evaluates to t. The integral of dτ between 0 and t is t minus 0, or just plain t. So looking at this, I see that I equals a constant times t. That's the equation of a line.

If I plot it on the I and t axes, 2 over 10 millihenries is the slope. To plot that line, what I need is two points on the line. I know one of the points already; I know one of the points is going to be right there at (0, 0).

What is the current at the end of this pulse? So at 2 milliseconds, let's figure out what the current is, and that equals at two milliseconds. It's 2 times two milliseconds divided by 10 millihenries. The millies cancel themselves out, and I get 4 over 10, or 0.4 amperes.

So at 2 milliseconds, the value of the current is 0.4 amperes, and I can draw a straight line between there. So now we have the value of the current during this first excursion, this first part of the voltage waveform.

Now we're going to switch again. Now we have a new I zero. Let's go to a new point here; let's go to a new point at now we have I zero equals 0.4, and we have V equals what? Now it goes to minus 2 volts.

So let's use our inductor equation one more time with these initial conditions. That tells us I equals 1 over L integral from what's the new point, 2 milliseconds to t, and V of t after two milliseconds is minus 2 volts.

So we have I equals minus 2 dτ plus now we have a starting current. Our starting current was 0.4 amps.

Let's continue on here. I equals 1 over 10 millihenries times minus 2 times the integral from 2 milliseconds to t of dτ plus 0.4.

I equals minus 2 over 10 millihenries times what's this? It's again t minus 2 milliseconds plus 0.4.

Again, we have the equation of a line. This is the equation of a line. I is a constant times t, followed by a constant. This is basically a sloping line; it has a negative slope, and the slope is negative 2 over 10 millihenries, which is just the opposite of what we had over here; we had 2 over 10 millihenries, so it's the same looking kind of line.

Let me mark out here, here's four milliseconds, and it's going to be a line that looks like this.

So the current is going to ramp up; it's going to turn around and ramp straight back down because the slope up and the time up is the same as the slope down and the time down, just with a negative slope.

So there's an example of the inductor equation in action.