ï»¿ Let's talk about AC circuits, and specifically how inductors and capacitors are going to fit into these AC circuits. So, one thing that we said before was if we just have a battery, V, and we have a resistor R, then we know that Ohm's law holds. V equals IR. The voltage drop around the circuit has to be zero, V minus I R equals zero, which leads to V equals IR. But let's say we go to an AC circuit. This is, of course, DC. If we switch to an AC circuit, what does it look like? Well we draw our voltage like this. It's an alternating current source, goes around, goes through the resistor, comes back. And now we have a current that is going to be alternating. So I equals some I naught times cosine of omega t. Current's gonna go one way, and it's gonna go the other way, it's going to oscillate back and forth sinusoidally. But what about the voltage drop across the resistor? The voltage drop across R it's the same thing, it's just I times R. But now that current is alternating. So it becomes I naught times R times cosine of omega t. And let's label this V sub R. The voltage drop across the resistor. Good. What it says is the voltage drop across the resistor goes like cosine omega t, it's in phase with the cosine. So if the voltage is going up and down like a cosine and the current is going up and down like a cosine, then the voltage drop across the resistor is also going up and down like a cosine. It is in phase with it. But what happens when we now take out the resistor and we put in an inductor? What happens to the voltage drop across the inductor? Well, again, we're going to have a current that goes like this: I equals I naught cosine of omega t. But what about that voltage drop? How do we deal with that? We go back to Kirchoff's laws, and Kirchoff's law says the voltage drop around a circuit has to add up to zero. So whatever voltage I have here in my source, I need to subtract the voltage drop across the inductor, and that has to equal zero. So let's just write that out. V minus V sub L is equal to zero. But I know that a voltage drop across an inductor depends on a changing current. If it's a steady current, there is no voltage drop across the inductor. The inductor only responds to changes in current. And that proportionality constant is what we called the inductance, L, of the device. And now we have to include the changing current. Delta I over delta t. If I make a big delta I, I get a big voltage drop across there. If I do that in a short amount of time, a small delta t, I get a big voltage drop. This is the equation that describes Kirchhoff's laws for this thing right here. Alright. Alright, so we just wrote down this equation right here. V minus L delta I over delta t is equal to zero. We can now solve this equation for delta I over delta t. And that's not too bad, right? What do we get? We move this over to the other side, and then we can just divide by L. So what's the changing current? It is that. It's V over L. The battery voltage, divided by the inductor. Now this is going to get a little bit confusing, what I'm gonna say next, but try to stick with me. If I think about the voltage drop across the resistor, Can we plot out what that looks like? Yes, we said it looks like a cosine. Okay, no problem. But how does that relate to the driving current as a function of time? The current also went like a cosine, so it looked like this. Pink one is V R, green one or yellow one is I. The resistor has a voltage drop that is in phase with the current. But when we deal with inductors, suddenly things change. If I think about V sub L, the voltage drop across the inductor as a function of time, And that current I, What does it look like? Current is still going like a cosine. I didn't draw that very good, let's try it again. I want to see if I can match it up with the one above it. This is our time, let's see if we can get it to line up a little better. There's the current I, goes like a cosine. But the voltage for the inductor doesn't follow the cosine anymore. It, in fact, goes like a sine or a minus sine. And to be consistent with your book, we want to make sure that we get the right one. And so let's look at it, it looks like it's a minus sine. So it, in fact, looks like this. And this means that the voltage drop across L is 90 degrees out of phase with the current. And if you look at your book on figure 2140 they have this curve, it's maybe a little easier to read in the book then I showed you here. But the idea is that when the current is changing rapidly, that's when you get a big voltage drop across L. So at the top here and the bottom, it's not changing very rapidly. It's only on these steep slopes where the voltage is change- where the current's changing rapidly and hence generates a big voltage. Okay. So, if you don't like my horrible picture here that's totally fine, there's better ones in the book.