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Inductors in AC Circuits

Patrick Ford
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Hey, guys, in this video, we're gonna talk about conductors and the role that they play in a sea circuits. All right, let's get to it. Now, remember that the current in an A C circuit at any time is gonna be given by the equation that by now we've seen a bunch of times I max co sign of omega teeth. Okay, this is going to tell us that the current is simply oscillating with a new angular frequency of omega between some value positive I, max, and some value negative imax. Now, the question is, how does the voltage across the induct er look Well, remember that the voltage across an induct er, which we saw during our discussion of Faraday's law, is the induct INTs times, Delta I over delta T, which is the rate at which the current is changing. Okay, now I can't show you how, but using calculus, you guys can arrive at the answer that the rate at which the current is changing looks like this. So if I multiply this by the induct INTs, then I get the voltage across and induct er in in a see circuit at any time. This is gonna be I Max times, Omega times L a times cosine of omega t plus pi over two. Okay, so once again, remember the voltage across a resistor the voltage across the resistor Looks like I'm max. Times are times cosine of omega T. So the angle that it operates at omega T is different than the angle that the voltage across the induct er operates at. This is some other angle fada prime, which is Omega t plus pi over two. Okay, so the current and the voltage across resists are in phase right there. Plots, lineup. But the current and the voltage across and induct er are not gonna line up. They're gonna be out of phase if we plot the current across an induct er and the voltage across and induct er you concede e that the voltage across an induct er actually leads the current by 90 degrees. Okay, what's happening here is that the voltage is deciding to go up at a time when the current is zero. Okay, then at a future time, the current starts to go up. But at this point, the voltage has already peaked. Then, at a future time, the current peaks. It's trying to match what the voltage is doing, but at this point the voltage is already decreasing. So then, at a future time, the current decreases. But the voltage has already bottomed out. So you see, the current is trying and trying and trying to match the voltage. But it's lagging behind. Or we can say that the voltage leads the current. Either one is fine. Okay, now the maximum voltage across an induct er is gonna look like I Max times omega l This looks a lot like OEMs law where we said that the voltage across the resistor was I times the resistance. There appears to be a resistance like quantity off omega l that resistance like quantity for capacitors. We called the capacitive reactant. Now we're calling it the inductive react INTs for induct er's. The units are still OEMs right, same unit. As for resistance. All right, let's do an example on a C Power source delivers a maximum voltage of 120 volts at 60 hertz. If an unknown induct er is connected to the source and the maximum current in the circuit is found to be five amps What is the induct INTs of the induct? Er Okay. What is the maximum current in an induct er circuit? This is just going to be the maximum voltage across the induct er divided by the inductive capacitance. Okay. And now we know that because the induct er is connected on its own to the power source that they both have to share the maximum voltage. That's what Kyrgios Lupul says. So this is just gonna be V Max the maximum voltage by the source divided by excel. And what is that? Inductive capacitance? Well, that is just gonna be Omega. L Okay, so plugging this into our equation, we could say that I Max is simply v Max over omega L in our unknown is thean duck tints. So what I want to do is multiply the induct, INTs up and divide. I'm Max over, and I have induct INTs Is V Max over omega imax. Before I can continue, though, we need to know what Omega is. We're told the linear frequency is 60 hertz. Okay, remember, this is linear frequency because the units are hurts. So the angular frequency, which is two pi f is gonna be two pi times hertz, which is gonna be about 377 in verse seconds. Now we can solve for the induct INTs and the induct. Ince's just going to be Z Max over omega Imax. The max is 120 right? Always look at it. Make sure it's a maximum voltage, not an RMS voltage divided by 377 which was our angular frequency. The maximum current in the circuit was five amps, so that's five and this whole thing is 0.64 Henry's the unit of induct INTs. Alright, guys, that wraps up our discussion on induct er's in a sea circuits. Thanks for watching.