31. Alternating Current
Inductors in AC Circuits
Inductors in AC Circuits
Was this helpful?
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.
Inductors and Graphs
Was this helpful?
Hey, guys, let's do a quick example. The voltage across and the current through and induct er connected to an A C source are shown in the following graph. Given the information in the graph answering the answer the following questions part A What is the peak voltage of the A C source B. What is the frequency of the A C source and see what is the inductive reactant since of the circuit. Okay, so part A what is the peak voltage of the A. C source? This is the voltage across the induct er, but we know that the voltage across the doctor is going to exactly match the voltage of the A C source in terms of the maximum voltage because they are connected together. And that's just what Kirchoff Lupul says. So whatever the maximum voltage of the sorry of the A c source, that has to be the maximum voltage of the induct er. So 10 volts is clear. The maximum voltage of this induct er so V Max, the maximum voltage of the source is just 10 volts, part a done. Okay, Part B. We want to figure out what the angular frequency is What this graph tells us about is that tells us about the time from the time we can find the period and from the period we confined the frequency. That's always how you want to approach these problems with time as we've talked about four in the instance of time tells you about the period and the period could tell you about the frequency. Okay. From here. Sorry. From here. Thio here, which is the time that we're given is a full cycle. So we know that this time is a full period. So the time is zero point. Sorry. The period of 0.1 seconds. And the frequency, which is just one over the period, is 1/1 seconds, which is 10 hurts. Okay, Very, very straightforward. Very simple. Part C. What is the inductive reactant in the circuit. Okay. Now, in order to do this, we need to figure out how to relate information on the graph. Thio. Sorry to the inductive react INTs. The only piece of information we haven't used yet is the maximum current. We know that the maximum current is 2.5 amps. That's the last piece of information we haven't used. And we know that this is going to be equal to the maximum voltage across the induct er divided by the inductive capacity ins. Once again that maximum voltage across the induct er is equal to the maximum voltage output by the a c source divided by the inductive react. It's okay, So if we want to solve for the inductive reactant So all we have to do is multiply this up and divide that down Pretty straightforward. The inductive reactant is gonna be V Max, divided by 2.5 amps, which is gonna be 10 volts divided by 2.5 amps, which is going to be for OEMs. Okay. Also very straightforward application off the formulas that we've used so far. Alright, guys, that wraps up this example. Thanks for watching
Will a frequency f = 60 Hz or ω = 75 s −1 produce a larger max current in an inductor connected to an AC source?
f = 60 Hz
ω = 75 s-1
Additional resources for Inductors in AC Circuits