Practice: Will a frequency f = 60 Hz or ω = 75 s ^{−1} produce a larger max current in an inductor connected to an AC source?

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Concept #1: Inductors in AC Circuits

**Transcript**

Hey guys, in this video we're going to talk about inductors and the role that they play in AC circuits. Alright, let's get to it. Now remember that the current in an AC circuit at any time is going to be given by the equation that by now we've seen a bunch of times, I max cosine of omega T. This is going to tell us that the current is simply oscillating with an angular frequency of omega between some value positive I max and some value negative I max. Now the question is how does the voltage across the inductor look? Well remember that the voltage across an inductor which we saw during our discussion of Faraday's law is the inductance times delta I over delta T which is the rate at which the current is changing. 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 inductance then I get the voltage across an inductor in an AC circuit at any time. This is going to be I max times omega times L times cosine of omega T plus Pi over 2. So once again remember the voltage across a resistor. The voltage across a resistor looks like I max times R times cosine of omega T.

So the angle that it operates at, omega T, is different than the angle that the voltage across the inductor operates at this is some other angle theta prime which is omega T plus Pi over 2. So the current and the voltage across resistor in phase, their plots line up but the current and the voltage across an inductor are not going to line up they're going to be out of phase. If we plot the current across an inductor and the voltage across an inductor you can see that the voltage across an inductor actually leads the current by 90 degrees. What's happening here is that the voltage is deciding to go up at a time when the current is zero. 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 trying to match the voltage but it's lagging behind or we can say that the voltage leads the current, either one is fine. Now the maximum voltage across an inductor is going to look like I max times omega L. This looks a lot like ohm's law where we said that the voltage across the resistor was I times the resistance. There appears to be a resistance-like quantity of omega L. That resistance-like quantity for capacitors we call the capacitive reactance. Now we're calling it the inductive reactance for inductors. The units are still ohms, same unit as for resistance.

Alright, let's do an example. An AC power source delivers a maximum voltage of 120 volts at 60 Hertz. If an unknown inductor is connected to the source and the maximum current in the circuit is found to be 5 amps, what is the inductance of the inductor? What is the maximum current in an inductor circuit? This is just going to be the maximum voltage across the inductor divided by the inductive capacitance and now we know that because the inductor is connected on its own to the power source that they both have to share the maximum voltage that's what Kirchoff's loop rule says. So this is just going to be V max, the maximum voltage by the source divided by X and what is that inductive capacitance? Well that is just going to be omega L. So plugging this into our equation we can say that I max is simply V max over omega L and our unknown is the inductance. So what I want to do is multiply the inductance up and divide I max over and then I have inductance is V max over omega I max. Before I can continue though we need to know what omega is. We were told the linear frequency is 60 Hertz, remember this is linear frequency because the units are Hertz. So the angular frequency which is 2 Pi F is going to be 2 Pi times 60 Hertz which is going to be about 377 inverse seconds. Not we can solve for the inductance and the inductance is just going to be V max over omega I max. V max is, always look out make sure that's the maximum voltage not an RMS voltage, divided by 377 which was our angular frequency, the maximum current in the circuit was 5 amps so that's 5 and this whole thing is 0.064 Henry's the unit of inductance. Alright guys, that wraps up our discussion on inductors in AC circuits. Thanks for watching.

Example #1: Inductors and Graphs

**Transcript**

Hey guys, let's do a quick example. The voltage across and the current through an inductor connected to an AC source are shown in the following graph. Given the information the graph, answer the following questions. Part A, what is the peak voltage of the AC source? B, what is the frequency of the AC source? And C, what is the inductive reactance of the circuit? So part A, what is the peak voltage of the AC source? This is the voltage across the inductor but we know that the voltage across the inductor is going to exactly match the voltage of the AC source in terms of the maximum voltage because they are connected together and that's just what Kirchoff's loop rule says. So whatever the maximum voltage of the AC source that has to be the maximum voltage of the inductor. So 10 volts is clearly the maximum voltage of this inductor so V max, the maximum voltage of the source is just 10 volts. Part A, done.

Part B, we want to figure out what the angular frequency is? What this graph tells us about is it tells us about time. From the time we can find the period and from the period we can find the frequency. That's always how you want to approach these problems with time as we've talked about before. Any instance of time tells you about the period and the period can tell you about the frequency. From here, sorry, from here to 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, sorry, the period is 0.1 seconds and the frequency which is just one over the period is 1 over 0.1 seconds which is 10 Hertz. Very very straightforward, very simple.

Part C, what is the inductive reactance in the circuit? Now in order to do this we need to figure out how to relate information on the graph to the inductive reactance. The only piece of information we haven't used yet is the maximum current. We know the maximum current is 2.5 amps, that's the last piece of information we have used and we know that this is going to be equal to the maximum voltage across the inductor divided by the inductive capacitance. Once again that maximum voltage across the inductor is equal to the maximum voltage output by the AC source divided by the inductive reactance. So if we want to solve for the inductive reactance all we have to do is multiply this up and divide that down. Pretty straightforward, the inductive reactance is going to be V max divided by 2 and a half amps which is going to be 10 volts divided by 2 and a half amps which is going to be 4 ohms. Also very straightforward application of the formulas that we viewed so far. Alright guys, that wraps up this example. Thanks for watching.

Practice: Will a frequency f = 60 Hz or ω = 75 s ^{−1} produce a larger max current in an inductor connected to an AC source?

0 of 3 completed

Concept #1: Inductors in AC Circuits

Example #1: Inductors and Graphs

Practice #1: Current in Inductor AC Circuits at Different Fr...

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