Okay, we've we've talked about this previously, but just a review. Notice I'm going to stress, pure inductive circuit. So in a pure inductive circuit, I lags el by 90 or I can say el leads i L by nine. He was saying the same thing we're playing with. We're playing with words. But But the point is, if you look at E L, and I l right at that point, you'll see that i L is as far as the sine wave or degrees and a sine wave.

It's at 90 degrees where i L is at zero. So we say he leads i L by 90 or I legs by 90 in a pure inductive service. And here's my graphical representation if I'm using some type of vector analysis, all right, where el I'm going clockwise this way. So el leads i L by 90 degrees right there is that's my 90 degree angle. All right. So with that said, let's clear off the slide and go to the next one.

Okay, on this one here, we're doing a pure. Well, we talked about the pure inductive circuit right there. That was on the previous slide. But what happens if I put an R in series with my X sub l? Well, if you notice before, I said a pure inductive circuit. Okay.

I lags L by 90 degrees. Well, if that's the case, it's not clear anymore, is it? So maybe, I L is going to lag el by something others Than 90 degrees. And here we go. inductive resistive circuit. phase angle is determined by the tangent of x L divided by R. All right.

And we'll do that in a minute. Also, I can find the angle even though I didn't say it right here with V to Z. All right. Now I don't really think that I've talked too much about z, okay, z stands for impedance. Okay. So z stands for impedance and impedance is the combination of our plus x sub l or X sub c. All right now, we already looked at SMC that's capacitive reactance.

But z is the combination of R plus x sub L or R plus x sub c, or r plus x sub l plus x sub c. And we'll see those and then in the next, the next course, the next module and when I do our x sub c and XML circuits Alright, but right here in this module, for this discussion, Z is r plus x about All right, now since R is resistive, and we know that in a resistive circuit, voltage, and current are in phase meaning What we're showing here the sine wave where I have my, my peaks, and right here, hit at the same time. When they're zero, they hit at the same time, zero here, they hit at the same time, my negative peaks are in the same time. So, the current and the voltage waveforms are in step with one another. All right?

However, if you look at vl Alright, and I, we're out of phase here, right? All right, we know that so it's not in step so they don't add directly. All right. We have to use this formula here. And if you took my math electronics that is Protagoras theorem with a right triangle and I explain the vectors in that, so forth. All right, but for this purpose here, okay, we cannot add are an XML directly.

All right, we won't get the correct answer. And because of the phase relationship in the circuit, we'll have to use this formula here, which happens to be Pythagorean Theorem. All right, and we're going to find z. So what do we got there? So each is 100. So we get 100 squared.

So we're looking for 100 squared plus 100 squared. We want to take the square root of that after I do the math. I get 141 ohms. Okay. And if you look at vt, what we do is we gave you 100, or 100 volts across R, and 100 core volts across the coil. All right, so now i square my hundred volts 100 volts squared plus 100 volts equals 141 volts.

Now, let me just stop here for one instance and say I didn't really show you all the steps in these two math problems right here. Well, that's because I've done a lot of them in the previous modules. All right, and I've also done a module on math basics or math for electronics. So I've kind of reviewed that I don't want to convolute these, these modules with with a lot of math at this point. I mean, I've gone over them. I've done them.

So If you need a little bit of refreshing, refreshing refresh on this refreshment, that whatever our review, I should say, then please look at math electronics, or look at some of the earlier modules, I really, really go into detail with them. And all you have to do is substitute these numbers in there and you'll get the answer. All right, or if if you can just give me a call and we'll, we'll try to fix you up via email or do a little love phone conference of something. All right, so anyways, with that said, here we go. So now all I do is plug let me stop here and clear the slide. So now all I do is use the inverse can function and I'm going to bring the calculator down and show that so with V to Z is one amp and with x l over x r that's also one.

Because look at excel in R. So, okay, so I x sub l over R is 100 over 100. Because X sub l equals 100 and r equals 100. I do the math, I get one. All right. Again, I t v t over z, that's going to be one also because we've got 141 volts divided by 141 ohms. That's one amp.

And now I'm going to take the calculator down and show you how to find the inverse can function to find the angle. Okay? All right, so let's do that. What? Okay, I bring my calculator down. It's the calculator.

Cane comes with the Windows operating system. And if you notice, there's a little up arrow here where I can change the functions. And what I'm looking for is it's the tan of theta. Okay, but I'm looking for the Actually, I'm looking for the inverse tan function, or its tan over one. And this says what angle? That gives me a ratio of one.

What's the degree? What is the phase angle, or the degree, the angle, the degree angle of that. And in this case, we happen to know it's 45. But this is how we get it on the calculator. So I go up, and I find what the tan function is. All right.

So we can we can do it the long way. We know that x sub l over R is 100 over 100. All right, so 100, which is XML divided by 100 ohms, which is r equals one. The inverse can function 45 degrees. Okay, we can also use vt over z it and we've got vt is 141 all volts, and z is 141 ohms. So we just plug in 141 divided by 141.

And that should come out. Let's see, what did I do wrong? Okay, I see. Hang on I made, I made a key mistake. Let's do it again. 141 divided by 141 equals one.

Now I hit the tab function. It's 45 degrees. So what we're saying here is the circuit current here, let me get rid of the calculator. The circuit current lags EA by 45 degrees right there. Alright, so the circuit current lags EA by 45 degrees. And we can look at the graphical representation here.

All right, we're here is EA. And here is my circuit current. So if you look, my circuit current is at zero. That's about 45 degrees right there. All right, we're, we're good. One thing here.

Let's not get mixed up. Bye. I call this vt. Alright vt and this one vt and VA. What are they use here? ea is the same. All right. So that that's good.

All right. So with that said, let's go on to the next slide. And you can see how the phase relationship between current and voltage will vary between the value of x sub L and R in a series circuit. Okay, I just put this problem up here. And I've done them for you just to go through another one here. So we've got an R and an allen series and we got a voltage source AC voltage source and we want to Z i voltage across our voltage across L and the phase angle.

So we first thing we do is we find z, I specify that this was Pythagorean Theorem. I go through the math and I get a Z of 50 ohms. Next thing I have to do is I have to find I, okay, and again, I in a series circuit stays the same. It's just that this is an AC current. Alright, so now I do my voltage divided by z. All right, my voltage is 100 volts AC.

We found z In this step, it's 50 ohms. I do my math and I come up with two amps. The voltage across the resistor right here VR, okay, which is i times R. It's two amps times 30 ohms. 60 volts AC vl voltage across the inductor. All right, I times X sub l, two amps times 40 volts. I do the math and it's 80 volts AC.

Now, I'm going to find the phase angle of the whole circuit. That's what we're looking for the tan of data, the tangent of theta. this is theta 40 over 30. All right, 40 over 30 X sub l over are 40 over 30. I do the math, it comes out to 1.3. I go back like I showed you before, I put 1.3 in my calculator for the inverse hand function.

And I come out with 50 to 53 degrees. Therefore I lags V in the whole circuit by 53 degrees. Now I can do a check right here. All right, VA, which is the supply voltage will equal 60 squared plus a t squared, which we found here in these two steps. I do my my mathematical calculation. And I'm looking for the square root of 10,000.

I go through the math, and it's 100 volts. And that's what I started out with. So it checks and again, I'll again along with my, my courses, I put some worksheets in problems for you to do with the answers. And if you look there, bear in there, alright, so I suggest you do them. Alright, we're going to stop here and go to the next slide. All right.

On this slide here, I just want to make a point in discuss something with you. If you look at that, that's that's what we solved in the previous slide. But really I want to look at at at this chart in the in the vectors that I've vector vectors that I put up here. All right, and this is my xo Bell axis, this is my yard down here. xo Bell, whoops, our extra bell in our Alright. So what am I trying to prove here?

Well, if you if you remember when we we started this section, I made a statement. I said What did I tell you? I said voltage leads. Current or AI lags voltage in a pure inductive circuit by 90 degrees. And notice I said I emphasize pure. So let's go down here and look.

Alright, let's look at this one here. Okay, R is one ohm. And XML is can XML is 10 times greater than R. So my face angle is going to be closer to 90. All right, right there. And I get at 4.3 degree phase angle. All right?

If we're equal, where this one here, where R is 10 in XML is 10. My face angle Is 45 basically right smack in the middle. All right, because a pure inductive circuit is has a 90 degree phase angle. A pure resistive circuit has a zero degree ache phase angle, meaning they're both in phase. So if I've got R and X sub l that are equal, my phase angle is 45 degrees, which is half of 90. Last one, where we reversed where r is 10 and X sub l is 10.

So now R is 10 times greater than x sub l. So that will be close close to zero. And if you can see it now clear the slide off. It's 5.7 degrees. So the point I'm trying to make is If I look actually let me let me stop here and clear the clear the slide off so you can see it better. Okay? If you look at at both these z values, okay?

Because z is impedance, and impedance is the phasor addition of X sub L and R in this. In this case, they're both attend. All right, and they're both going to have the same opposition to AC current flow. However, what changes the phase angle? If it's more resistive, we get closer to zero degrees. And if, if it's more inductive, we get closer to 90 degrees.

And that's and that's the point I'm trying to make with these vector analysis where we start Hear, go here and then go closer to zero. Okay, where this is 90 and zero, 90 and zero, 90 degrees and zero degrees. Okay. So I just wanted to bring that point up. And it's good to know, especially when we get into complex impedances, where we have an X sub l and an X sub c in the circuit. All right, if we need, this is how we adjust to know what the different phase angles are.

All right, so with that said, Just Just hold on to that, be aware of it. And we're going to stop and go on to the last slide in this series here. Okay, and this circuit here we've got an inductor And a resistor in parallel. All right, here's a circuit right there. And we have a voltage supply of 100 volts, R is 10 ohms. x of L is 10 ohms.

We've got 10 amps of current flowing through each leg. So the phase angle here to find the phase angle, it's I sub l over R. Okay, we're looking for the tan of that. But we can't really get that. Well, we can, I guess, with that would given us there we could, but let's go through some of the other rough. The other values here that we can find we certainly can find it. It is.

Again, it's we have to use Pythagorean Theorem, because the current involved voltage urata phase in a inductive circuit, so we just can't add them. directly, that 10 out there, 10 amps apiece. So we need to use Pythagorean theorem and go through that. So it, it would be 10 squared plus 10 squared, the square root off. And when I do my math, I get 14 dot one, four amps. All right, if I want to find the Z of the circuit, which is the impedance, which we introduced a couple of slides ago, z t, or total impedance would be E, which is what you've given here, divided by i t, which we found over here.

Again, we just do our math and the total impedance of that circuit or the total Z of that circuit is 7.07 ohms. Okay, now I want now I mean, I want to find the phase angle and the phase angle is i L over IR. Okay. We happen to know that that's 10 over 10. And that's one. And we know by using the inverse tan function would try showed you in the last slide.

If I use that inverse tan function, I come up with 45 degrees. All right, so in a parallel circuit, the phase angle is between the line current and the generator voltage. So if I look up here, here's the waveform for my supply voltage. And here is my eye T. If you look, they're out of phase by 45 degrees, and the voltage el leads the supply current by 45 degrees right there. All right. Now, what we're showing here is a vector diagram where E and R are at the zero axis lines.

And they they lead they lead the current and i L is down here. And if it's if it's a pure inductive circuit, then this will lead i L by 90 degrees. But we know that we've got a resistor in there. So now, this would represent I TI L is down here it is here, and IR which is the current flow through the resistor are in phase. That's what that means. All right.

Okay, with that said, this is the last slide in this this course. And I hope you enjoyed Write it. And I certainly hope that you've learned something. Again, if some of them some of these principles, or some of the things that I've done, are a little bit fuzzy. The good thing about this is you can start from the beginning and go over it again. And I'm sure that you'll get it.

You may have to go through this two or three times to really grasp it. I mean, I did when I was learning I, we weren't lucky. We didn't have a video system. Basically, we had to study I mean, I would, I would read the book two or three times and do the problems two or three times so I so I would get them and that's what I suggest you do. Alright, with that said, we'll see you in the next course. This is Al take care now.