Okay, welcome to advanced circuit analysis, part three. And this will be the last part of this section and also the last part of this course. So I hope you enjoyed it. If you have any questions or comments, you can get ahold of me or communicate with me through this platform. And if you have any questions, feel free to reach out and I will help you. So with that said, We're right now we're talking about Naughton feminine conversions.

And basically, here's my circuit. All right, and if I feminize it, I get this here and we know how we we know How to do that. And then we have our Norton equivalent here. I know what an equivalent circuit there. So if we want to go from like Norton to seven, and seven into Norton, we'll notice on both of them here that are th equals Rn, and Rn equals r th. So if I find, if I find my note and equivalent resistance, and I want to convert to feminine, it's the same value.

All right, it's the same value, that configuration in the network is going to be different. Okay, but it's the same value. Now, if I'm in Norton, or if I find my Norton equivalent, and I want to find Vth, all I do is multiply i n times r n, and I get Vth here. And if I have my Norton equivalent I'm sorry if I have my feminine equivalent and I want to go to Norton, all I do is divide Vth by R th, and I get my values now. As I stated previously, Devin is a the feminine voltage source in series with my feminine equivalent resistor. All right.

And in this in Norton, okay, I have a current source which is i n the value of i n. And then my Norton equivalent resistance is in parallel with that current source. All right, and on both circuits A and B is where I will place my load resistor. And I've been calling it my resistance of interest and I'll just expand upon that a little bit. Okay? It doesn't have to be a passive resistor. Okay?

That can be a black, we call that a black box. All right, that can be a black box. And from ohms law from ohms law, we know the voltage across that is going to be the current through times the resistance. And from previous courses that I've given or or maybe you've taken elsewhere, you know that if I know two of those parameters, I can find the difference. So even though feminine, and Norton, we say, okay, we're gonna put up my arm We don't really go through this to solve this, this resistor we could, and it's done. I'm not saying it's not, but usually that RL is a load of some sort.

Maybe it's an amplifier. Okay, maybe we're driving an amplifier, okay. Or maybe some type of analog to digital circuit, or some type of electronic circuit, a light bulb, a bank of lights, many, many things that can be attached to a and b and we want to know what's going to happen there. Okay, so even though I say RL, we're using a very simple term, it can be usually it's a black box of some sorts. We take that circuit and we convert it to a black box. We'll see as we go on when we get more into semiconductor theory transistors in amplifiers and that type of thing.

You'll see what I mean. And I'll bring that up again. And if it's a little fuzzy now, it's okay. Once we go along, and and your understanding increases, because every time you take a course, your understanding of that of that material gets a tad bit better. All right, you may not realize it then. But you're getting a tad bit better.

Now, I've been doing this for about 35 years or so, show my age, right? And I look at some of the things that I teach now. And when I went many, many years ago, when I went through my training, or I went through my courses, and I went to a university in Boston, Northeastern. It was hard. Okay, I could not some of that I could not understand but all of us Sudden, as my experienced increased my knowledge increased ice I finally got an understanding of these and that will happen to you too. Okay, you guys are not any different than I am.

It'll happen to you. Okay enough with the lecture lecture. We got up one more Rotherham to go over and we're going to end this. Alright, this is the last network theorem that we're going to look at. And again, this will end this portion and this course we have Okay, we have resistor networks here. And this one here is obviously it looks like a why and this one here looks like a Delta but look at the A, B and C. Okay?

I did point those out. All right. So what that means is if and let's, let's just talk about here, first of all, if I've got a Y, okay, some resistors that are in a y configuration or in a very complex circuit. Maybe if I convert that to a delta, I can understand the resistance or the voltage across these two points a little bit easier. Okay, so, we have a method where in this example, we can go from a y to a delta, and it's just a formula right here. So for instance, r one times r two plus r two times r three plus r three times r one, all over r one will equal Ra.

Same thing With our B over here, we do the same thing, but we divide our two by that, and we get our B. And then if you look at the last one, our C, we divide our c by r three, and we get the value of RC. So what we're doing here is we have so we have a mathematical formula, which allows us to find the equivalent ra in a Delta circuit between points, A and B, A and B. All right. Now, the point I want to make here and let me let me clear the sleep screen Is that this this formula really isn't as complex as it looks out to be. If you look, okay, well multiplying here, times this r one times r two, and then r two times r three, and then i three times r1.

And we do that for each resistance in my wide network in the numerator, a one, r two r three, and then we do the math. And we find the value of Ra RB RC, Ra rb, C, Ra RB RC. Alright, so So that's it. So what we're doing here is we're going from a y network to a delta network. And the reason we're doing it is we want to know maybe what the resistance is from point A and B. Or maybe the voltage between points A and B.

So when I find these by go from like a to b, and I do my math and I find my equivalent resistances everything should be the same. Now if you look, right, isn't it easier to kind of digest this than it is this? Yeah. Okay. So we go from y to delta, and on the bottom here, we're going the opposite way. We're going from delta to y and everything I set out on the top is exactly the same in the bottom only.

My how I find my equivalent resistances in my Y is different. But if you look this is the second My denominators the same on each one. When I'm looking for r one, r two r three, r one, r two, r three. So here, here and here, alright. But look at what we're doing. All right?

Our B times RC, RC times RA, and R A times RB. So I'm going around around multiplying the two adjacent resistances are the values of the two adjacent resistances to get my sum, or I'm sorry, get my product because these would be a product up here. All right. Okay. Let's go to the next slide. All right.

Now we're, what I've done here is I Va given you an illustration of delta to y, and if you look over here and i delta, okay, Ra is four ohms, our C is 10 ohms. rb is six ohms. And I just plug into the equation, and I do my math and I've got r1 is three ohms right over here. And same deal, r two is two ohms right there, and our three is 1.2 ohms. All right. So what I've done is, is superimpose them.

In other words, I've superimposed my y into my delta as you can see here, alright, and the points A, B, and C, a common And if you look, okay, there's a, there's my er a BNC. Alright, this, okay? And I define them over here in this. And then I put my calculated values for r one, two and three here. So let's, let's look at our C. Okay? And from C to A.

We've got what we've got 10 ohms, right. All right, so what do we have? Okay, all right, let me regress here, okay, if I put a no meter on a to see, okay, what would be what be what would be my resistance? Well, I using my delta now, I've got RC and then I've got two other electron paths. Here and here, don't I? Okay, but if I'm just if I use my why, and I put my own meter on point A to see I've only got one path, and that path is this way.

All right? Because this point here, which is B. All right, is open, so I'm not going to get any current flow. So what would be my resistance from A to C? Using my y? Well, r1 is three ohms.

Alright, and r two is two ohms. So if we add them up, all right, from let's say from from A to C in my y. Okay, that's going to equal one. Two plus three is five ohms. All right. Now, if I've got my delta and I've done my calculations properly, and to prove this theorem, that's also going to be five balls, but look at what I got here.

All right, I'm gonna redraw the Delta. Okay, I've got RC Okay, I got RC, which is 10 ohms. I'm not gonna, not gonna write down the ohms. But I gotta say that's RC, and then I've got RA and RB. But they're in series over here. I'm thinking So I've got Ra.

And I've got our D, I, and ra is four, and RB is six. So really, what do I have? I've got 210 ohm resistors in parallel. And if you remember from my circuit theory class, what does that equal to? That equals? Five ohms.

So the theorems work. Right? That's what I wanted to show you here. The theorems work. All right. So from from Why?

From A to C, I get five ohms. And from Delta from points A to C, I get five ohms. All right. Okay, so, right here, we've gone to delta to y and on the next slide, we're Going to go from y to delta. Okay, now we're going to go from y to delta and we want to find Ra. I mean, I've got the formulas up there.

You can you can follow it. We've done so much math here that if at this point, I think you should be able to kind of follow what I'm doing here. So we've got the values of Ra, rb and RC. So, again, I've superimpose them one into another. And if I want the resistances from A to B, and let's say I'm using my Y, what do I have? I've got r1 which is three ohms.

I've got our three which is 1.2 ohms. If I've got an ohm meter, and I put my own meter right across there like this, what am I going to read? I'm going to read 4.2 ohms. That's what I'm going to read. Because now I'm using I'm using my Y, okay? I'm using the Y, because why?

Because C's open, and there's no current flow. All right? So we're saying, ideally, I should get that same resistance of value of resistance with the Delta. Alright, so let me clear the slide off the close. So I've just cleared the slide. And now I'm going to redraw the Delta.

All right, so my delta would be there's my points A, and my points C. So we've got RB which is what, which is six ohms. I'm just going to do this six. And then we've got Here's B, let me put B. So then we've got our, our ABC here, we got Ra. And we've got our C. So we've got RA, and we've got our C. And what do we have? We've got four ohms.

And we've got 10 ohms. So basically what I have is two resistors in parallel, this one six, and this one's 14. Okay, it is a, it is B. So what do I do? Well, I'm going to have to see what the value is. And if I've got six and six in parallel with four genomes and if I do the calculations, and again, you can do that I get 4.2 ohms, just the way I got over here.

All right, 4.2 ohms. So the theorems work. If you if you want, you can you can go through the derivation of of six, and 14 ohms in parallel. All right, you can do the math or I think in a couple of previous courses, I gave you a couple of websites where you can just plug in the value and get the equivalent, so it's good and they work. And that'll pretty much wrap up this, this section in this course at Advanced circuit analysis. So we'll see you in the next course, which is going to B, AC theory, we're going to talk about AC current AC voltage.

And then we're going to introduce resistance, capacitance and inductance. So that's coming up next. I hope you enjoyed it. As always, if you have any questions, please contact me through this platform. And I will get back to you. And keep on plugging guys and gals.

You'll get it. You'll get it, you'll get it. Everybody does. You just gotta stay with it. Take care. Bye bye