Okay, and welcome to part two of this course advanced circuit analysis. And what we're going to be covering here is superposition, seven and theorem, Norton's theorem, how to convert from seven into Norton or seven and Norton conversion, and t and delta networks. So, without any further ado, let's go on to the next slide. All right, now, the first one we're going to look at here is superposition. And what we're looking for is V out in this circuit here. So what we do here with super positioning is we remove one voltage source and connect that to ground So let's, let's look here.

So if you look at the original circuit, I've got two voltage sources, one here, one here, all right? v one is 240 volts, and v2 is a minus 90 volts DC. So what I'm going to do is I'm going to shout out one at a time. And not you can't really do this physically, you can't shot the battery out of the voltage source, you'll have problems. So this theoretically, so as we say, here, remove one voltage source and connect to ground. So ideally, if you want to do this in a lab setting, you want to remove the voltage source and then go to ground don't shut the voltage source out.

You'll have a problem. Okay, so what I'm doing over here is I am going to remove one of the voltage sources up here and that happens to be 90 volts, DC, and I'm going to go to ground. So now I'm going to find the voltage at V out, which I call V out a. So that would be 240 over 60 because there's my 60 K, and the case cancel out here. That's why I don't show it on the show it on the formula here, both case cancel out. So 60 over the sum.

So 60 plus 30 is 90. So 240 over 60 over 90. That's actually two thirds of 240. My answer is 160. All right. So now the next thing I do is okay, let me stop and clear the slide.

Okay, the next thing I do is I shot out this one meaning I remove it and shout it out. So now I have this circuit here. And again, the Wiis, I'm sorry, the case cancel out. And I have 30 over 90, because I'm going referencing to groan here. So I'm actually taking the voltage drop. If you look at that, I'm actually taking the voltage drop across this 30 K resistor.

So it's 30 over 90. All right, this should be a minus 90 volts here because of this hike I brought up here, but I didn't show it there. So that should actually be a 9999. It's 90 volts DC and when I do the math, one third of 9030 over 90 is 30 volts minus 30 volts. So now all I do is I add these up 160 plus a minus 130. And so my Samad V out when I bring this circuit in, I normalize the circuit, okay V out is going to be 130 volts DC.

So basically if you if you look at this again what we're doing is, in this example, we had to supply voltages one here and one there. Okay, we saw the voltages or the voltage at the point of interest in this example the point of interest would be V out. Okay. And then we took two voltages and we add them algebraically and we can Got our, our rough solution right there. Now I kind of like this one of all of them, I kind of like this one the best. Because to me, it's you may have to do a couple of more steps.

But it's simpler. Right? It's a simple method as far as math and simultaneous equations and that type of thing that we were doing in the in the previous section here. So I kind of like this one a little bit better for that fact. All right. So take a look at it again, and see if you can digest it.

All right, going on. All right. The next theorem that we're going to talk about is feminine stem. And what we want to do here and feminine stem is, is we want to simplify a circuit into a theremin. In equivalent voltage and a seven equivalent resistance, and then what we will do is put our, in this case, our l across that and solve for current and voltage through RL. Alright, so here's my original circuit here.

And I have RL which is two ohms. I also said, Okay, I want to find the voltage across VA B. All right, so notice that RL is a rheostat, meaning it's variable. So I want to you know, I don't want to keep, you know, finding the parallel equivalent and I'm just looking for something simple to find that so I can use 70 Plus it's a good way to illustrate it here. Alright, so the first thing I want to do is okay, I want to move remove RL, which I'm showing you here. And I want to find Vth or Vth is a tenant of equivalent voltage.

Okay, that's what I want to do. So, that's my first step. My second step on the next slide is I'm going to find rth which is the Deven in and I'm going to abbreviate that resistance. All right, so that's what I want to do. Okay, two step process. I want to find Vth, the feminine equivalent voltage.

And then I want to find what we call rth. The feminine equivalent resistance. All right. So let's do that. We're going to clear the slide. Oh, here's what we do, we moved our L. And basically, all I want to do is find the voltage drop, where the load was, which I call VA B right here.

So I want to find the voltage drop. So how do I do that? Well, basically, I've got two series resistors and I got a voltage source v one, which is 36 volts. Isn't that a voltage divider? All right, so basically, I've got a total resistance of nine ohms because six ohms plus three ohms is nine ohms. And I want to find the voltage across six ohms.

So it's actually two thirds of 36. So when I do the math, I come up with 24 volts DC. So we know V th, is 24 volts DC. We know that. All right, so let's stop and go to the next one and the next slide, we're going to find our th. Okay, so I put my original circuit here.

And now we're going to find we're going to find rth. And what we need to do is we need to remove the voltage and shot the circuit. So I'm going to remove this and put a shot here. Okay, so what what what do I have when I do that? I've got this over here. All right, that's what I am.

And really, what do I have there? Okay, if you've taken basic circuit analysis, you notice that we've got two resistors in parallel, right, what I show you right here I want in parallel with our two that's what those slashes mean. It means r1 is in parallel with r two. And now I just use my parallel equation for, for two resistors product over sum. So I got three ohms times six is 18, and three plus six ohms is nine, I do the math. And what do I have?

I've got two ohms. So are th equals two ohms, don't I. All right. So now, my last step, what I tell you that we wanted to do with 70, we want to redraw the circuit, I want to have my feminine equivalent voltage which is there in series with that I want my feminine equivalent resistance which is here, which is two ohms. And now I put my resistor of interest In this case RL and I just saw my voltage and current through that. So vrl is going to be 24 volts DC divided by two voltage divided by Mueller, two ohms over forums because RL is two ohms r th is two ohms we use the voltage divider formula and now we got 12 volts DC.

I solve for current, I know the voltage across RL and I note the value of resistance that RL is made up of. So when I do that, I get six amps. Alright, so basically just to capsulize this with feminine we want to do we first we want to find the feminine equivalent voltage. And then we want to find the feminine equivalent resistance. Let's go back okay. Find the feminine equivalent voltage.

And then that's step one. Step two, let me clear the slide. Step two is we want to find our th, and then we saw for voltage across the resistor or the load, and then we solve for current. That is evidence there. So we've got one more that we're going to look at in the next slide. And that will be what happens if I have two supplies, so two voltage supplies.

We'll look at that on the next slide. So now what we're going to do is we're going to use super positioning. So what do we do? We remove one of the voltage sources. I'm going to do this one first. We shot it out.

So what do we got? We got this circuit here. All right. So now again I'm looking for Vth what I call Vth one, which is the voltage across here where I'm remember I removed my load resistor. So it's actually the voltage drop drop across here and I gave it the idea of AB or voltage AB, so Vth one, which is right here, okay, is going to be three ohms. Okay is actually going to be the voltage drop across three ohms.

So I can use the voltage divider formula right there, which is three ohms over the total which is 15. Because 12 plus three is 15. My voltage choices A minus At four volts DC, when I do my math, one third, because three, three over 15 is one third. Okay? And now it's one fifth I'm sorry, one fifth. So I come up with a negative 16.8 volts DC.

Okay? And that is V th one. All right. Now what am I going to do, I'm going to do the same thing. And what I'm going to do now is I'm going to shot out this supply I'm going to remove in shot. All right, and now what am I left with?

I'm left with this circuit over here. And now what I'm going to do is I'm going to find what I call VH Vth number two. So what really do we have, we've got a We've got a voltage divider here a series voltage divider are actually finding the voltage drop across our one which is a 12 volt resistor. So we have 12 ohms divided by 15 times a minus 21 volts DC. When I do my math, I get a minus 16 eight volts DC. Now what do I do?

Well, I add them up because Vth is going to equal Vth one plus Vth two. And when I do my math, I get 33.6 volts DC. Okay, so now we found in the previous slide, we found our feminine equivalent voltage. Now what we're going to do is we're going to find our feminine equivalent resistance. So I is, here's my circuit, I pulled the resistor out of it. interest.

All right. So now I want to find my equivalent looking from these two points here. What is the resistance? Well, if you can see, okay, because what we do is we remove both voltage sources and shot them out like that, which if I redraw the circuit, I get something well I don't get something I get this. Basically what I get is two resistors in parallel, okay, three ohms and 12 ohms. And then I use my voltage divider formula which we've talked about the find the equivalent resistance product over some.

So I got 12 volts 12 ohms times three and 12 plus three ohms. I do the math, and my equivalent resistance or my feminine equivalent resistance is 2.4 ohms right here. So now What is feminine feminine is feminine is a feminine equivalent voltage in series with a feminine equivalent resistance which I show you here. And now I place my resistor of interest back into this circuit. And now I do my math and we see that the voltage across vrl if I do the math, alright is 24 volts DC because basically, all we have here is the is the is a voltage divider, and we've got six ohms divided six ohms over 2.4 ohms times the feminine equivalent voltage which is 33 dot six. We do our math and we get 24 volts DC across that resistor.

So we have 20 volts DC across this resistor, I can solve for my current IRL, or the current through the resistor is vrl divided by RL 24 volts DC divided by six ohms. The current through the resistor is 64 amps. That's it. So that's my feminine just to capsulize this gear. What we want to do with feminine is we want to find the feminine equivalent voltage. So to do that, we remove the resistance of interest and look back and find the feminine equivalent voltage like we've done in the two examples here.

And then I want to find the feminine equivalent resistance. And then I redraw my feminine circuit which is the feminine voltage in series with the feminine resistance that I found. I then placed my resistance of interest in there and I solved for the voltage across it and the current through it. Okay with that said, let's go on to the next theorem which is Norton