All right, we're looking at another rough advanced circuit analysis technique from curvature off. This one is called node voltage analysis right there. And we've we've kind of got this well, we don't kind of we do have the same circuit up. And we still have this relationship with current right there. All right, I one plus i two equals i three. And again, I one is over here.

I two is in this loop here. Okay, and the node that we're looking to find is where it says node and it's actually the voltage across our three and if you look at it, if if we find the voltage across our three, we can find all our currents. And if you'll notice, we've got resistor values assigned to all three resistors. So if we find the currents, we can solve for anything. All right? So that's what we're trying to do.

And again, he mentions or he calls this node voltage analysis, because what we're going to do here is find the voltage across R three. And as we go through, you'll see that he calls that vn. So, let's, let's stop here. Let me clear off the slide and we'll start Alright, so now we know that this relationship right here is correct. I one plus i two equals i three. But if you remember on slaw, where, what do we have I in ohms law i equals E or The over the resistance, right?

And that equals I Well, can I write I won in that in those terms. So if you look here, I one can be written as v one over r one, the voltage drop across this resistor divided by the value of that resistor. And Can't we do the same thing with AI to where we have the voltage drop across this resistor divided by the value of that resistor and that's what we kind of doing here. All right, so up here, we've given you the equation the the actual equation and then we're starting to plug in the values which we're doing down here. All right, so v one over 12 ohms v two over three ohms. Alright, equal What?

V n over V six, alright? Because that would be i three, okay? So I three can be that but that's also again, the n, which is the voltage across our three, which we're looking for he calls that the voltage node, okay divided by the value of that resistor which is six ohms. All right. So now we what we're doing here, what he's doing here or what we're doing here is setting up the equations. All right.

Now if you look, we really don't know what the voltage is across r1 Dewey. However, this is what we do know we know that we have 84 volts over here. And we know that we have a node voltage, which we're looking for called v n. So if we find v n, and we subtract our supply voltage over here, minus the voltage across our three, which is designated as vi n, and we divide that by 12. Can we not find I one? Yeah, we can. Okay?

We can, because that reduces to the voltage across r1 divided by the value of r1. Right here. All right. And it's the same thing with this one over here. Okay, we have a supply voltage of 21 volts and again, there's v1. All right, if we're looking for the current flow through our two again, Wouldn't that be 21 volts minus this value here, this voltage drop here across our three, the difference would be the voltage across our two.

Alright? And those numbers when I solve those equations should equal vi n divided by six. So again, with setting up the equation to up to kind of find our answer, in this case, vn, so I'm gonna stop clear off the slide. We're going to continue. So in this this step here, okay, what we're doing is what kind of what I call cleaning up the equation. So we're cleaning up the denominator.

So if I divide this by 12 over here, This whole thing right there. All right, what do I get? Well, 12 goes away, and so one there, so I get 84 minus vn. And then if I divide three into 12, I get four. So now I can multiply four times 21 minus v n. And when I do on one side of the equation I have to do on the other side of the equation. So if I multiply by 12, that becomes one.

This becomes two. And so I reduce my equation here. All right. All right. So now on this next step, and I'm going to play the slide off again. We're going to combine terms.

So let me stop, clear the slide off and we'll go on. Okay, so here I go. I am going to start today. combine terms. So I still have 84 minus vn right here 84 Plus Now I multiply here. So four times 21 is 84.

Four times vn is minus four vn. All right, and that's still equals to vn. Now on here, I actually start combining terms right there. And 84 plus 84 is 168 minus vn plus a minus four vn is five vn. So now I've reduced my, my fraction or my equation so I shouldn't say fraction my equation there. And now I'm what I'm trying to do is, is isolate VPN, I've got to find one VPN.

So let's stop and clear the slide. So again, going back a bit, we found 168 minus five equals to vn. What I want to do is get a single vn on one side of the equation. So what do I do I add a plus vn here, a plus five vn. All right, what I have what I do on one side of the equation I have to do any other. So if I add a plus five vn over here to to vn, I get seven vn.

So now I'm 168 equals seven vn I need to find one vn 168 divided by seven equals vn and when I do the math, vn equals 24. And then basically it's it's gonna, it's gonna fall apart. Because again, when I come up here, if I'm looking for v1, which is this guy, v1 equals 84 minus v n while we know what vn is. All right. So So I actually made a mistake here. That shouldn't be vn that should be v1.

I'll change that. So v1 equals 60. So I got 60 volts across here. All right, and v two is 21 minus vn. So v two is a minus three volts DC. And then we can just find our current so I want is 60 volts DC divided by 12 ohms.

I do the math, I get five amps. I two is a minus three volts DC divided by three, so that's a negative one amp. Okay, and that means currents flowing in a different direction. It's flowing away, as you can see here, and then i three equals five amps. Minus one amp is four amps. Okay.

And what I meant by what I should have said not up here, when it's in a different direction, it's a different direction through this leg of the circuit, which is our three. Okay. So that's it. That is it. And that's just another way to analyze a circuit. All right, so we got one more to do.

And then we've got we'll put one up where we can just look at it and take some of these advanced circuit analysis has kind of determined the currents and the voltages through them. I mean, you don't have to always use one you can kind of match like current law and voltage law. And it it all should come out. I mean, it should, it should come out, you should be able to figure out the voltages on the circuits on the circuit. Okay, let's Stop here and we'll go on. Okay, I've got my two mesh current equations right here.

And what we want to do is we want to eliminate one of the variables so we can solve for either IAA or IB it looks like we're going to solve for ROP, ay ay ay ay first. So, one of the things that I want to do is probably bring them down to the lowest common or simplify the equation. So I can divide this one here by two. And if I do that, I get nine ay ay ay minus three IB equals 42. That should not be a negative there. I'll correct that but that should not be a negative.

Okay, so we've done that. Here's my simplified equation. And we can also do it on on mesh current IB and i divided by three here. So I get a minus two i a plus a minus three IB and a minus seven volts. All right. So now I'm going to come up here.

All right, what do I want to do? If you look, I'm trying to get rid of one of the variables so it looks like if I add these, I will get rid of three IB. Alright, so if I do the math, three IB plus a minus three, IB is 09 plus a minus two is seven is a 42 minus 42. And I add a minus seven is 35. So I can go through the math and i equals five amps. Okay, so now what I can do is, I can use this equation here.

And substitute ay ay ay ay for five amps. And when I go through the math, I get three, IB is three. So IB, there's one AP. So what we're saying here is I have five amps flowing in here in this mesh, and I have one amp flowing in that mesh. Okay. Now you can see it's very easy to calculate the voltage drops across R one, r two and r three, which will, will look clean the slide off and we'll we'll do it on the next slide.

All right, just just to say we did It all worked out. All righty. Let's stop here. I'm going to clean the slide off and then we'll finish this up. All right, now basically what we've got to do, we found the currents in the previous slide. Now we're going to just solve the voltages.

So VR one, which is this guy, we got five amps flowing through it. Do the math, 12 ohms 60 volts DC. Notice my polarity, that's very important. Notice my polarity. Okay? VR to one amp flowing through at times three volt, three ohms is three volts.

Notice my polarity very important. All right now, this this resistor here because this resistor is common to both meshes, both mesh current IAEA and mesh current IB, okay because we subtract the current When I go through that through this derivation here, five amps times six ohms minus one amp times six ohms. Again because the current subtract, okay, when I go through the math, I get 24 volts DC. Notice the polarity, okay? Notice the polarity. All right, now before we go on, notice the polarity on our two, okay, even though this is 21 volts, and this side here is on the negative side, or the negative terminal of the battery, because this two voltage sources in this circuit and quite honestly, if you look at the one on the left here, that's a higher magnitude or a higher voltage, even though that's negative This is forcing current flow this way through that resistor.

That's why we have a negative terminal there. Get a polarity. All right. So that's why now let's clear the slide. And we're going to just we're going to prove it. We're almost done here.

We're going to prove that we did it works out because voltages in a closed loop should equal zero. And you'll find out that this this, this will do that. So let's stop. Let me clear the slide off. We'll go on. Okay, so here we go on mesh Ha.

All right. So basically, I'm going to start here. And what's VR? One? It's 60 volts minus 60 volts, plus going this way. VR three.

What's that? It's 24 volts, but it's a in this mesh, it's a minus 24 volts. And we come over here and we hit a plus 84. If I do the math 60 plus a my minus 60 plus a minus 24 plus at four, zero, you can do the math, get your calculator out, so that works out. So mesh a works out. Let's sum.

Let's look at the voltages around mesh IB. And we're going to start right here again. So what do I have? Well, I've got the br two, which is three volts, but again, we talked about that polarity, it's a minus three volts DC plus my voltage source over here, which is a minus 21 volts DC coming up this way. What's the voltage drop across our three? It's 24.

What's the polarity it's plus. And you can do the math, a minus three plus a minus 21 is a minus 24 plus 24 is zero. So we've done it right. The tricky part here is that resistor in the middle, okay? Because the currents are flowing in two different directions plus, we have two voltage sources here. Okay?

And, quite honestly, the larger voltage source wins. It's supplied. So my it's 84 volts. Notice that our polarities are the same, but this guy wins because it's a larger magnitude. So it's going to kind of force current this way. All right, through the circuit.

So that that's it. We're done here. We're going to go to the I'm going to stop it here and then the last section. All right, we're going to talk about applications and then we're done with kerchief slock Alright, see you next