All right, well, on this slide here, we've just put a circuit up there and we're looking at circulate applications of kerchief law, or some of the techniques and laws and exercises that we've done previously. So let's let's look at this circuit here and and see what we've got going on. Basically we want to find the currents and the voltages across the resistors. And probably the value of voltage at that supply v one. And let's look at what we got. Well we've got some current ay ay ay ay ay ay is nine amps and we know that it's flowing through r1.

We have a node which we call nodes. Today and look at what we got here we've got IAA three which is three hams falling down this, this side or this branch. Looks like we could probably find what ay ay ay ay ay. Two is, wouldn't you think? And let's look on the other side here. We've got the IB, we've got six amps flowing this way.

And IB two is four amps. And I'm sure we could find IB three, don't you think? And obviously, if we've got six amps flowing this way, according to our diagram, we could probably find the voltage drop across our four. Wouldn't you say that? So let's let's look at let's look here. I'm going to stop here, clear the slide off.

So we could find a two right here because I've we've got a question. Question mark. So what do we got we've got ay ay ay ay ay. is nine amps. minus i a three which is three amps. So ay ay ay to equal six amps.

Okay, according to this diagram and the way that DEP these arrows and showing me the current flows. All right, what else do we want to okay? I be three. All right, well, if we've got six amps right here, and we've got four amps flowing up this way, we should be able to find IB three, shouldn't we? That's, that's two amps. Now we have a voltage across these two parallel resistors here, one we know which is r three and one.

We don't know Which is are two but those two resistors are in parallel. And what do we know about that we know that the voltage is the same across two parallel branches of resistors. So we call that V A B. So if you look here, VA B so we've got to subtract ay ay ay ay ay ay ay three minus IB three. Because the currents are flowing against each other, and quite honestly, the larger one wins So, ay ay ay ay three is the larger current because that's three amps. So we subtract them.

And so we know that the net current flow is one amp. So as I we say here, the current flow Through r three is from left to right this way. And the total current flow is one amp, three amps from the left to right and two amps from right to left. So again, the logic current wins. So therefore v AB is going to be 10 volts divided by six amps minus four amps which is oh I'm sorry, we went ahead. So the voltage across our three right here is going to be 10 volts.

Okay because I have one amp times 10 ohms equals three. Okay, now we're going to find the voltage across our two. All right, so what do we have? Well, we know that ay ay ay ay ay two is six amps. Actually, let me stop here and clear the slide off. Okay, I've cleared the slide off.

So we look in the for the voltage across our two and we know that the voltage is the same across parallel resistors. So we found the voltage across our three which was 10 volts here. So the value of our two is going to be 10 volts. Six amps minus four amps because I a two is six amps and IB two is four amps. All right, the larger current is going to win. So when I go 10 volts divided by two amps, I get five homes So that's the vote, that's the value of r two.

Now, we can find the value of cross v r1, which is this guy here. All right, so we know it's six ohms. We've got nine amps flowing through it, we do our math. We got 54 volts DC, over here, VR four. All right, we've got six amps flowing through that resistor. So six amps, times two ohms, which is the value of r four is 12 volts.

There we go. All right. Now what we need to do is we need to find the value of v one. And we know the voltages around a closed loop will equal zero or will equal a supply voltage. So This voltage drop, that voltage drop that voltage drop. And this voltage source.

When I take the algebraic sum of all those voltages, it's going to equal my v one, which we've done here. So we've got 50 for VR one, a minus 10 volts, which is the A B plus 12 volts minus 20. And when I add that all up, I've got 72 volts, so that should be a minus there. And that value is 72 volts for that battery voltage source. So that's it. We've done it.

So what have we done here? we've, we've looked at Kurt chops was And we had a circuit. And we needed to find, okay, I've cleared off the screen. So as I was saying we need to, we need to find some of the currents, we needed to find some of the voltages. And basically, the only thing we did is we use the appropriate rules for current shops law, define the elements or the properties, either current voltage or resistance in the circuit. So we could find what we needed.

And that's pretty much it. Now that we're going to wrap up. Kurt chops law here. What I am going to do, and I'm hoping to do it in the end of this course is do some labs have another circuit we'll, we'll calculate the values and then we'll go in there we'll build a circuit will be three or four resistors. We'll take a meter and go in there and verify that the voltages across the resistor is what we calculated. And the current flow through the resistor is what we calculated just just to give you guys a little bit more of a confidence builder that that we're doing things right here.

But what we're going to save that to the end. All right, and the next thing we're going to talk about is network theorems. All right. So with that said, I'm going to stop here. We're going to start a new six section called network theorems, and see you there.