All right, the next method we're going to talk about is branch currents. And before we start, you'll notice that we've got some current flow here, and I labeled them I one, we have I two. All right? And we have a current right here are three, which is the sum of i one plus i two. Okay. Now, if I find these currents then what can I do?

Well, I can figure out the voltage drops across these resistors Can I all right? So let's clear off the slide. And what we're going to do here is, again, we're going to, we're going to write some loop equations. But we're going to do what they call a simultaneous way that we're going to solve for one current. And when I solve for one current with these two equations, and you'll see as we go, once I get one current, I can get the, I can get the other current by subtraction, and I can get the current that flows through our three by by addition, so let let me clear the slide off and we'll start going through this. Okay, so I cleared off the slide and let's look at this here.

I've actually got three loops, but we only do what we need to do here. We only need two loops. So I've noticed I called it loop one and we've got loop two Over here. Okay, I've got a point A and point B. And again, I just picked those points arbitrarily, we could start any way you want. It just said, it seemed like that was a convenient place for me to start.

So I'm going to write my first loop. Okay? And we're going to go around this way, as I'm showing you with these arrows. Okay, so we go through plus 84 up around and a minus voltage across our three. That's what I mean by VR three here. All right, so the voltage across this resistor, which I call VR three, and then the voltage across r1, which I call VR one right there.

And you know from our discussion on current shops, voltage laws that if we had go around a closed loop, that's got equals zero. We know that. All right, so now let's look at the second loop. And again, we're going from point B right here. And I'm going to go this way. Actually, I messed up we went, we went counting this way, we're going to go clockwise.

All right, so we're going clockwise right there. Again, it doesn't matter, I can go clockwise, counterclockwise. As long as I observe the polarities, it all should work out. It's just that it's, it's, it's more convenient to go clockwise in this example. So here we go. We go plus 21, right there, up a minus VR three, voltage across our three, and then a minus VR two, right there, voltage across our two and when we do that, we should get Zero.

Okay, I cleaned off the slide. And let's look again at this equation here. Okay, and I'm expanding this one. All right. All right. And what I'm doing is, you'll notice it says v v r three a V times r three v times r one or voltage times r three, or I'm sorry, the voltage across, we're looking for the voltage across our three.

All right, so what does voltage equal with loans law, V equals i times R. So what I'm doing here on this equation is I'm taking this term right there and I'm presenting it In a mathematical equation, voltage in terms of i times R. Okay, so let's clear off the slide and go again. Alright, so with that said, What am I doing here? Well, I'm looking for the voltage across our three. What is the current? And again, we're looking at loop one here. Alright?

What's the current that flows through our three? It's I one plus i two. So if I want to find the voltage, it would be r three, which we have a value four times I one plus i two. All right. All right. So now I've gone up this way, I'm going to come down here and I'm looking at VR one.

All right, VR one in terms of ohms law with V equals i times R. All right? It's r1. And what is the current flow through r1? That's I one, so the voltage here, v r1 is actually r1 times I one, and again, that has to equal zero. So 84 minus r three times the sum of i one plus i two minus r1 times I one has to equal zero from current shops, Voltage Law. All right.

All right. So we're looking at the This part of it now. So what am I doing? Well, what's the value of i three is six ohms right there. So what I'm doing is multiplying a minus six by I one plus i two. So since we have a negative value there, minus six times I one is a minus six I one minus six times a plus side two is a minus six.

I two right there. All right, so all I'm doing is multiplying the value of our three, which I have, which is six ohms. There, and I'm finding I'm expanding this expression, right down here. Alright, now, I'm going to The next one minus r1 times our r1. Well, r1, the value of r1 is 12 ohms. So that comes out to be 12 by one.

Now, when I combine the terms, I get 18 I ones and six I twos, alright. 12 plus six is 18. And then I'm just left with six I two. All right, so we know from my purchase Voltage Law, if I go around a closed loop, it's going to be zero. But what I want to do here is I want to find the value of, of my current so what I need to do is get my currents on one side of the equation and what's at My voltage on the other side, which I've done, so if I subtract an ad for here, all right, it goes away on this side of the equation. But we know from mathematics what I have to do from one side of the equation I have to do to the other.

So if I subtract a minus 84 over here, the zero I gotta work, subtract a minus 84 over there. So I'm left with a minus 18. I one minus six I two, and that's going to equal 84. All right, so let's stop here and go to number three. All right, well, we're doing the same thing here for this loop equation that we did for my first one, and I've brought it up there. So we're starting from point B again, and again, we go plus 21.

All right, and we go from a month. Yes. And again, my voltage drops and where we're presenting our voltage drop in the equation with the relationship of our times I, because we know that V equals i times are. So if I look at our three here, the voltage drop across our three in this example is I one plus i two. Okay, we go around again to the next component, we have a minus r two times the current that flows through it. I too, because V equals i times R. All right.

All right, so now, well, let's do it again. I'm gonna clear the slide off. Okay, again, we're starting with this side here. We're starting from B again. So I got plus two 21 And now, what's the value of minus resistor is six, the current flow is six times I plus I two. Over here we have a minus polarity on our component, the value of the component is three ohms.

So three times I two, that should equal zero. So all I'm doing again, right here is I'm multiplying I'm expanding this term up here. So six A minus six times I won is a minus six. I won a minus six times I two is a minus six is two and minus a minus three, I two right here. This one just drops down. All right, now I come back bind my terms.

All right. So, six itu plus three. itu is nine itu, and I'm left with a minus six I one. All right, I've got 21 minus nine, eight to minus six, I one equals zero. And again, what I want to do, I'm actually looking for current. So I need to get my current value isolated on one side of the equation, which I'm starting to do here.

So I subtract the minus 21. What I have to do on one side of the equation I have to do to the other. So I've got zero over here. So zero minus 21 is a minus 21. So I'm left with a This one right here, minus six I one minus nine I two equals 21. All right, right.

Now what we got to do is we're going to reduce our two equations here. Okay? And the first one, which is this here, I got a divide that by a minus six. So I have to do the whole equation, both sides, and I don't change the value. So I divide minus 18. I won by a minus six, I get three I won.

Because I'm dividing by a minus six. All right, two minus signs, either multiplication, or division will give my answer will be plus so I get a three. I won that I take a divide minus six by minus six I two right here, and I get a positive one I two, then I take my minus 84 and I divide that by a minus six. And I come up with 14 right here. So after I do that, I get a three I one plus a one, I two equals 14. Now I've got to do this one up here.

All right, I'm gonna divide that by a minus three. So a minus six is one divided by A minus three is a two I one minus three. My excuse me a minus nine. I two divided by A minus three, gives me a positive three. I one In a minus 21 divided by A minus three is seven. So I've reduced those two equations to the lowest term.

Okay, give me a minute to clear off the slide. We'll go on. So now here on Step five, what am I trying to do? Well, I'm trying to just to get one variable. So I need to subtract the two equations. So I've got to do something with my equations, so I can get rid of one of them.

So if I multiply the top one here, by three, I will be able to subtract and get rid of what I two. So that's exactly what I want to do. So again, it's just a mathematical mechanics, which I call three times three times. I want 9913 times one I two is three I two and three TV. You know, you can't see it. Three times 14 is 42.

Right there. All right. All right. So now if I subtract them, all right, I get this. Nine I one minus two, I one is seven, I one, these go away. Because three is two minus three, I two is zero, so they go away.

42 Minus seven is 35. All right, what do I want to do? I want to find I one. Alright, so now, I divide seven by both five. I should have put it here. I'll let me do that here.

I divide by seven. So Seven I one divided by seven equals one I 135 divided by seven is five amps. So I one is five amps. So we solve for one current here. So I know that I want is five amps. So now what do I do?

Well I can go back up to here and say two I one plus three equals seven and substitute the value that I got from my one, which in this case is right here. So we know that I want is five amps. So two times five amps plus three I two equals seven right there. So I didn't show it, but two times five is 10. I bring this over on this side of the equation. All right?

So seven minus 10 equals three I two. So what happens when I'm when I subtract x minus 10 plus seven is what? That's a minus three. I divide by three. I get 1am. All right.

Okay. Now, what I want to stop here and say this, okay? The math on Electronics is math. All right, I kind of skipped over them. I, we're going on here. I can't show you stop and do every little math exercises because these trains videos are these videos on electronics will become much, much larger.

What I have done is I've put math electronics up, right? It's, it's, I think it's 10 hours, maybe a little bit longer, maybe a little bit less, I don't remember, but there's a lot of information. These are simultaneous equations, I go through them. Alright, take them bone up on your math skills. If you've know what I've done, you're fine. All right.

Or if you don't want to take my course, that's fine. Go get a math book, go to YouTube, find some math exercise, do something. But you need to be comfortable with math, especially when we're going to do advanced circuit analysis. Alright, with that said, we solved it. It's one amp. So let's go on to the next slide.

Okay, um, what I've done is I've expanded the What I've done on this slide here, I found the voltages across these resistors. We solve for the current through R three. So we know from the previous slide I want is five amps. I two is a minus one amp, so I find a three, which is the current flow through that resistor there. All right, then we solve for the voltages. So VR one would be 60 volts, VR two, which is this guy here would be a minus three volts.

And v r three, which is this guy here will be 24 volts I didn't put let me get volts DC. I forgot to put that in there. And I mean, basically we did this on the previous slide, how I solve for current. So take a look at this. So it's a little we've repeated a little bit the only again, the only thing I really different is probably the step scare where I saw for the voltages across the resistors and and solve for this current right there. Alrighty.

Okay, let's stop here.