Okay, before we get into a Norton, I think we're going to not stop, but I want to show you one more thing, or one more issue called feminizing. A bridge circuit. And basically, here's my bridge circuit. It's a bridge circuit, the way that these four resistors are configured, it's a bridge. And we have a resistor in there and we want to know what the current is and what the voltage is across it. So, I mean, it's, what's the first thing we do, we want to find what we want to find the 707 equivalent voltage.

So I pull it out. I pull a resistor in Interest out which we call RL, which is two ohms right here. And I open that up. And what do I want to do? I want to find the voltage between A and B, which is my feminine equivalent voltage. So if you look right there, what do I have?

I've got two branch resistance. And each branch has two resistors in series, right? So basically this 30 volts across here, and 30 volts across here because my supply voltage is 30 volts. Alright, so let's let's go on to the next slide. And what I what I've done is I've went ahead and I calculated the voltage drops across here. Each one of these resistors vi three v r for VR one and VR two.

Alright, so now what we want to do is we want to find the feminine equivalent voltage, which is actually the voltage from A to B, so I'm going to go this way. So basically what I do is minus VR three plus VR one, and that will be if I took a meter and placed it across a and b, that's what I would read the three. We've done the calculations right here, so vi three is 10 volts, but I have a negative right there. So it's a minus 10 volts, VR. One is 18 volts. And I'm just going to add that so a minus 10 volts plus a plus 18 volts Alright, is what it's eight volts DC it's not negative.

That negative is wrong. It should be a plus eight volts DC from A to B. So let's go on. Okay, so now I've redrawn it. Okay? So, if you look, this was my original circuit, and I've redrawn it over here.

What did I do? I removed and shot it out the supply. And when I do that, okay, from the ANP point of view over here, I have four resistors. I'm sorry, two branches. Okay, two branch resistors one here, and one here. All right, and We have our three and our four in parallel.

And r two and r one in parallel. All right, and we find the equivalent resistance because these two, these two parallel branches are in series. So when I find the equivalent for my three and r four, it's two ohms. And when I find the equivalent of r two and r one right here, it comes out to 2.4 ohms. I just add them up. I get 4.4 ohms.

All right, so now I have found the AR th. All right, so VR th is right here. All right. We know what Vth is. It's what it's eight volts. So now I put my resistor of interest back in and I can find the voltage and the current.

So for instance vrl is two ohms volt using again using the voltage divider formula, two ohms over 6.4 ohms because 4.4 plus two equals 6.4. So two over 6.4 times eight volts DC which is my Vth, which we found in the previous slide vrl is 2.5 volts DC. And then I can just do my math and get 1.5 1.25 amps through that. And and that that is pretty much it. So you know, we've done I mean, basically notice the same thing over and over again. Find the Norton equivalent voltage Vth.

Then we find the Norton equivalent resistance. Okay, we have our Norton equivalent voltage in series without Norton equivalent resistance, then we place the resistor R. I'm going to call it the load now whatever that may be. All right, and you'll see as we go on the load in series with that, and then we can determine the current flow through it, and the voltage drop across it. All right, thanks. Bye bye. All right, now we're going to talk about Norton's theorem.

And its name for our rock. Excuse me el norte. He was a scientist at Bell Labs. All right, if you if you look at Devin, feminine was used to simplify the circuit in terms of voltage. Where Norton is you To simplify the circuit in terms of current, instead of voltage, alright, so Devin, we use voltage equivalence and ignored and we use current equivalence. So it's kind of the same with just a little bit of a twist meaning Norton we kind of do the same things with a little bit of a twist.

Okay, so in feminine we we determined Vth but in Norton, we we determine the Norton current i n. Okay, then we're going to determine the Norton equivalent resistance. And we did that for Theremin and also you'll see that it's, it's, it's the same and then we're going to solve for RLS current and voltage which we did with Devin and also. So if you look at my circuit here, very similar. All right. Um, Got a load resistor RL, I got a voltage source at 36 volts and I've got two resistors. In series three ohms are one and six ohms are two.

So we're going to find i n. So let's stop here and go to the next slide. Okay, so now the next thing we do like with the seven and what do we do, we we shot out, we took the load out and we shot it out the well we didn't shut out we, we determine the the feminine equivalent voltage. Now we're going to shot this resistor out here. Again, here's my original. Okay, we're going to remove that resistor. Okay, and we'll get a shot.

Ah anb out. That's what We're going to do so we're going to shout out our true okay and this will allow us to determine i n. So now I N is just v one over r one and v one when we shot this out, we just we just have a we got all our voltage across this resistor r1. So my current is going to be 36 volts divided by three ohms is 12 amps. So basically let me just draw what we got equivalent when we shot that out. And this is my three Ohm r1 and this is 36 volts DC right here. Okay, so when I shot that out, that resistor three ohms is right at Cross the voltage source and then I solve a current 12 amps.

Okay, so now we've done that we we found i n the right there. Now what we're going to determine is we're going to find the Norton equivalent are. So I put the original circuit here. So basically just like Theremin and we need to remove the voltage source and shot that. Alright, and when I do that, I find that my two resistors here in parallel, so I use my parallel formula here. And the RN or the Norton equivalent resistance is two ohms.

All right. All right. It's almost the same thing as Norton, alright, almost the same thing, a little bit of a twist, like I said earlier, but almost the same thing. So now in this, what we've done so far is we find the Norton equivalent current, which is 12 ohms and the Norton equivalent resistance, which is two ohms. Let's go to the next slide and finish this up. All right, well, here we go.

If you remember with seven in my head, a feminine feminine voltage on my feminine equivalent voltage, and I had my feminine resistance, and we put our feminine equivalent resistance in series with the feminine voltage now with Norton. Norton is a current source, which I show you right here. We found i n, which is the Norton equivalent current that goes in goes in parallel with my Norton current source. All right. And we found out that that current sources 12 amps can deliver 12 amps. And now what I do is I place my resistor of interest in parallel with my feminine equivalent resistance R n. All right, and now it falls apart.

So if I want to find vrl right there, I mean basically what I did was I said that there were two resistors in parallel. I did the math and basically, the equivalent these, these two resistance have an equivalent of one Ohm All right, I just say 12 amps times away One home is 12 volts DC. So I know that I have 12 volts DC, across this whole circuit. Let me stop and clear out the slide. So again, if I know I've got 12 volts across this element All right, then I can find my current through RL, in this case, which is two ohms. And I just do my math, and I find out it's six amps.

So again, we use we use feminine and we use feminine with a voltage source to simplify the circuit, and Norton uses a current source. To simplify the circuit. We both find equivalent resistances feminine rth Norton is RM But the system or the process in which we find these equivalent resistances are the same. It's just that we, we use them a little bit differently at the end when we go to solve the circuit, feminine, that feminine equivalent resistance goes in series with the voltage source. And then Norton, that equivalent resistance goes in parallel with the current source. And that's pretty much it.

So let's stop here. The next slide, we're going to talk about the convergence between Norton and Devin in and I've already kind of mentioned it here, but we'll go through that again. So let's stop here and go to the next slide. Okay, let's look at one more Norton's theorem. I said we would go on to conversions but let's look at this one here. What happens when I've got a series resistor and then I've got my load off here like that?

Well, we do the same thing. We shot this out. Alright. And now I N is the current flow through our two. All right, and we can calculate that. All right?

If you come over here, I've got six and 12 ohms in parallel. So when I find the equivalents for ohms, okay, my supply voltage is 48. So when I do my math, I've got 2424 volts DC across right here. So then I can just I can just solve for the current in that resistor which is two amps. All right. So now I found I n which is two amps.

All right. And now I want to find Rn, well, Rn now we shot this out over here. So now I've got four, and six homes in parallel. And again, when I find the equivalent of that it's 2.4 ohms. And then I just add, add 2.4 plus 12. And I get 14.4 ohms.

And that would be my rn. So I found my i n, two amps, and my rN 14.4. Okay, in the previous slide, we found I N, and we found rn. So here's my Norton equivalent circuit right there, my current source and Rn which is in parallel with my current source. So now I go over here. I put my resistor in question RL.

And now basically I just do the calculations. To find the current through RL, I can use this voltage divider formula where it would be AI n over RT divided by R L. Okay? Now RT is going to be the parallel equivalent of RL in Rn. All right. And I went ahead and did that and that's 5.9 ohms. So it would be the ratio of this 5.9 divided by 10 because RL is 10 ohms.

All right, my i n is two amps right here. I do the math. I get 1.18 amps flowing through RL. So that flowing through that resistor there. All right, and I said make a note RTC equivalent note is the parallel equivalent of RL and R n, then I do the math. And I find out that there is a lot of 11.8 volts flowing through that.

Ideally it should be 12. But I, but there's a lot of repeating decimals in our calculations. So we get a little bit of an error right there with that. All right, so with that said, we're done and I thank you and we are going on to the feminine and Norton conversions. See over there.