Okay, looking at this circuit, we have a x sub l x sub c in a resistor in series, we have more than one, we've got two resistors, two capacitors, ah, I'm sorry, here's the resistor here, two inductors and two capacitors. So what we need to do is we need to combine them. So for instance, 30 ohm resistor 10 ohm resistor that equals 140 ohm resistor, a 30 Ohm capacitance inductive reactance plus x l two which is 30 ohms. That equals a total of 60 ohms. And then we add up the reactance is on the capacitors and that is 90 right there. All right.

So I go to the next step here. All right. Take my equivalent reactance element of this circuit. So if I look, we got x empty x, a total of capacitive reactance of 90 ohms and a total of inductive reactance of 60 ohms. All right? They're equal.

I'm sorry, they are opposite each other. All right, I take the difference. And I get 30 ohms. So my, my react my net reactance in this circuit is capacitance, and it's the total is 30 ohms. All right. So now I can find z. z is r squared plus my reactance total squared.

We found out that again that the properties in this circuit is primarily capacitance and it has a value of 30 ohms. So now I plug in here, I do my math and we get 50 ohms. So as far as the source is concerned, looking into the circuit, the components that are in series that are connected to it, it looks like a 50 Ohm impedance. Now I can find it where in a series circuit I stays the same. So current is the same in a series circuit. All right, I just cleared off the slide.

So current is the same in a series circuit. So I have two amps of current flowing in the branch in the branches, okay, well, it's actually one branch one path but current flow. Alright. So now I just find the voltage drops across each one of my components via one is it times r1. Two amps times 30 is 60 volts. The voltage across see one right here is it times XC one two amps times 70 volt 70 ohms is 140 volts.

Right here we have two amps times 30 which is 60 volts. Going down to the next one, VX two is two amps times 20 which is 40 volts. We have VR two which is two amps times 10 ohms, which is 20 volts. And then the last one, V XL two is two amps times 30 and that's 60 volts. So we just found all all the voltages Across the components. Now one of the things I can do to make sure I've done it right we'll see it on the next slide is I can do a check.

So let's go to the next slide and we'll show you that all right, if you look here, we combine the voltages across the common components. What I mean is I combine the voltages across the resistor 60 and 20 is 80 volts. I combine the voltages across the capacitors 140 plus 40 equals one ad and then I combine the voltages in this example across the inductors and I get 60 plus 60. Is 120. Now to find the T because if you remember my react ances are out of phase with with my resistance And then has to do with the voltage also. So the voltage across the resistor are out of phase with the voltages across my reactive elements.

So right here I find my total voltage across my capacitive element minus the total voltage across my inductive element. And these brackets mean absolute so when I take the difference, I'm always gonna end up with a positive number. So we've got a t squared, I take the difference here. All right, and it's 60. I do my math. And then I get 10,000.

Here, the square root of 10,000 is 100 volts. So I did a check. And I know it's right Okay to find the phase angle right here, v x CT or the voltage across the capacitor right here at at the very end over the voltage across this resistor, alright will give us the angle, we calculated the voltage across here to be 60 because up here at minus 120. All right, that's what we want. So 60 divided by 80 gives me a ratio when I use the, the tan minus one function, I get 37 degrees right there. All right, so that's my phase angle in the circuit.

And the properties of the circuit. After all is said and done is capacitive and we know that current leads voltage, so the current will lead the voltage by 3072. And the voltage that I'm talking about is the voltage across are generated by the voltage source. All right. Okay, Nuff said. Let's go on.

Okay, in this circuit here we've got multiple elements in parallel. So for instance, we've got two resistors inductors, two capacitors and what we need to do is we need to combine them. Now, we've already given you the reactance is of the reactants components. You can see them here. Okay, Excel one is 33.3 ohms. Excel two is 100.

We've got x One of 200 Xc to have 200 ohms. We've also given you the current. So ideally what we need to do is we can calculate the current up here by taking the value of either the resistance or the ohmic value of the reactive component. All right, and divide it by the source voltage here, and then we'll find the current. So for instance, looking at IR one, what do we have? We've got AI equals V over in this case R. So what do we have?

We've really got what do we got 100 volt 100 volts divided by 50 ohms And that is two amps. All right, that's what we have. So I've done that for all all the branches. You can you can do that if you'd like. Again, we've done a lot of these, and you should be familiar with them. So now what we'll do is for instance, we'll take this element here my resistance element, and I'm going to combine the current.

Alright, so just to do an example. Let me stop here and clear this. So for instance, right here, we've got a total of four amps. And I've added this with this because that's the current that is flowing through the resistive element. Here it is there. two amps, plus two amps equals four amps.

And I do Same thing with my inductive portion here. So what do I have? I've got three amps in one amp. three amps plus one amps is four amps. And again the same thing with my capacitive element here. We've got a half an AMP plus a half an amp.

So we've got one amp that is has the one amp of capacitive current I'm going to use that term Okay. So now, we know from previous discussions that the voltage and current across a resistive element, in this case a resistor and phase However, my current and voltage across my reactive elements are 180 degrees out of phase or their opposite. So since in a parallel circuit voltage is the same, what's equal and opposite in my reactive circuit is current. So if you look and again, let me stop and clear the slide off. If you look, I've got four amps of current through my inductor, and one amp of current flow through my capacitor. Again, we're talking about AC current flow, not DC, AC current flow.

Remember I have an AC source right there. Okay. So what did I tell you? current flow through reactive elements in a parallel circuit are opposite So I take the difference, and the big one wins. So for instance, I have four amps minus one amp. I take the difference, which is three amps.

My inductance is larger than my capacitance. So my inductance wins. All right. All right. So now again, going back, all right, I want to find the equivalent resistance. So what do I have?

So I have IR, which is the current through this leg here. Let me stop here for a minute and clean that off. So over here, all right, I want to find the resistance equivalent. So what was That'd be that would be V divided by AI. And that will equal R. So in this example here, I have 100 volts AC divided by four amps. And when I do my math, I get 25 ohms.

All right, so those two resistors up here. As far as the generator is concerned, that has a resistance of 25 ohms. All right, let me get that in there. Okay. All right, let's do the inductor. Let me clear the slide off.

Again, the same thing with the inductor like I said one or two slides back, it's the same dance. Okay, so what are we looking for? Excel? I'm going to call it Excel total. All right. What do we got?

All right, we have V, which is 100 divided by x, l total. And we know that that's going to have inductive properties because we have more current flow through our inductor, and now we just do our math. 100 volts divided by three amps. I do the math, I get 33.3 ohms. That's it. All right.

That's it. So I've solved my circuit. I've gone from this giant thing, okay, with what do I got? I've got six, six components in there, and I reduced it to two. All right, we know we when we reduce it, we have an R l circuit in parallel or an equivalent RL circuit. So we go from here to here.

And that's an equivalent RL circuit. All right. Let me stop the slide clean it off for now and I'll explain some of the formulas and in theory on this side of the slide, so let's stop. All right, right here, we can think of this as basically a check. Or we can actually look at what the source current is. It's IR squared plus i x squared.

Okay. And again, we we went through this we've got four amps flowing through all my resistive elements. And over here, I have again, I have to take the difference between my inductive components, one amp minus four amps. Okay. This bracket again inside here means absolute value. So we get a three.

We do my math, four squared is 16. three squared is nine, I do the addition and I find the square root of 25. My answer is five, five amps. All right? So that's it. I want to find z. Well, z can be V source divided by it.

We just found it up here. So it'd be 200 or 200 volts AC divided by five amps, I do the math and as far as the generators concerned, and its supplying current out, it sees a 20 Ohm component that is inductive. That has an inductive property. Okay, we find out phase angle here. I have four amps on my zero access. Because what have we got?

We've got four amps flowing through my resistive element. And we know that current and voltage are in phase and a resistive element. All right. We also know that when we went through the derivation that when we combined everything, my my L and my C, all right, l wins means inductive. So I have an inductive property. So here's my i, t, right, which is three amps.

All right. And then again, I do the math and I still find, find five amps. But here is the issue here right there, the phase angle, all right, the phase angle is 37 degrees. All right. So we know that voltage leads current In an inductive circuit, right voltage leads current. So my phase angle between my if I monitored, let me let me explain this if I monitored my supply voltage, because this is a parallel circuit.

All right, my current, the waveform of my current would lag by 37 degrees. Right. And at some point, we're going to show you that we're going to be doing some labs. I'm not sure when that's going to come probably a couple of months down the road. Just to give you a little bit of a heads up, I've got this course here, and I'm doing one more and that that is what I'm going to call my course section. Alright, my my first course section, this will give what I believe will give you the basics of electronics, you will understand the components and the principles.

All right? And then on the second phase, we're going to be putting some circuits explaining the circuits. And this will come, this will be a little bit nicer. All right, because I couldn't put a transistor amplifier out there and say, Okay, put a voltage in here and I put a voltage out, this, this thing will give me five volts, the roll off will be 33 db, it will kick in at 10 kilohertz a kick out at 100 kilohertz, okay? You're going to say, you know, hey, what's he talking about? But then I can say okay, because of this RC circuit here.

All right, and this, this filter over here, or because I have a larger value resistor here, this is what it does. You guys will understand it. You can go in there and you can troubleshoot, competitive. Once instead of replacing boards, or you could replace a board, but at least you'll get a better feel for that. And that's, that's really what I'm trying to do. When I started in this industry many, many, many, many moons ago, when we sent out field service guys out, okay, sometimes we went down to the component level, okay, some of the ICS or transistors, they were in sockets.

Okay, we could pull them out and put a new ones. Okay. Yeah, I know you can pull a board out. But, but to me, you don't really have a good feel of what really is going on. So we'll leave it at that. So anyways, let's stop here.

I'm going to clear the slide off. I expand upon this in the next slide. We'll go in a little bit deeper, and then we'll go on. Alright. see you over there.