Okay, we're going to talk about RLC circuits now and I should say RLC circuits series parallel. So I have a series up to two resistors in parallel in series with two series legs that are in parallel. Okay. So the first thing we want to do is we want to convert, convert these and like start from, from the farthest point from the battery. And right here I've got two inductors they're in series A add so I, I can say that I have two 100 Ohm inductors I add them up and I've got one 200 inductor right over here. I have a capacitor and an inductor.

And we know from previous discussions that their reactance is opposite of each other. So what do I do with that? I subtract, and the larger one wins. In other words, we take on the proper properties of the larger inductor, or the I'm sorry, the larger reactant. So, right here. I have 600 ohms for my inductive reactance and I have 400 ohms for my capacitive reactance.

What's the bigger one inductive reactance? I take the difference. The difference is 200. I told you that the larger one wins, so it has properties of an inductive reactance right here. Okay, what do I have? I've got two resistors.

In parallel, and if you've taken some of the previous courses, I'm not going to go out, but I'm going to use the product over some equation to get my equivalent resistor. Or if you remember what I said, if I've got two resistors with the same value in parallel, all I need to do is take one resistor and divided by two, so I've got two 200 ohm resistors. If I divide that by two, I get 100 ohms right there. I suggest that if that's a little fuzzy, you go back and review that. Okay. Now, what do I do is I just, well go over here, that's pretty much in its simplest form that hundred ohms but right here, I have two inductors In parallel, the same deal as what I spoke about over here, I have two inductors.

And each inductor has a reactance of 200 ohms. Since I have two in parallel, I can say 200 divided by two, and I get 100 ohms. Or I can use the product in some equation that we spoke about earlier. Now, I'll add one thing because the reactance is and the resistance are equal, I can use this you know, two divided by two business that we spoke about a few minutes ago, but if they're unequal Okay, different values like 75 and 150 or 80, ohms and 90 ohms. I have to use them This. Alright, so my suggestion is, you go back and review that now if you look at the prerequisites of this course one of the things I stated, you should have a familiarity of electronic circuits and electronic concepts.

Okay, so this is somewhat of an advanced course. That's why I'm not going back to square one and explain everything. All right. So let me just stop here. I want to clear the slide and expand upon one or two more things. All right.

So since we if we go from here to here, all right. Now I can find my z. And as we stated before, z stands for impedance. And z is what the generator sees. All right. All right.

This generator sees one combination of reactance and resistance of 141 ohms. That's what that generator sees. It doesn't see the two resistors in parallel, it doesn't see this network. As far as that generator is concerned. It sees one element. It's what we call a complex impedance.

Okay because it consists of resistance and reactive elements. And when we when I did my calculation right there, that's 141 ohms. All right. Now when I solve for it, it is vt over z. I do my math I get point seven one amps AC. So they That's the amount of current that flows out of this battery sorry out of the voltage source All right, all right. So point seven one amps, AC current flows out of the voltage source.

Okay I can now calculate vrt and vrt is the voltage across this, which is the voltage across that element vrt. Alright, and that would be point seven one amps times 100 ohms because the hundred ohms is my resistance equivalent right here. I do my math and I get 71 volts AC across this network. Now since my resistors are the same current that flows in this branch has to equal the current that flows in that branch. So what I've said was IR one equals IR two, all right. So v r t divided by r one is 71 volts AC divided by 200.

And here we are here. Point 355 amps. Or another way that I could have done it is I solve for it. We know that i t is zero dot seven one amps, all I need to do is take since regress here since the resistances are equal. Okay, I know that I'm going to have equal current flow in each branch. So I can just take the 071 amps divided by two and I get that Okay, get the voltage drops across each each element and a couple of other things.

And then we've got one or two more slides to go and that finished up on this module. Alright, see over there. All right on this slide here, we're going to we're going to look at the voltage drops across each one of the components. And we're going to start off right here, where we're looking for VR t. Now VR T is actually the voltage drop across that parallel branch have resistors in series with the rest of the circuit here. All right, we found the equivalent right there. All right.

What do we know we know that we have an i t of zero dot seven one amps. Well, that has to flow through this leg here. All right. And so basically what happens is the current splits through these two branches and I have vrt across those two resistors. So basically, what I do is I say okay vrt is it times RT, which I mean by this here, point seven, one amps times 100 ohms is 71 volts AC, and that's across this. And across that, all right, as I stated earlier, these two resistors are equal.

So what I'll do is I'll say okay, I won, which is the current flow through resistor one equals I to one Which is the current flow through resistor two equals v RT over our 171 volts AC over 200 ohms, which is the value of the resistor. I do my math and I get zero dot 355 amps are another way that I could have done this is okay. I know that zero dot seven one amps flows through that. That resistance combined, I can just divide that by two, which I've done here and I get the same answer. All right. Okay, let's go down to the next three.

Let's stop right here. I'm going to clear off the slide. Okay, so now I'm looking at VX t which is the voltage drop across this combined element, which is this, which I go back to my original circuit is there So let me find v x t XT is it times x t equivalent point seven one amps times 100 ohms. Okay, so I solve for that I get 71 volts AC. So 71 volts across that equivalent 71 volts AC across this equivalent and 71 volts across that equivalent. Well not this is not I'm sorry, this is not the equivalent.

This is the original circuit right here. Alright, so now we're looking, we want to find the voltage drop across each one of these inductors. So I can say V x l one equals v x l two, because they're both 100 ohms. I can say say V x t, which is 71 volts AC and I can divide that by two And I get 35.5 volts AC across each one of these inductors. Let's stop here. Let me clear the slide off.

All right right here now we're looking for the current through l one which equals the current through l two which is this current here. Basically, all I do is saying 35.5 volts AC divided by the value of one. The reactance of one inductor, I do my math, and I get zero dot 335 amps. Now I want to find the current flow through here. I see ix equals I XL three, right there. Okay.

That's going to equal point seven one amps, which is my it total minus zero dot 335 amps which flows here which we've, which we calculated from. And now I get my answer zero dot 355 amps. Let's clear off the slide, stop and clear the slide off. All right, now I'm going to find the voltage drop across each one of these reactive elements. This is where it's not going to add up and you'll see what I mean. Alright, v XC equals x MC times IXL two.

So I have 400 ohms times the current equals 142 volts. Okay, V XL l three equals ix l three times ix. That would be zero dot 335 amps times 600 ohms right here. And I have a voltage drop of 201 volts AC. Let's stop. You say, Well, geez, how can I get that I've only got 100 volts across the generator, right?

The generators only supplying 100 volts. Plus I've got a voltage drop here. So I know that's less than 100 volts. Yeah, you're right. But the reactive elements, that's the point I'm trying to make the reactive elements in because the current and the voltage are out of phase with each one by by 90 degrees. All right, and actually, the current is out of phase.

With with the two references, like 180 degrees, you're going to get these voltages, alright? And they're not going to look like they really add up. But what we can do here is we can do a check And we can we can see that they do. So for instance, VR voltage across our T squared. Let's let's let me stop here and clear that slide off for a minute hang on the voltage across VR x t squared, which is the voltage across that element plus v x t squared, which is the voltage across this equivalent element. All right, both 71 volts.

I go through my math and I get that 100 dot 42. Okay, the error here probably is due because of rounding. We may not have gone on gone down enough of decimal points. It's no big deal. That's less than point 5%. So it works.

Okay, it works. So that's really the only thing that you got to be careful about. This was a, this is tedious. Do you always have to do this? No, you really don't. It's nice to know, it gives you an understanding of how things work.

And maybe when you're when you're checking voltages, and you get look at something and say, well, geez, that really doesn't make sense. And then you look at it, oh, you're too reactive voltage, and I got a reactive element over here. Yeah, that that's why that doesn't make any sense. And then you look at it, you can understand things. But do we have to really do this all the time? No.

It's a problem solving exercise. It gets you to think and get you to understand the component. properties. All right. So with that said, I'm going to stop here I think we've got either one or two more slides and we've ended this section on RLC circuits. So let's stop here and go to the next slide.

Okay, in this slide here with talk going to talk about real power. On the previous example, I explained to you that you remember we had those reactance is voltage and they were I mean what across the capacitor was like 170 volts and across the inductor in one series, it was over 200 and my my supply back then could only was only supplying 100 volts. Okay, real power is actually the actual power that the that the source is required to supply is actually i squared times Are. So what's the actual current times the resistance? Okay, that's real power. And I show you that right here.

I squared R. All right. Another way that we can find real power is by using what we call the power factor. And the power factor is the cosine of theta, or the cosine of my angle. And I give you an example here, okay? If I'm working with a circuit, like I show you here, all right, I have an X sub l of 173 ohms and a resistance of 100 ohms. I find my cosine of theta and we've done that previously.

And so what I would do is say okay, real power equals e times L or the voltage times current. times the cosine. So in this example here, it's 400 volts AC. I've got two amps of current flow. So 400 times two times point five is 400 watts. When I go up here and do i squared R, I also get 400 watts.

So those are my that's my real power. So when I'm calculating my supply, my voltage supply, I take into consideration the real power element, not the apparent power, which in this case, is I times v. All right, which would be 800 volts. It's not 800 watts. It's 800 volt amps. All right. So that's it.

That's it. So, and this here is I just went here and say, okay, the power factor is a cosine of theta. All right? So again, it's real power that we got to watch out for it's real power, that will generate the heat, not the apparent power. And remember, real power is measured in watts, and apparent power has the label volt dampers. Alright, let's clear off the slide.

And let's look at the last one. Okay, this is just a little chart I put up here for you guys. And I should say guys and ladies and guys and guys and gals. And resistance in phase opposition to alternating current direct current. All right, the effect on frequency are is the same no matter what the frequency is. All right, inductive reactance XML measured in ohms.

Okay, we have a 90 degree phase shift like I spoke to you before, it increases with higher frequency. So if you remember, two pi fowl, right equals some constant, whatever that is. What happens when f increases? That increases what happens when f decreases, that decreases so that's what we mean. So, as frequency increases, XML increases as frequency decreases, XML decreases, okay. Okay, and this one here, extra X sub c, again and 90% lagging opposition to alternating current.

This decreases with higher frequencies because X sub c equals one over two pi fc. Right? So when this gets bigger x MC gets smaller. So it's the inverse operation. All right. One of the things I want to point out is this phase three relationship right there.

I l legs vl by 90 in a VC legs IC by 90. So that's the phase shift I was talking about. And that's it except for z. We spoke about z right here is my formula. Okay, again, Excel components increases, but z component decreases on the effect on frequency. Okay, the phase angle is the tan of x over r, quite honestly.

Define that the way I do we need the inverse can function. And I call it 10 minus one over x R. All right, and that will give you the actual degrees. This will give you a ratio and then you got to go find the degrees minus can goes right to the degree. So I use, I use this one, it's easier was saying the same thing. All right, and then in parallel, it would be i x over IR. That would be the facia.

All right. There's a little chart I put it in there, it kind of gives you a breakdown of what we talked about. And I really, really hope you've you've enjoyed this. This one was an advanced topic. Again, if you haven't taken any of my courses, and you jumped in here, I you know, if you've got a background in electronics and electronic circuit, you should have followed If if you're pretty much new and you're saying wow, I don't think I can handle this, it's not that you can't handle it you probably don't have a built the foundation to understand this yet. So what I suggest is, you look at my my earliest courses are understanding voltage, current resistance, DC circuit analysis one.

I have a math course out there. And then I have a very, very basic course, a brief history of electronics with electronic basics and and we we really start our journey there. It's really, really a very basic course. That is for a basic basic understanding of what we talked about here. So anyways, with all with all that said, Take care. We'll see in the next course.

Bye bye.