All right. Now on this slide here, we're looking at capacitive reactance and inductive reactance in series. All right, so everything's the same how we do calculations. But here's the important part. Okay. They subtract my inductive reactance is our equal and opposite what I'm showing you right there.

All right. So the bigger one will dominate the circuit, meaning when I subtract them, I'm left with 20 ohms. But it is in inductive reactance, meaning the characteristic is inductive. reactance so therefore, my phase shift between my voltage and current will be inductive. All right. So if you look here, we have we just do the math.

It is XML minus x sub c 120 120 volts over 60 ohms six amps. So I got six amps flowing through the circuit like I show you here. All right, and then again, just just to get the voltage across each component against ohms law, six amps times 60 ohms 366 amps times 40 ohms to 40 all right to find the total now, all right, I do or what I should say I subtract them and so I got 365 Se c minus 240 volts AC 120 volts AC. It checks. That's my voltage. I mean, that's it.

All right. So let's hang on here for a minute and we're going to I'll do the phase angle real quick for you. Okay, what I did was I put the phase relationship right there I added on I was delinquent, I probably should have done it when I did the original slide and as I went through this with us as all I need to do this, but basically we have a because of the difference here and in the in the inductance. The inductances of both components, X sub l x sub c, I take the difference. Okay, x Bell is bigger, it's 20 ohms. So I have a neck again, I have an extra Bella 20 ohms.

Then we have this phase relationship. Ship where voltage leads current by 90 degrees. All right, because it's more inductive. If it was more capacitive, which I'm sure we've got a slide, then it would be down this way. All right? So take a look at it.

Remember, like I'm showing you here. Okay, the inductive reactance and the capacitive reactance cancel each other out and the bigger one wins. Alright, the bigger one wins. That's why I put these little brackets in here. That means the absolute value, okay of the difference, which will always be positive. All right, so that's it, and we'll see on the next slide.

All right, and this, this slide here, we have a capacitor and inductance in parallel Right there. And we've assigned the reactant says Excel has 60 X sub c has 40. We've got 120 volts AC, as my source, they're in parallel across these two components. And again, I mean which game? It's like the same song over and over again, the same song. Alright, if I want to find the, the current through this branch here.

All right, it's V over x MC one, I do my math, three amps, the same thing over here. I do my math. I get two amps. All right. But here's the difference. Now, because of the phase relationship currents are opposite.

So I take the difference, and the larger one wins. So I've got three amps, I see current, and two amps of i l current flowing. I take the difference. And it is one app. The bigger one wins. So my net reactance is X sub c. All right, and here's my phase angle relationship right there.

So current leads voltage by 90 degrees. All right? My it from the source again, I get it up there. Let me stress that. That's actually the current, the real current that flows from the source. All right, is one amp.

Alright, that's it. Let me clear the slide. We'll go to the next one. Okay, now we have an RLC circuit. Okay, and then we have them all in series and XML and CNR in series. And basically the song is the same, just a little bit of different rules.

All right? My reactance is because they're in series subtract. So we call it x sub c total, meaning the the net result. And it's, we give you that X sub c is 90, and XML is 60. We take the difference of 30. All right, so now I want to find my total current, because that total current is what flows in the circuit.

Okay, remember, that's a series circuit, the rules stay the same. I stays the same in a series circuit. So now I do my math 100. Okay, over 50 Z, and we find z over here, where we have 40 squared, which is the resistor there. My net reactance is 30 squared. I go through the math and I get 50 ohms.

I do the math again, I do the math over here. 50 volts AC divided by 50 ohms. I have a current flow of two amps. Now I find the voltage drop across each component. All right, I T times r 180 volts, two amps times 90 ohms for X sub c, and the voltage across the capacitor is 180 volts. And my inductive reactance is 60 ohms.

And I do the math I get 120 volts. All right. Now remember, the voltage across each of my reactive elements X sub c and X sub l are different. They cancel each other out. The bigger one wins. All right.

So now come Over here. So I'm just going to scoot over here for a minute. 80 volts, which is the voltage across our and this is my difference. This is my x t, my equivalent on my net reactive element between the inductor and the capacitor. And again, I didn't go through every step, but when I go through that, that I'm finding my v source, which is right here. That'll do a check.

I get 100 volts, right? So it checks my phase angle. All right, is my equivalent reactance over VR. Alright, so what is my equivalent reactance it's 60 ohms divided by via I'm sorry, it's v x t over VR. So v x t or V total is. Let's stop here.

Let me look. Okay, VX T is the difference between these two voltages 180 and 120, which is 60. The voltage across my resistor is 80. So when I find the inverse tan function, it's a minus 37 degrees. Okay, that wraps it up here. So, I'm going to clear the slide off and go to the next one.

I do have some problems for you to do. I've got a worksheet and that same worksheet that's, that'll be embedded I believe after this month. also download it and do the problems. Okay, and this is slide here we're going to talk about a parallel circuit, where we have a capacitor and an inductor. And again, we give you the reactance of each component, C has 40 Ohm reactance and X sub l, which is inductive reactance has 60 ohm, right there. And basically, we're going to find the current through each branch, as we show here, and again, it's the same deal.

I see one and i l one is V over x sub c one for IC one or V over XL one for i l one. All right, we solve for the currents. three amps through it. See one and two amps through II l one. And because the capacitance and inductance are opposite we take the difference. So I see is three amps and I LS two amps we take the difference and that's one amp Alright.

So I show my equivalent circuit here, when these two are in parallel I have a net XC Alright. So now I need okay found it here. So now we we now we know that we are we have one amp we have one amp of current that actually flows out of the source and that is Ideally is X sub c. So it's the relationship is is with current, so that current is 90 degrees ahead of the voltage right here, which was okay, before we go on to the next slide, I just, I should have added this and I put it in previously, but I got it in now. If I want to find the value of my net reactance in this case, it would be a net x subsea, that's going to in this case, it's 120 ohms. Because what we'll do is it's vs divided by the supply current, which is one amp, I do the math I get 120 ohms.

Now the first thing you should say is because I like that, all I learned here before, in a parallel resistance, it's got to be less than a smallest resistor SMC is is party and That's because we're we're approaching the resonant effect. And that's all I'm going to say. We're doing a section on residence, okay. And we have something called resident when, when to when to react and Sue's are equal in a parallel circuit. We have a very, very high Z or high impedance, and we'll talk about that when we get there. But we're not quite there yet.

If you look at the ratio, it's x sub c is 40. And XML is 60. They're in parallel. We're getting in we're not quite so that's why when I go through the math, as far as the supply is looking at those two components, it's seeing 120 Ohm reactive component that has properties of capacity because the net result is is capacitive reactance as we showed you previously. All right, that's what I wanted to say. So we'll end it here and go to the next slide.

Okay, in this slide here, we're looking at a parallel circuit with an X sub l, an X sub c and an R. Here we go right here, right here and right here. All right, we have a supply voltages of 100 volts AC. And we were doing the same thing with going to find I. Alright, so I solve for my current through my resistor. It's 100 divided by 25 ohms. I get four amps, I see is one amp and i L is four amp. Now if you remember, the currents in my reactive elements, we take the difference.

So what do we have? We've got a total of three amps. And the biggest one wins. So who wins? I L. So my net inductive reactance is inductance. All right.

So now I'm going to find it. And I have to use my Pythagorean Theorem, i r squared plus i x squared, which is, which is the total. All right, so I got four squared, which is the current flow through the resistor. I take the difference. And I come up with the 16 plus nine. I do my math and when all is said and done, I get five amps.

I solve for z. All right, 100 volts AC. My total current that I found up here is five vamps, I do my math and I get 20 ohms. All right, now the only thing I need to do is find the phase angle, the phase angle to find the phase angle. It's, it's the inverse can function over IL plus IR, I L is the equivalent current that flows through the inductor that I found down here. I do my math, I get my calculator.

And when it all is said and done, I find that I have a phase reversal or a phase shift of 37 degrees. All right, so what we're saying is it's an inductive circuit. So voltage leads current or current lags voltage with the source by 37 degrees. That's it. That's it. And this is the final slide in this particular section and we'll