 In this video, we're going to do a walkthrough of some coils, a parallel RLC circuit, but this time in my previous video, I did like pure R, pure L, pure C. Now I've got myself a situation here where I have some inductance and some resistance in series, but then those guys are in parallel with that coil. So again, it might look a little daunting, but it's not that bad. We just have to remember how much we love triangles. So what I'm going to do is I'm going to sign some values for this, and we're going to work through this together, so don't worry. So here's the values I've given. I've given 240 volts as a source voltage. I've given myself an inductive reactance on this coil of 15 ohms, a resistance of 10 ohms, an inductive reactance of 20 ohms here, and resistance of 18 ohms on that coil there. Now the first step, whenever I'm working through one of these, is I draw triangles up. I draw a triangle below each branch, and then we'll eventually have one triangle over here, one larger triangle. So let's take a look at what happens. Our first step is to draw those triangles. So now we've got our triangles drawn. Our next step is we need to figure out power, because we know that the best way to go about working out these circuits is to determine power. Well, I don't really have anything to work with at this point. So what I have to do is I have to figure out what the Z is of each coil. I have reactance, and I have an ohmic value, so I've got to figure out the Z. So let's go ahead and do that. So I take 15 squared and 10 squared, and I get 18 squared. At this point, you should understand how we get that. It's just an impedance triangle. So that's a Z of 18 ohms for this side here, for this branch. Then what I can do is I can take this 240 volts, divide it by the 18 ohms, and I get 13.3 amps running just through this branch right here. Just this guy right here. So our next step is going to be to work out this branch. So 20 squared plus 18 squared gives me 27 squared, 240 volts divided by 27 gives me 8.9 amps. Now I've got current through this branch, and I've got current through this branch. So I can go ahead. Now I've got current, and I've got these values. I can easily build myself power triangles down here. So let's take a look at our first one here. Using the I squared method, I go I squared times XL gives me my bars of 2.7 kVar. I squared times 10 gives me 1.8 kilowatts down there. Now one thing I never do is I never work out the VA of the branch, only because I don't need that number ever. And if I have it there, it just gets confusing, and then I'm going to want to add it to something, and I can't. What I can do is I can add up my bars, and I can add up my watts. I can never add up my VA's from the triangles because they're heading in different directions, so I can't do that. I move over here to the next branch, and I do the same thing. 8.9 squared times 20 gives me 1.6 kVar. 8.9 squared times 18 gives me 1.4 kilowatts. So now we've got our power triangles down here. What's next, you ask? I get my large triangle up there. I jokingly say to my class all the time it's one triangle to rule them all because it is going to be the sum of all these guys put into one giant triangle, which will determine what my line current is. So let's go ahead and add this. I'm going to go 1.8 plus 1.4 to get my kilowatts down here. 2.7 plus 1.6 to get my kVars here. So I've done that. I get 3.2 kilowatts on the bottom. I get 4.3 kVar on the side. Again, using the power of Pythagoras and how much we live triangles, I get 5.4 kVar there. We're getting close to the end here. It's not that hard. See, once we start building these triangles up, it gets a lot easier. Now I can easily determine what my current is, my eye line, because I have a source voltage over here. I have my kVar. I can just take my kVar divided by my source voltage. 5,400 divided by 240, and that gives me my line current, which is 22.5 amps. So that is my line current. Now to determine my Z, all I have to do is I've got source voltage here. I've got my line current here. So I can take 240 divided by 22.5, which is what I've got written here. And that determines the impedance of the circuit. See, there's no need to build impedance triangles. There's no need to do current triangles. Again, power triangle solved everything for me. Power is my friend. So my overall impedance now is 10.7 ohms. The only thing left really to figure out now in this circuit is the power factor of the circuit and the theta or the phase angle of the circuit. So let's take a closer look at this bigger triangle here. Now I've got the big triangle. I've got 3.2, 4.3, 5.4. I've got my triangle here. If I want to determine my power factor, all I have to do is go 3.2 divided by 5.4, because power factor is just a ratio of watts to VA. That gives me a power factor of 0.59 or 59%. And it was mostly inductive, so we say it's lagging, because the current is lagging the voltage. Now the nice thing about this whole power factor thing is I can go this guy, which is this ratio, a ratio of this guy over this guy. If we're working out the phase angle, which is this fellow right here, that is adjacent over hypotenuse. It's close. So all I have to do is take my power factor of 0.59, inverse-coast that, and that will give me my angle of the entire circuit. So there we go, 53.6 degrees lagging. Not too bad, because once you have this one triangle to rule them all, you've got everything else you need. So looking back at our circuit, I got rid of all the mess of the triangles. We've determined what our line current was. We've determined what our z was. We've determined our theta or a phase angle and our power factor, all using power triangles. Please, I'm begging you, do not try to build yourself impedance triangles. Please don't try to build yourself current triangles. It can be done, but then you end up with a ton of vectors and it's super-duper confusing. And it's not fun. It takes away from the whole magic of these circuits. That's it. If you need help with the next one, the next video is going to be on power factor correction capacitors.