 So this is going to be a walkthrough of a series RL circuit and what I mean by the RL is we have Resistance so there's a resistor there and we also have this resistance here So that's the R portion of the circuit and then we have the L which is the inductance there Now you might notice I've got this little dotted box around this one over here What that signifies is that this is a coil. This is the whole coil It has inductance and it has resistance in it Now as we know a coil is literally that it's just a coil of wire that is wrapped around a core All wire has resistance. So we need to signify that somehow. So that's why we do this We have the resistance there the inductance there we put this box it signifies that that is the whole thing So what we're going to do is I'm going to sign some Some things we're going to try to figure out here and we're going to go from here We're going to walk through the step of each one of these. So let's take a look at what I'm talking about So here we go. I've assigned values to the circuit I've said that this has got a source voltage of 120 volts 60 Hertz that this resistor here is 75 ohms The coil itself has a 300 millihenry inductance and a 25 ohm resistor in it And what we're going to try to do is we're going to try to figure out the impedance of the entire circuit We're going to try to figure out the current through the entire circuit We're going to figure out what the voltage is across the coil. So from that point to that point Then we're going to try to figure out what the power is of the entire circuit The theta of the circuit or the phase angle of the entire circuit the theta or the phase angle of the coil itself We're going to figure out what the reactive power is of the circuit And then we're going to figure out what the apparent power is of the circuit So let's take a look at what's next Before we go any further we need to quickly go over the formulas that we're going to be using in this walkthrough So here we have the formula for inductive reactants, which is XL, which is measured in ohms It's 2 pi times the frequency times the inductance. So we'll be using that Then we're going to get into a power formulas We have I squared R and it's important that it is the R because that determines the watts of the Circuit so that is the resistive power or the true power or the active power of the power being dissipated Then we also have the I squared XL that gives us our reactive power and then we have the I squared Z Which is going to give us our VA So let's take a look at what's happening in the circuit now with these formulas in mind Now whenever I'm working on an RL circuit or any kind of circuit that has inductance and reactants I love triangles. So I get myself a triangle drawn up here. I've got this triangle drawn up here It's a right triangle. I add up my total resistance in the circuit. So I've got 75 ohms here I've got 25 ohms here. These guys are heading in the same direction so I can add them Arithmetic 75 plus 25 equals 100 ohms of total circuit resistance The next thing we've got to do is figure out what is my reactants in the circuit my inductive reactants So what we're going to do is take that formula 2 pi FL, which is going to be 2 pi Which is 6.28 times the F which is 60 Hertz times the L which is 0.3 or 300 millihenries And then we're going to figure out our next step We punch the numbers in there and we get 113 ohms That means that we have 113 ohms on this side of our triangle Now we've got 100 ohms of circuit resistance We've got 113 ohms of circuit reactants. Now we can figure out what our impedance in this circuit is Using Pythagoras' theorem, we just use 100 squared plus 113 squared gives us the square root of and that will give us what our hypotenuse is. In this case if you're following along it ends up being 151 ohms That is our total circuit impedance That is the total opposition to current flow in this entire circuit because we've added these resistors Up with this guy, so we have to add them vectorially. We get 151 Everything is unlocked. So we are ready to go on this So let's plug in that 151 ohms of our impedance Now and I've got that all here as well. Now we need to figure out what our current is going through this circuit So let's do that. All we have to do is take our source voltage and divide it by our overall opposition to current flow And that will give us our total current that is flowing through this circuit In this case that works out to be 795 milliamps. So I've got 795 milliamps there as well and to keep in mind This is a series circuit. So the current stays the same if I have 795 milliamps through here It means that I've got 795 milliamps across this inductor across the resistance of the inductor And down here and across this resistor here as well and back 795 milliamps. So let's talk about the voltage of the coil. So we're looking for the voltage across this entire thing Now what we're going to do is we already worked out a little impedance triangle for this guy. Let's take a look at what that looks like So what I've done is I've taken 25 ohms of the circuit resistors not the circuit but the inductor resistance I've taken 113 which is what this guy was of this one That squared plus that squared gives me this squared. So that's a hundred and sixteen ohms of impedance just across the coil we worked up before across the entire circuit Now we're working out this the impedance across this coil. So we do that that ends up being a hundred and sixteen ohms Now with that in mind all we have to do now is take current which remains constant Multiply it by the impedance of the coil and we should be able to get the voltage of the coil So point seven nine five times a hundred and sixteen ohms gets me ninety two point two volts and That gives us that voltage of the coil. So we're off and away Now we need to figure out what the power is of the entire circuit Well in order to figure out what power is power is the power being dissipated across this resistor and this resistor So the easiest way to do that is I've got this resistance Which is 75 ohms and this resistance which is 25 I can add those two together and I can use that formula we looked at before which is I squared R So I'm going to take point seven nine five squared and multiply it by 100 to get my power being dissipated in this entire circuit Again point in the seven nine five squared Which is the current times a total circuit resistance gives me sixty three point two watts of power being dissipated That is only across the resistive element watts are only dissipated across resistance not across inductance Now let's talk about what the Theta is of the circuit Now we're going back to this triangle from before where I've got a hundred ohms of resistance this plus this I've got a hundred and thirteen ohms of reactants, which is this I get my overall impedance Which is over here at 151. This is a circuit triangle. It's my impedance triangle I can figure out what the theta is and I like to just go 100 divided by 151 and then I inverse coset to get my angle Which in this case works out to be 48.5 degrees So we put that in there now we want to figure out what the theta is just of the coil alone So in order to do that we need to go back to our impedance triangle of just this coil So here I've got just the coil alone. I've got 25 ohms of resistance of this coil I've got a hundred and thirteen ohms of reactants, which gives me a hundred and sixteen ohms of impedance of the coil I'm going to use cos again twenty five divided by a hundred and sixteen and I inverse cos that Boom I get seventy seven point six degrees just for the coil alone. That is my theta just for the coil So there we go. We've got seventy seven point six degrees there for the theta of the coil next up The bars of the circuit Well, the bars of the circuit if you notice we only have a hundred and thirteen ohms of reactants in this entire circuit We only have that one coil here So to determine the bars again, I can use that I squared formula But I'm using I squared times XL to figure out what my bars are because bars are reactive Which is just across the inductive component of the circuit Let's take a look at what that looks like I squared point seven nine five squared times the reactants, which is a hundred and thirteen ohms gives you seventy one point four bars and That gives me my total bars for the circuit next up is my VA Now when I did my VA before I had my impedance triangle I worked out my total circuit impedance is a hundred and fifty one ohms All I have to do then is go I squared times a hundred and fifty one ohms to get my apparent power of the circuit So I punched that in there. I got point seven nine five squared times a hundred and fifty one and I get Ninety-five point four VA now another way you could go about it is VA is volt amps I know we always call it apparent power, but I always look at this as VA volt amps I could take this current and multiply it by a hundred and twenty volts I'm gonna get the same answer. So that's the awesome thing about these circuits is you can double-check your work So there you have it. We have a VA of ninety-five point four VA So these circuits once you break them down into the components. It's not that difficult to work through So the next step is we're gonna talk about power factor in our next video And then we'll be putting these things into a parallel circuit later on down the road