 In this video, we're going to be discussing the currents in a delta configuration. If you want to understand how currents work in a y or star configuration, you can go back and look at the video on that. Or you can just remember that a phase current and line current are the exact same thing in a y. That's not true, however, when we discuss delta configurations. So here we have ourselves a delta configuration. Why is it called a delta? Well, I'll tell you. It's because it's shaped like a triangle and the triangle is a, the Greek symbol for delta is a triangle. So there you go. You're a little smarter in your Greek now than you were a couple of seconds ago. Now the difference between a y configuration and a delta is that we have the currents coming to a node here, two points to a node. And if you remember Kerchoff's current law, that you have to add currents that come to a node. So let's discuss how we're going to do that because we can't just do it arithmetically because these phases are not directly in line with each other. They're 120 degrees out of phase with one another. So first off, let's just give these guys a name. We'll call this one a phase, this one b phase and this one c phase. As I always love to do, I try to get my polarities drawn in here. So I've got a negative and a positive and a negative and a positive and a negative and a positive instantaneous polarity. Again, as I said in my y configuration video, I understand that these are not all going to be happening at the same time. They happen 120 degrees out of phase with one another. But for the purposes of our video and for the explanation for now, we're just going to say that we're going to stop time and say that this point is negative. That's positive, negative, positive, negative, positive. So with that, we've got this being my a phase. If you follow this down here, that's my b phase. And this is my c, actually line. So that's a line, b line, c line. Now we've got this. These are coils, which means that this is a source. We're going to say that this is a three phase alternator or generator. What we end up having here is the current flows from negative to positive. Remember, or sorry, positive to negative through a source. So that's why I've got the vectors drawn from positive to negative, positive to negative, positive to negative, which I do this just because it helps me tell my, teach my students how to set this up so that we can see how it looks. Now I'll get into that in a second, but here I'm just going to assign 10 apps. Let's say that for whatever reason, these lines are drawing a situation that causes 10 apps to flow across each phase. Now here we are. I've got a phase heading this direction. So we're going to say that that's a phase. B phase is heading down this direction. So that's b phase. C phase is heading up that direction. That's c phase. This just helps me add this stuff up a little better. It helps me visualize the 120 degrees out of phase with one another. Because when you look at it this way, it doesn't necessarily look that way, because you've got these small angles on the inside. But when you place them as the vectors go with the current, it actually ends up looking more like a y, just for mathematical purposes. All right? So we've got our a, b, and our c. Now I've got assigned them 10 apps at zero degrees because it's that phase there. B phase is 10 apps at 240 degrees, which it is if you follow it the whole way through. And then c phase, which is going up this direction, is going to be 10 apps at 120 degrees. And if you want to learn more about how we add vectors, because that's what we're about to get into, you can go back to the trigonometry section in this channel and see how you add vectors. But I'll discuss it a little bit as we go here. Okay, so the first thing I do, as always, is set up my x, y chart. Because what I have to do is 10 apps at zero degrees and 10 apps at 240 degrees. I can't add 10 apps and get 20 apps because they're not heading in the same direction. So I have to find parts that I can work out that are heading in the same direction. So what I end up doing is finding x and y coordinates for each one of these. So 10 apps at zero degrees is going to give me 10 on the x, zero on the y. What I'm going to be doing is adding this current to this current. So what I have to do here is I'm going to take negative 10, because I'm heading from up here to here. So I'm starting at this point, negative 10 times the cos of 240 degrees gives me 5. If I go negative 10 plus times the sine of 240 degrees, I get 8.66. Again, if you want to know why I'm using sine and cos, go back and watch the vector videos. So I end up with 15 and 8.66. That is my x coordinate and my y coordinate for my vector. What I have to do now is figure out what my resultant is just by using Pythagoras. I end up with the current being 17.3 amps. So if I take this current plus this current, A ends up getting a line current of 17.3 amps. Again, not 20, as you might think, but 17.3 because of that vector displacement with the angles. So we'll try it again. We're going to try another A phase to C phase here. So what we're going to do is I'm going to go back this one because I'm adding this current to this current. So I'm going to go negative 10 on the x and 0 on the y because it's all flat lined here. Down here, I'm going to go positive 10 times the cos of 120 and I get negative 5. I go positive 10 times the sine of 120 and I get 8.66. I end up with negative 15 and 8.66 using Pythagoras again to get my resultant, 17.3 amps. So let's take it for one more spin here. I'm going to be adding the C to the B phase here. So if I go negative 10 times the cos of negative 1, or sorry, negative 10 times the cos of 120 end up with negative 5. Negative 10 times the sine of 120 ends up with negative 8.66. Positive 10 times the cos of 240 gives me 5. Positive 10 times the sine of 240 gives me negative 8.66. Again, this is why polarities are so important. Negative 10, positive 10. I add those all up and I end up with negative 5 plus 5 is 0 and negative 8.66 plus negative 8.66 ends up with negative 17.3. So I end up with 17.3 amps. So there you go. We have added all three of these up in every configuration to figure out what the A line was, the C line was, and the B line was. And it always worked out to be 17.3 amps, which means if I take 17.3 amps divided by 10 amps, I get 1.73 or root 3. So for current in a delta configuration, I line is equal to I phase times 1.73. I could have used any different curve. I could have used 246,333 amps per phase and then figured out what the line was based off of this here. So remember, if all that didn't make sense, at least if you can remember this, I line is equal to I phase times 1.73. You're going to be good.