 So in this video, we're going to walk through how to calculate power flows for a three-node network. So our network is going to look something like this. So we're going to have two generators, one at node 1 and one at node 2, and then we're going to have a load at node 3. And we're going to have some lines connecting our generators and our load so that our network looks like a triangle. So the process of figuring out power flows in this three-node network consists of three steps. The first step is that we have to calculate the distribution factors for each of the three lines for the power injected by G1. So we're going to calculate the following things. So we're going to calculate the distribution factor on line 1 to 3 from G1. We're going to calculate the distribution factor on line 1 to 2 from G1. And we're going to calculate the distribution factor on line 2 to 3 from G1. The second thing we're going to do is we're going to calculate the distribution factors for generator 2. Okay, so we're going to calculate the distribution factor on line 1 to 3 from g2. The distribution factor on line 1 to 2 from the power injected by g2. And the distribution factor on line 2 to 3 from g2. The third step is we're going to use our flow equations using the power injected by g1 and g2 and the distribution factors to calculate the power flows. So the first step is we're going to have to calculate the distribution factors for the power injected by g1. So the first step is to calculate the distribution factors for g1. So to do this, we need a set of resistances. So for this example, I'm going to assume that the resistances on all three lines are the same. So line 1 to 3 has resistance r, line 1 to 2 has resistance r, and line 1 to 3 has resistance r. So if we think about how the power injected by g1 gets to the load at node 3, there are two possible paths. Some of the power injected at node 1 is going to go straight from node 1 to node 3. So this is a path with resistance r. The rest of the power injected by g1 is going to go from node 1 to node 2 and then through node 2 all the way to node 3. So the second path is going to have a total resistance of 2r, r plus r. And so we can use these resistances to calculate the distribution factors. So first we'll do the distribution factor for g1 for line 1 to 3. So we're going to have the distribution factor for line 1 to 3 from g1. And this is going to be equal to the inverse of the resistance of the path that line 1, 3 is a part of. And so this is this shorter path with resistance r. So we'll take the inverse of that divided by the sum of the inverse resistances of both paths. So we have one path that has resistance r, that is the direct path. And then we have a second path, the indirect path, which goes from node 1 to 2 and then node 2 to 3 that has total resistance 2r. So all of the r's are going to cancel here and we will get 2 thirds. Next we're going to calculate the distribution factor for line 1 to 2. And again the distribution factor is going to be the resistance of the path on which line 1 to 2 lies. So this path is the indirect path with resistance 2r. So we're going to have 1 over 2r in the numerator. And then we're going to have the sum of the inverse resistances of both paths in the denominator. So the same thing as the denominator for the distribution factor on line 1 to 3. So we're going to have 1 third. And then the distribution factor on 2 to 3 from g1 is going to be exactly the same as the distribution factor for line 1 to 2 from g1 because lines 1 to 2 and 2 to 3 are part of the same path from g1 to the load. So the distribution factor on line 2 to 3 from g1 is going to be equal to 1 over 2r divided by 1 over r plus 1 over 2r. And that's going to be equal to 1 third. So the next step is to calculate the distribution factors for g2. So from g2 there are two paths by which the power injected at g2 can flow to the load at node 3. First is the direct path from 2 to 3 which is going to have resistance r. The second path is the indirect path where the power injected at node 2 flows through node 1 and then to node 3. So the indirect path is going to go from node 2 to node 1 all the way to node 3 and this path is going to have total resistance 2r. So the distribution factor on line 1 to 3 from g2 is going to be 1 over 2r because that's the inverse of the resistance of the longer indirect path of which line 1, 3 is a part divided by 1 over r, the inverse of the resistance of the short path and 1 over 2r, the resistance of the indirect or longer path. And so we are going to get 1 third. Now the distribution factor for line 1 to 2, we have to be a little bit careful with this one because remember we often define that flow from node 1 to node 2 is in the positive direction and flow from node 2 to node 1 is in the negative direction. So the distribution factor on line 1 to 2 from g2 is actually going to have be a negative number so there's going to be a negative sign there and so that's going to be equal to 1 over 2r divided by 1 over r plus 1 over 2r is equal to negative 1 third. So we see that in the case of g2 the distribution factor on line 1 to 2 is equal in magnitude as the distribution factor on line 1 to 3 but opposite in sign because flow going from node 2 to node 1 is in what we call the negative direction. So the last one, the distribution factor on line 2 to 3 from g2 is going to be equal to the inverse resistance of the short path so that's 1 over r divided by the same denominator as in the other distribution factors 1 over r plus 1 over 2r and so we are going to get 2 thirds. So our last step is to use the distribution factors that we calculated and the assumed production at g1 and g2 to calculate power flows. So for this example I'm going to assume that g1 produces 50 megawatts and g2 also produces 50 megawatts. So remember that the flow equation says that the flow on line j to k is going to be equal to the distribution factor for line jk with respect to g1 times the output at g1 plus the distribution factor on line j to k with respect to g2 times the output at g2. So we're going to use our outputs of 50 megawatts at each generator. The distribution factors that we've already calculated to calculate power flows on all three lines. So first we'll do line 1 to 3 so the flow on line 1 to 3 is going to be equal to the distribution factor on line 1, 3 from g1 which was 2 thirds times the output of g1 which is 50 plus the distribution factor on line 1, 3 from g2 which was 1 third times the output of g2 which is 50 and so we get 50 megawatts of flow on line 1, 3. On line 1, 2 we're going to have the distribution factor on line 1 to 2 from g1 which we had calculated was 1 third times the output of g1 which was 50 plus the distribution factor on line 1, 3 from g2 which we had said was minus 1 third. Remember it's minus 1 third because the flow goes from node 2 to node 1. So 1 third times 50 plus minus 1 third times 50 is equal to zero so the flow from g1 on line 1, 2 and the flow from g2 on line 1, 2 exactly cancel each other out. And finally on line 2, 3 we're going to have the distribution factor on line 2, 3 from g1 which was a third times 50 the output from g1 plus the distribution factor on line 2, 3 from g2 which we said was 2 thirds times 50 and so that's going to give us 50 megawatts. So in sum we find that flow from node 1 to node 3 is 50 megawatts. Flow from node 2 to node 3 is 50 megawatts and flow from node 1 to node 2 is 0 megawatts.