 In this video, we'll be doing a complete walkthrough of calculating out our circuit here with three motor loads in parallel, which is something that would be very close to a real-life situation. And then we're going to be discussing this capacitor over here as well. We'll be determining what the I line is, your overall current in the circuit with these three loads in parallel, as well as what the overall power factor of the circuit is with these three loads in parallel. And we'll be determining the size of capacitor needed to correct the power factor to 95%. You can look right now and see that we have a 56% and a 71% power factor. So we're going to be definitely below the 95% that we're looking for. So we're going to size this capacitor in k-bars to get to that. And once we have sized the capacitor, we're going to see what our new line current will end up calculated out to be. We'll probably see a significant change in current, hopefully, going down. We're going to start with our five horsepower motor. First off, we're going to be getting rid of this five horsepower. This five horsepower is the output power, but we actually need to figure out what the input power is in this circuit. And we'll do that utilizing our efficiency. We're going to use our efficiency formula, output over input equals efficiency. So what we'll do first off is we're going to convert our five horsepower to watts. So five times 746 equals 3,730 watts. That's our output power. Then we take what we know and we put into the formula. We know that our output power is 3,730 watts. We don't know what our input power is. We're trying to calculate for that. But we do know our efficiency is 79% or 0.79. We're going to cross multiply and get 0.79 times input is equal to 3,730. We'll lay that out there. And then we need to get input alone, so we're going to divide 0.79 out of there. So our input is equal to 3,730 divided by 0.79, which calculates out to be 4,721.5 watts. This is our input power, which is the power that we are concerned with. So what we're going to be doing is taking each branch and we're going to build a triangle for it. So we calculated out that the true power being used at this motor is 4.7 kilowatts. We see here that we have a power factor of 56%. If we take that power factor and change it into a 0.56, which is the same as 56%, and we inverse-coast that, we're going to get an actual angle. In this case, it's a coincidence, but we end up with a 56% power factor actually gives us an angle of 56 degrees. This is not always the case. This is just in this example. It worked out that way. Then we go ahead and we use tangent and we figure out that we have 6.9 kV on our side. So we have a reactive power of 6.9 kV. We don't need to calculate out the VA, but we do need to calculate out the VARS and the watts. Next up, we build another triangle for motor B. We do know that we have 13 kilowatts on the bottom. We do know that we have 60 degrees as an angle. Again, using tangent, we can calculate out that the reactive VARS are 22.5 kV. And for our third triangle, we have 21 kV, which is what calculates out that this 71% ends up actually by converting that 0.71 inverse-coasting that we get an angle of 45 degrees. Using those 45 degrees, we can calculate out that we have 21 kW at the bottom. And that rings true because if you have a 45-degree angle, both your opposite and your adjacent side will be the same. Now we're just going to go ahead and we're going to add up all of our watts together because they're all heading in the same direction. They're all in the bottom of the triangles. So 4.7 plus 13 plus 21 gives us 38.7 kW on the bottom. 6.9 plus 22.5 plus 21 gives us our total kVar of 50.4 kVar. Using Pythagoras' theorem, 38.7 squared plus 50.4 squared, the square root of that gets us 63.5 kVa. And then we can take 38.7 kW divided by 63.5 kVa inverse-coast that and we get an angle of 52.5. So this gives us the triangle that we have right now. It's got a kVa of 63.5 kVa with a total kVar, a reactive power of 50.4 kVar, with a total true power of 38.7 kW with a phase angle of 52.5 degrees. We can use that phase angle to calculate out what the power factor is by taking the cos of that. The cos of 52.5 degrees is equal to 0.61, which means in this case that our power factor is 61% lagging, meaning that the current lags the voltage by 61%. We can take our total kVa because it's our kVa that determines line current and we take 63.5 kVa or 63,500 va and divide that by 480 to get our overall line current. In this case that works out to be 132 amps. So that is the triangle that we have. We calculated out all these values. We calculated out that we had a 61% power factor. This is the triangle we have. Now what we need to do is calculate out the triangle that we want. We know that our kW don't change because they're based on the resistive element. So the kW move over, so we have 38.7 kW, but now we know that our power factor has changed. This 0.95, that's completely changed from the 0.61 because this is what we had or have. This is what we want. Then we have enough that we can calculate out different sides here based on the two values that we have. What we had, what we want. We can take 38.7 and divide that by 0.95 as a shortcut to get our overall va of 40.7. Then we can use Pythagoras, 40.7 squared minus 38.7 squared gets us to 12.6 squared. So in this triangle we have 50.4. In this triangle we want 12.6. So in order for us to get that, we're going to take 50.4 and subtract 12.6. That will tell us that we need a capacitor that is 37.8 kW to be able to take this from the 50.4 down to the 12.6. So we'll go ahead and plug everything that we have over there. 37.8 kW is the size of the capacitor we need. Here's the triangle that we wanted. Now it becomes the triangle that we have. So again, this is not to scale this 0.95. It should be actually a lot smaller down there. But you get the point. We have a 12.6 kW, 38.7 that never changed and it definitely a new kVA. So we're going to use this kVA to calculate out the new current. We do that by taking 40.7 kVA or 40,700 divided by 480 and that will give us our new iLine of 84.8 amps. And if we compare that to our old current, it's a substantial change. Our old current, of course, being 132 amps. So now we're running the same system. Motors are running exactly the same that they did before. But our iLine current is now 84.8 amps as opposed to the 132 amps. It's a significant change. That means that our feeder lines can be smaller because they have to handle less current. It means that the i squared R loss is being lost in those feeder lines are substantially smaller because we are running at a substantially lower current. So you're starting to see the value of a power factor correction capacitor and the significant changes that it can make.