This is the second half of section 6.3. We'll be doing some Phasor Calculations. This will be nothing but calculations and if you are very familiar with this trigonometric functions you may want to just skip over this portion. If you're not you may want to pay special attention. We will be solving figure 6.27, 28, 29, 30 and 31 and these are all the practice problems at the end of section 6.3.
Okay, we start out with figure 27 and we are given this triangle. We are given the opposite side of 15, the adjacent of 25 and we are asked to solve for the hypotenuse and the phase angle. So, what we could do if we look at our sine functions, we'll see if we are given the opposite and the adjacent, we'll probably want to use the Tan function. The Tan function=Opposite/Adjacent. We are not solving for the ratio here. We already know the ratio.
It's going to be 15 over 25 so we could say we are solving for Tan=15/25 but like I say we have the ratio but we want the angle so actually in this one we would use the arctan so let's grab our calculator and first of all let's say 15 divided by 25 gives us the ratio. Then we want to know the arctan so we'll go shift and arctan and that gives us the degree: 30.96. So, that's going to equal 30.96 and that is our degree so that is our.... We know that the angles is 30.96 degrees.
So now we want to solve for the two, or the unknown over here and we could do this in a number of ways but we want to have one that includes the hypotenuse. The sine, if we say Sin=0/H and we know what the opposite is, it is 15. And we need to know, let's see, we know that the degrees are 30.96 but we do need to know what is that ratio for Sin and so we could go in. Let's go again with our calculator and let's say, let's go in with our value that we already got 30.967, let's get the Sin of that and that is going to be 0.514.
We'll say 0.514 is going to equal, let's see. We are given the opposite is 15/H and we are solving for H. Let's see, we could just say that H is going to equal 15/0.514 but let's actually solve it algebraically. So what we could do, we could say, we could multiply this side by H/1 and this side by H/1 and this would eliminate the H over here so then we'd have a H*0.514=15 and then we could divide each side by 0.514 and that would say H=15/0.514 and so if we got our calculator and we said 15/0.514. We would get 29.183. And that should be our solving for H right here. So we said our angle was 30.96 and we said our hypotenuse was 29.183. Now in the end of this chapter, all of the answers, now you could check your answers.
But we could do it another way. We could use the Pythagorean Theorem and that would say that the square root of the adjacent squared plus the opposite squared will equal the hypotenuse, so the square root of this. So let's do that and use that as a check, so we could say the adjacent square, let's see that's going to be 252 plus the opposite squared which is 152 equals this value and then we said the square root of that would equal the hypotenuse so the square root of that and there it is 29.1. We're very close and we had a little bit of rounding because we had several decimals that we didn't include. But 29.15 is very close and that confirms our answer for figure 27.
Let's go to the next one. Figure 28. In figure 28, we are given 75 degrees as the phase angle, the adjacent angle is four and so we need to solve for B and C. Now we could address this a number of ways. Let's see, we want to address, either the opposite or the hypotenuse. Why don't we take the tangent. Let's do the Tan and Tan=Opposite/Adjacent. And we have the adjacent value which is four, but we don't have the degrees, so we can solve and get the ratio for the tangent. So let's do that. Let's go and say the angle is 75 and the tangent is 3.732. So we will say that 3.732 equals the opposite over four. And if we were to solve for it, we could take this side times four and this side times four and we could find out that the opposite is going to equal 3.732 times four. And why don't we grab our calculator and we take that value times our and we get 14.928.
And we were solving for the opposite, so that is this number right here, 14.928, and now we need to solve for, let's see we need to solve for the hypotenuse, and we could say… Let's say we could use the sine function to do that. Sin equals the opposite over the hypotenuse. The opposite side is 14.928. The hypotenuse is unknown. The sine function, let's get our sine function. We have 75 degrees, and we take the sine of that, that is 0.965. We have 0.965 equals 14.928 over H. And we could solve for this and we're going to find that H is going to equal 14.928 divided by 0.965, and if we do that we are going to use our calculator to see what that is. We have 14.928/0.965 and that equals 15.46. And that would be our hypotenuse. And so again we could test our values. We say the opposite is 14.928, hypotenuse 15.46, and the adjacent is four. So if we go in with our calculator, we could say four squared plus 14.928 squared equals… And then if we took the square root of that, that is 15.4, and that is what we would expect. Again the answers are in the back of the chapter.
Okay, work figure 29. Here we have 10°, we're given as the phase angle, the opposite is 1000 and we're solving for a and c. Let's see, we need to solve for either the opposite or the hypotenuse and we are given the, excuse me either the adjacent or the hypotenuse, so let's see. We're given the opposite one that we use at the sine; sine = the opposite over the hypotenuse. The opposite we're given is 1000 and we're solving for H. Let's see, our sine is 10. We have a 10° phase angle. Let's get our calculator, and we need to get 10 and the sine, and we get 0.173. So we say 0.173 is going to be 1000 over H. And if we were to solve for this, we would find that H is going to equal 1000/0.173. So let's go ahead and do that. So we will say 1000/0.173 equals 5780. And that will be our hypotenuse.
Now we need to solve for the adjacent angle and the adjacent angle. How shall we do that? We know the opposite and the hypotenuse. We want to know the adjacent. We could do a tan function here, which is the O/a, we have the opposite which is 1000. We would need to solve for the tangent of 10°.
Let's go up here and say, 10 tan and we get 0.176 so, 0.176 would equal 1000 over the adjacent, and we could go ahead and solve that. But the adjacent angle will be 1000/0.176, so let's solve for that. We have 1000/0.176=5682, and that will be the adjacent angle, 5862, remember right here to 5682. Again we could do our Pythagoras theorem just to confirm that. Grab our calculator and we'll say, 5682, or we already have it there. Let's square that (5862^2) plus 1000^2 and that equals… Then we'll take the square root of that and 5716, that's close to this value, we did a little bit of rounding because we didn't use the decimal points but we're quite close to that value and that will be satisfactory.
Okay, solve figure 30. Figure 30, what have we got here? Here we have a phasor and we're looking at, if we look at this, we can draw some relationships to what we have just been doing. We have the VP, this is our phase from the beginning of this chapter and this is the hypotenuse that we have been talking about. We also have the opposite angle, and this is the instantaneous volting, and this would be 10 volts. If we look at our, sine, cosine and tangent, we have the opposite and the hypotenuse. That would equate to the sine. We have the two values, the tan and the 25. What we're solving for here is, what is the value of the angle? We don't know what the value of the angle is.
The sine function, now remember when we do sine, we say sine of degrees and that gives us the angle or the ratio, but we have the ratio, so in this case we're going to do the arcsine of 10 over 25 and that should give us the angle. Let's go in and again grab our calculator and first of all we will say tan divided by 25 gives us our ratio. Then we're going to take the arcsine of that and that gives us 23.57. That would be 23.57 degrees and that would be our answer for what is the angle, given 10 volts instantaneous over 25 volts peak.
Then our final one here, we're going to be looking at a different part of the phasor and in this case, we have the voltage peak is 25 volts, the instantaneous value is 45, but again, we have the same two variables. We have the hypotenuse and the opposite and again we would be using the sine function and the, again, but this time we have a -45 volts, -45 and the peak value is 100. And again we have the ratio so what we would want to do here is the arcsine. Again, let's grab our calculator and first of all we need to say, 45 and it's going to be a -45. We need to keep our signs accurate here, divided by 100.
That would give us this value, then we need to do the arc sine of that, so we go shift and arcsine. We're going to get -26.7°. That will be our answer for this. Keep in mind, all of the answers to these practice problems are at the back of the chapter. You have many more problems and the regular problems at the end of the chapter and those are in our part of the assignment. This was just a practice session with phasors and we looked at figure 31, 30, 29, 28 and figure 27.
Video Lectures created by Tim Fiegenbaum at North Seattle Community College.
by Aaron Hanson
by Robert Keim
by Robert Keim
by Robert Keim