# Working with Phase Angles

## Alternating Current

Working with Phase Angles

Video Lectures created by Tim Feiegenbaum at North Seattle Community College.

A phasor consists of a vector that can be rotated around a central point. Here is our phasor and it could be rotated around the central point which is right here. The length of the phasor corresponds to the peak value of the sine wave. So here we have… this phasor is going to represent the peak value of the sine wave, so if we had our sine wave, okay this is not a well-drawn sine wave, but we would have a peak value right here and we would have another peak, the negative peak, would be here and the length of the phasor corresponds to this peak value. The angle of the phasor corresponds to the instantaneous phase angle of the sine wave.

The value that we will be considering here is going to be the instantaneous value and that is going to be based on the phase angle right here. Remember that we started out at zero degrees and this was 90 and 180, 270 and such. Remember if we want to calculate the different values at the different phases those would be representative of the instantaneous values. Let's say we pick one off right here and let's say that is 45 degrees so the angle would be 45 degrees and we would be interested in what is the instantaneous voltage at 45 degrees given a peak value of whatever the peak value is.

### Example 6-14: Phasor

Okay, so here we have a phasor. The question is: What is the peak voltage? Looking at this we can say the phasor represents the peak voltage so peak voltage will be 100 volts. What is the instantaneous voltage? Here we have this value 50 volts, and that could be drawn down like this. That would also be 50 so the instantaneous value would be 50 volts and the phase angle here would be 30 degrees and that's just reading what do we have in this block of information.

### Right Triangle Relationships

A dotted line is drawn from the point of the phasor to the horizontal axis. This forms a right triangle. From right here, the point of the phasor, down to the horizontal axis and we had a right triangle form by that relationship right there. Side A is a portion of the horizontal axis. Side A here is a part of the horizontal axis. Side B represents the instantaneous voltage of the sine wave. That would be this right over here. This would represent the instantaneous voltage at a given angle. The angle of the sine wave is represented by this symbol right here, which is the Greek word Theta, and so that represents the phase angle. Side C represents the peak voltage of the sine wave. This right here, this represents C is the peak voltage.

A triangle drawn this way will always have a right triangle 90 degrees right here where B intersects the horizontal axis. Basic trigonometry is used to solve this type of problem.

Here we are going to look at right triangle relationships. Okay, the first one here we have a triangle and the hypotenuse is the longest side and across from the right angle. Okay, so here is the right angle right here and the hypotenuse is directly opposite the right angle. The angle represents the phase of the sine wave. Okay, so here we have this represents the angle of the sine wave.

The two sides are named based on the relationship to the phase angle. Here we have the phase angle right here and the adjacent angle A is so called because it is adjacent to this angle then the opposite is so named because it is directly opposite of the phase angle. Sine cosine and tangent these are calculators functions are used to solve for these values.

Phasor problems are solved using trigonometry and this is getting into right triangle relationships and we have ... there are three basic equations used to solve the phasor problems and they are the sine, cosine and the tangent. The sine angle has to do with the relations or the ratio of the opposite side to the hypotenuse, cosine has to do with the adjacent side to the hypotenuse and the tangent has to do with opposite side to the adjacent.

One of the mnemonics that you can use to remember that is this little phrase right here; Sally Can Tell, and that's for SCT. Oscar Has A Hat On Always. Okay, you can make up your own if you don't like that one but that is a mnemonic that would help you to remember that relationship.

### Calculator Operation

Scientific calculators contain tables for calculating sine, cosine, and tangent. If you key in the angle and the appropriate sine, cosine, or tangent your calculator will yield a corresponding code. Right after in your text right after trigonometric functions, there are a few practice problems. Let's do just a few of those to get a feel of this.

Let's see the first one they mention is the cosine of 28 degrees. What we will do is we would enter in 28 and would say the cosine and we get this value 0.882. Then we have the next one is actually in radians and if you are going to do that you will need to go to options here and change to radians and so here we have the sine is 3.6 radians so let's go 3.6 and we will do the sine and the value is -0.442. Now, if we had done that in degrees that would be a different value.

The next one is the tangent of 150 degrees. Now, what you need to do, it didn't specify radians, so we need to back and change this to degrees. Then our value is 150 and we're looking at the tangent so that's going to be -0.577. We'll just do a couple more here. Sine of 20 degrees let's see let's key in 20 and sine and that would be 0.342. We'll do one more tangent is going to be 175 and the tan function and that will give us 0.087. Now, there's five more. I'm going to encourage you to do these and the answers are at the very end of the chapter, so please do those.

### Inverse Trigonometric Functions

Now, this is the case where we know the sine, cosine or tangent but we do not know the angle. The function we want here is the arcsine, arccosine, and arctan. Arc means “angles whose” so the expression arcsine 0.8 means the angle whose sine is 0.8. There are some examples and I just want to do a couple of these down at the bottom there let's see let's bring our calculator back for a moment. The first one is the arcsine of 0.7. So let's go over here and first of all we will enter in 0.7 and then we are going to press the shift key and then we see the arccosine. You'll see that it is 45.573.
Now, with different calculators, there might be a second key or something to indicate the second function and usually you will press the same key only it will yield the second function and in this case, it is the arc function.

Okay, the next one is the arccosine and the value we have is 0.633 and this is a negative value and we want to get the arccosine and the value is 129. The next one is the arctangent. We are going to be doing 0.466 and shift and we want the arctan and that value is 24.9. Let's do just one more, the arcsine, and this is going to be of the value 0.0872 and we want the arcsine and so here is the arcsine and that is indicating five. And these would all be degrees that we're looking at.

### Phasor Calculations

In the next session, I'm going to stop here, the next presentation I want to walk through the practice practical problems at the end of this particular section so this will conclude. We've looked at inverse trigonometric functions. We did some inverse calculator functions. We looked at the sine, cosine, and tangent. We looked at a mnemonic to remember them, we looked at right-triangle relationships, and we considered phasors, and the subject is working with phase angles, and the next section will be actually solving some problems with them.

Video Lectures created by Tim Fiegenbaum at North Seattle Community College.

1 Comment • james7701 February 11, 2023

at about time 9:28 when doing the inverse trigonometric function of ARCSIN 0.7 (bottom of page 173) you used the ARCCOSIN of 0.7 instead of using the ARCSIN of 0.7

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