Comparing phase angles between different A.C currents

In summary, the phase angle between i1 = -4 sin (377t + 25) and i2 = 5 cos (377t - 40) is 335 degrees. This was determined by converting i2 to sine form and making i1 positive as well, leading to a difference of 335 degrees between the two. It is important to pay attention to signs and use a calculator for confirmation, as small errors can lead to incorrect results.
  • #1
NewtonianAlch
453
0

Homework Statement


Find the phase angle between i1 = -4 sin (377t + 25) and i2 = 5 cos (377t - 40)


The Attempt at a Solution



I first made i2 in terms of sine: 5 cos (377t - 40) => -5 sin (377t + 50) => 5 sin (377t + 230)

Now to make i1 positive as well: -4 sin (377t + 25) => 4 sin (377t + 205)

Comparing these two now, i2 leads i1 by 25 degrees.

Although the answer in the lecture notes say i1 leads i2 by 155.

What has gone wrong here?
 
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  • #2
5 cos (377t - 40) => -5 sin (377t + 50)
Confirm this using your calculator. For convenience, set t=0. :wink:
 
  • #3
NewtonianAlch said:

Homework Statement


Find the phase angle between i1 = -4 sin (377t + 25) and i2 = 5 cos (377t - 40)


The Attempt at a Solution



I first made i2 in terms of sine: 5 cos (377t - 40) => -5 sin (377t + 50) => 5 sin (377t + 230)
There's your mistake. Do it over!
Now to make i1 positive as well: -4 sin (377t + 25) => 4 sin (377t + 205)


Comparing these two now, i2 leads i1 by 25 degrees.

Although the answer in the lecture notes say i1 leads i2 by 155.

What has gone wrong here?[/QUOTE]
 
  • #4
NascentOxygen said:
Confirm this using your calculator. For convenience, set t=0. :wink:

Hmm. I redid it.

Paying attention to the signs properly now, I got 5 cos(-40) =>

5 cos (-40 + 90) = -5 sin (-40) = > 5 sin (140)

So, i2 = 5 sin (140)

and i1 = 4 sin (205), here i1 leads i2 by 65 degrees, which is still not 155.
 
  • #5
NewtonianAlch said:
Hmm. I redid it.

Paying attention to the signs properly now, I got 5 cos(-40) =>

5 cos (-40 + 90) = -5 sin (-40) = > 5 sin (140)
:frown:
 
  • #6
OK.

So converting i2 = 5 cos (377t - 40) to sine:

5 cos (377t - 40 - 90) = 5 sin (377t - 130)

Now converting i1 to positive:

- 4 sin (377t + 25 + 180) = 4 sin (377t + 205)

205 - (-130) = 335 degrees.

So, i1 leads i2 by 335 degrees.
 
  • #7
NewtonianAlch said:
OK.

So converting i2 = 5 cos (377t - 40) to sine:

5 cos (377t - 40 - 90) = 5 sin (377t - 130)
:frown: :frown:
 
  • #8
The only way I can think of is that -4 sin (377t + 25) -> 4 cos (377t + 115)

And now 115 - (-40) = 155

Although I don't really understand what's wrong with what I was doing.

If I wanted to turn a negative sine to positive cos, don't I add 180 to get positive sine, and then add another 90 to turn that into cos?

Although the book seems to have ignored -sin, and not added 180, but just added 90 straight away to get cos.
 
  • #9
You can check each step using your calculator. For convenience, set t=0.
 
  • #10
NewtonianAlch said:
Hmm. I redid it.

Paying attention to the signs properly now, I got 5 cos(-40) =>

5 cos (-40 + 90) = -5 sin (-40) = > 5 sin (140)
Tha's still wrong.
Where did you get this from: 5 cos (-40 + 90)?
And whence this: -5 sin (-40) = > 5 sin (140) ?

Try this: cos(θ) = sin(θ + 90)

BTW the answer in your lecture notes is correct.

So, i2 = 5 sin (140)

and i1 = 4 sin (205), here i1 leads i2 by 65 degrees, which is still not 155.

- rude man
 
  • #11
NewtonianAlch said:
The only way I can think of is that -4 sin (377t + 25) -> 4 cos (377t + 115)

And now 115 - (-40) = 155

Although I don't really understand what's wrong with what I was doing.
You were writing equations where LHS ≠ RHS
5 cos (377t - 40 - 90) = 5 sin (377t - 130)
Does cos(-40 - 90) = sin (-130) https://www.physicsforums.com/images/icons/icon5.gif [Broken]

Apparently you are not using the assistance of a sketch on the x-y plane? Try it. Sketch x-y axes, and draw a vector in quadrant 1 from the origin at, for example, 20° to the x axis. The sine of an angle is its projection onto the y axis, and its cosine is its projection on the x axis. Now draw a vector of the same length in the 2nd quadrant, so it makes an angle of 20° with the +ve direction of the y axis. The projection of this vector onto the x-axis is a short length along the -ve x axis. Estimating by sight, you can see this length equals that of the projection onto the y-axis of the vector in quadrant 1. So we can say sin (x) = –cos(90°+x). The beauty of this is that we need no equations pulled up from memory!

It takes practice to become adept with this, but you can use it in both maths and science, so is well worth the investment of time to learn. Practice makes perfect. :smile:

You can even use the technique to derive those trig equations that are easily forgotten!
 
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1. What is the purpose of comparing phase angles between different A.C currents?

The purpose of comparing phase angles between different A.C currents is to determine the relationship between the currents and to analyze how they interact with each other. This can be useful in understanding the behavior of electrical systems and in troubleshooting any issues that may arise.

2. How do you measure the phase angle of an A.C current?

The phase angle of an A.C current can be measured using an oscilloscope or a phase meter. The oscilloscope displays the waveform of the current and allows you to measure the time difference between two points on the waveform. The phase meter measures the phase difference directly and displays it in degrees.

3. Can phase angles be compared between different frequencies of A.C currents?

Yes, phase angles can be compared between different frequencies of A.C currents. However, the comparison is only valid if the frequencies are the same or if they have a known relationship, such as being multiples of each other.

4. What is the significance of a phase angle difference of 180 degrees?

A phase angle difference of 180 degrees indicates that the two A.C currents are in phase opposition, meaning that they are completely out of sync with each other. This can be caused by a variety of factors, such as a faulty connection or a malfunctioning component.

5. How does the phase angle affect the power factor of an A.C circuit?

The phase angle is directly related to the power factor of an A.C circuit. A power factor of 1 (or 100%) occurs when the phase angle is 0 degrees, meaning that the voltage and current are in sync and there is no lag or lead between them. A higher phase angle results in a lower power factor, which can lead to inefficiencies and higher electricity costs.

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