Comparing phase angles between different A.C currents

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NewtonianAlch
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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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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]
 
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.
 
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.
 
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.
 
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
 
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

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.[/color] 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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