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Comparing phase angles between different A.C currents

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the phase angle between i1 = -4 sin (377t + 25) and i2 = 5 cos (377t - 40)


    3. 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?
     
  2. jcsd
  3. Apr 19, 2012 #2

    NascentOxygen

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    Confirm this using your calculator. For convenience, set t=0. :wink:
     
  4. Apr 19, 2012 #3

    rude man

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    There's your mistake. Do it over!

    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]
     
  5. Apr 19, 2012 #4
    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.
     
  6. Apr 20, 2012 #5

    NascentOxygen

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    :frown:
     
  7. Apr 20, 2012 #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.
     
  8. Apr 20, 2012 #7

    NascentOxygen

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    :frown: :frown:
     
  9. Apr 20, 2012 #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.
     
  10. Apr 20, 2012 #9

    NascentOxygen

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    You can check each step using your calculator. For convenience, set t=0.
     
  11. Apr 20, 2012 #10

    rude man

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  12. Apr 20, 2012 #11

    NascentOxygen

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    You were writing equations where LHS ≠ RHS
    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!
     
    Last edited by a moderator: May 5, 2017
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