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Homework Help: Phase difference

  1. Sep 17, 2007 #1
    I can't believe I'm asking this, but for some reason I cannot remember this.

    Here is the basic question from my power class.

    Q: What is the phase difference between:
    [tex] v=V_{max}\sin \omega t [/tex]
    [tex] i = I_{max} \cos( \omega t - 30) [/tex]

    So we simply rewrite [itex] v [/itex] as: [itex] v = V_{max} \cos ( \omega t + 90 ) [/itex]

    Now my first thought to find the phase difference is the following. The voltage is shifted left by 90 degrees, and the current is shifted right by 30 degrees. Thus the difference is simply, 90 + |-30| = 120.

    Then I thought for a second, well if we draw the unit circle, the voltage would be rotated 90 degrees counter clockwise, and the current would be rotated clockwise 30 degrees. So the angle between them would be 240 degrees.

    Would someone please tell me what I'm missing here. Thanks.
  2. jcsd
  3. Sep 17, 2007 #2


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    cos(x+90) is the negative of sin(x). I think you want cos(x+270).
  4. Sep 17, 2007 #3

    Woops. Yeah. My bad there.

    Well for the sake of argument, say I need to find the phase difference between cos(x + 90) and cos(x - 30). Would it be 120 degrees as outlined in the method I discussed above, or 240?
  5. Sep 17, 2007 #4


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    You could say the voltage is leading the current by 120. Since cos(x-30)=cos(x+330), you could also say the current is leading the voltage by 240. 120+240=360. Phase difference is ambiguous unless you specify which is ahead or behind which. Depends on your conventions.
  6. Sep 17, 2007 #5
    "Phase difference is ambiguous"

    Good stuff. That makes sense.

    I appreciate it. Have a good one! ;)
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