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Electrical: Reduce the expression to the form V_mcos(wt+theta)

  1. Apr 23, 2013 #1
    1. The problem statement, all variables and given/known data


    Reduce the Expression:

    15sin(wt-45°) + 5cos(wt-30°) + 10cos(wt-120°)

    to the form Vmcos(wt+θ)

    2. Relevant equations



    3. The attempt at a solution



    My theta value at the end isn't coming out right.

    My first step was to put make sure each term was in terms of cosine, hence:

    15cos(wt-45°-90°) + 5cos(wt-30°) + 10cos(wt-120°)

    Which in phasor form is:

    15∠-135 + 5∠-30 + 10∠-120

    Then I used cos and sin to find the real values and the j values

    15cos(-135) + j15sin(-135) + 5cos(-30) + j5sin(-30) + 10cos(-120) + j10sin(-120)

    = -11.28 -j21.77

    Which in a phasor format is:

    24.52∠62.6

    24.53 = (11.282+ 21.772)1/2

    62.6 = arctan(21.77/11.28)

    But apparently the solution is:

    24.52∠-117.39°

    What am I missing?

    Any help would be appreciated. Thank you
     
  2. jcsd
  3. Apr 23, 2013 #2

    gneill

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    Staff: Mentor

    Note the signs of the complex components of your phasor sum. In what quadrant will the phasor lie?
     
  4. Apr 23, 2013 #3
    Are you saying that because I have -11.28 - j21.77 as negatives that the phasor will be in the third quadrant and consequently the angle made with the postive x-axis will be 180-62.6?
     
  5. Apr 23, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    The phasor will be in the third quadrant, yes. You can either add 180 or subtract 180, but generally the convention is to choose the result with a magnitude less than or equal to 180. Alternatively, employ the atan2(y,x) function rather than the atan function (if your calculator has it); it automatically sorts out the quadrant from the given y,x data.
     
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