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Phasor addition

  1. Feb 19, 2013 #1
    simplfy 5cos(wot+90°)+5cos(wot-30°)+5cos(wot-120°) to form Acos(cos+Ø)

    i did it this way
    5cos(wt) = 5e^j0 = 5+J0
    5cos(wot-30°) = 5e^-j30° = 5*(√3/2)+J5/2
    5cos(wot-120°) = 5e^-j120° = -5/2*(√3/2)+J5*(√3/2)

    i have only done partially the problem just missing to add them up however when i look up the solutions this portion where i convert it to Cartesian it doesn't match the solutions but i quite positive i am right, could anyone confirm?
  2. jcsd
  3. Feb 19, 2013 #2

    The Electrician

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    Gold Member

    I think the part in red needs to be changed; it should be:

    -5/2 - j5*(√3/2)
  4. Feb 19, 2013 #3
    yes that is correct it was a typo but the rest looks correct right?

    than adding these up 5+5*√(3)/2-(5/2) = 6.83
    j0+j5/2-j5√(3)/2 = -j1.83

    than magnitude would be √6.83²+1.83² = 7.07
    and tan^-1 = (-1.83/6.83) = -14.999° = -.08pi
  5. Feb 19, 2013 #4
    nevermind i found the issue but thanks
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