Converting phasor to polar form

Discussion in 'Homework Help' started by TsAmE, Mar 17, 2011.

  1. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    For the following three phasors


    A = 6 – j8
    B = 60 + j80
    C = 30 – j4


    Perform the flowing operations and give the final answers in Polar form:

    A + B – C

    Attempt:

    A + B - C
    (6 - j8) + (60 + j80) - (30 - j40)
    36 - j8 + j80 + j40

    M = (A^2 + B^2)^0.5
    =((6 - j8)^2 + (60 + j80)^2)^0.5

    I carried on simplifying this then got the following:

    (3636 + 9504j + 6464j^2)^0.5

    I am not sure what to do next
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,145
    1,791
    Not quite
    The real part of A + B - C is (6 + 60 - 30) = 36
    The imaginary part of A + B - C is (-8 + 80 - (-4)) = 76
    So the resultant phasor is 36 + j76

    |A+B-C| = √(36)^2 + (76)^2) = √(1296 + 5776) = √(7072) ≈ 84.1

    The resultant phasor is in the first quadrant so the angle will be
    > 0° and < 90°

    The angle is arctan(76/36) ≈ 64.65°

    So in polar form it would be:

    84.1\angle64.65^{\circ}

    If C is actually 30 - j40 you should be able to follow my example for 30 - j4 above
     
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