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Balanced Three Phase Load System

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data

    b17d9a210e.png
    f = 60 Hz
    pf = 0.8 (lagging)

    2. Relevant equations

    Z(L) = 2 (pi) * f * L
    Given z = x + iy
    tan(angle) = y/x
    pf = cos(angle)

    3. The attempt at a solution

    I changed the Wye-connected resistors to an equivalent Delta-connected resistors:

    36f2d8c63b.png

    The new resistors become 3R in the equivalent Delta-connected resistors

    Next step is calculating Z(L)
    Z(L) = j*2 (pi) * 60 Hz * 2 H = 754j ohm

    1/Zphase = 1/Z(L) + 1/Z(R)
    1/Zphase = 1/754j ohm + 1/3R ohm = -j/754 ohm + 1/3R ohm
    1/Zphase = (754 ohm - 3Rj)/754 ohm*3R ohm

    Since V is not given value in this problem, I choose to use an arbitrary value of x with angle of 0 degree, thus making it a constant.

    Iphase = Vphase/Zphase = x*(754 ohm -3Rj)/754 ohm*3R ohm

    It's known that power factor is 0.8

    cos(angle) = 0.8
    angle = arc cos(0.8) = +/- 37

    Since known that the current is lagging, the angle is lower than 0 degree, thus making the appropriate angle is -37 degree.

    Since I know the angle of phase I, I put it in the formula to find the angle of a complex number:
    tan(angle) = Im(complex variable)/Re(complex variable)
    tan(-37 degree) = [(x/754 ohm*3R)(-3Rj)]/[(x/754 ohm*3R)(754)] = -3R / 754
    - 0.75 = -3R / 754
    R = 754/4 Ohm = 188.5 ohm

    Am I doing it right?
    Thank you in advance for helping me to solve this!
     

    Attached Files:

  2. jcsd
  3. Sep 25, 2014 #2
    Sorry for not using the latex:

    Calculating total impedance in a phase
    [tex]{\frac{1}{Z_p}} = {\frac{1}{Z_L}} + {\frac{1}{Z_R}}[/tex]
    [tex]{\frac{1}{Z_p}} = {\frac{1}{754 j\Omega}} + {\frac{1}{3R}}[/tex]
    [tex]{\frac{1}{Z_p}} = {\frac{-j}{754 \Omega}} + {\frac{1}{3R}}[/tex]
    [tex]{\frac{1}{Z_p}} = {\frac{754\Omega - 3Rj}{754 j\Omega * 3R}}[/tex]

    Making the i = V/Z
    [tex]I = V*{\frac{754\Omega - 3R}{754 j\Omega * 3R}}[/tex]
    [tex]I = {\frac{V}{754 j\Omega * 3R}}*(754\Omega - 3Rj)[/tex]

    Finding the angle of I:
    [tex]\tan{\theta} = ({\frac{V}{754 j\Omega * 3R}}*-3R)/({\frac{V}{754 j\Omega * 3R}}*754\Omega)[/tex]
    [tex]\tan{\theta} = {\frac{-3R}{754 \Omega}}[/tex]
     
    Last edited: Sep 25, 2014
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