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Unbalanced Three-phase circuit.

  1. May 7, 2010 #1
    [SOLVED] Unbalanced Three-phase circuit.

    1. The problem statement, all variables and given/known data

    This comes from "practice problem" 12.10 from Alexander's and Sadiku's Fundamentals of Electric circuits 3rd edition, chapter 12.

    The problem statement is:

    "Find the line currents in the unbalanced three-phase circuit of Fig. 12.26 and the real power absorbed by the load."

    http://img413.imageshack.us/img413/8428/circuits.png [Broken]

    2. Relevant equations

    - LCK

    3. The attempt at a solution

    The problem is I can't get to the shown answer. I have followed two methods:

    Getting I_AB and I_CA from the known phase voltages:

    I_AB = (220|0º) / (5|-90º)
    I_CA = (220|-120º)/(10|0º)

    And then doing I_a = I_AB - I_CA.

    The second method is to construct a 3x3 matrix with complex coefficients based on the three meshes shown in the circuit. For those of you familiar with mathematica notation my matrix is:

    LinearSolve[{{-5 I, 5 I, 0}, {5 I, 10 + 5 I, -10 I}, {0, -10 I, 10 I}}, {220, 0, -110 + Sqrt[3]*0.5*I}]

    where I is the imaginary number and the first solution would be I_a.

    Both methods yield 11 + 44.866i, far different from the listed result.


    Any idea of how the authors got to (64, 80.1º)?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 7, 2010 #2
    I just did exactly what you said using my beautiful vector-capable TI-89 and got the book's answer. I assume you did something wrong with your vector math.
    I_AB = (220|0º) / (5|-90º) =(44|90º)
    I_CA = (220|-120º)/(10|0º) = (22|-120º)
    I_a = I_AB - I_CA = (44|90º) - (22|-120º) = (64|80.104º)
     
  4. May 7, 2010 #3
    Hmm it was indeed a problem of my vector algebra ^_^u. I was doing them by hand and when getting the imaginary part of I_CA I did only sin(-120º) and not 22sin(-120º)...

    I made the same mistake when constructing the matrix, the matrix should've been:

    [tex]\begin{bmatrix} -5j & 5j & 0 \\ 5j & 10+5j & -10j \\ 0 & -10j & 10j \end{bmatrix} \begin{bmatrix} I_a\\ I_b\\ I_c \end{bmatrix} = \begin{bmatrix} 220 \\ 0\\ -110 + 220\frac{\sqrt{3}}{2}j \end{bmatrix}[/tex]

    Since sin(120º) = 0.5sqrt{3}.

    Thanks! :D
     
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