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3 phase system electrical question

  1. Sep 21, 2014 #1
    1. The problem statement, all variables and given/known data

    https://www.physicsforums.com/attachments/question-jpg.73386/ [Broken] the current go from a A N and n
    the 2nd current go from b B N and n

    the first thing, i make the matrix
    27-4j -3 i1 230 <0
    -3 27-4j x i2 = 230<0

    x means the matrix multiplied to the next
    < means the degree
    then with determinant from the first matrix
    i find 767.04 <16.35

    then to find i
    i1 = 230<0 /767.04<16.35 27-4j -3 1
    i2 x -3 27-4j x 1

    is this right?
    how to count 230<0 / 767.04<16.35
    i dont understand

    2. Relevant equations



    3. The attempt at a solution

    the first thing, i make the matrix
    27-4j -3 i1 230 <0
    -3 27-4j x i2 = 230<0

    x means the matrix multiplied to the next
    < means the degree
    then with determinant from the first matrix
    i find 767.04 <16.35

    then to find i
    i1 = 230<0 /767.04<16.35 27-4j -3 1
    i2 x -3 27-4j x 1

    is this right?
    how to count 230<0 / 767.04<16.35
    i dont understand
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 21, 2014 #2

    gneill

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    Staff: Mentor

    Your matrix looks okay (although its formatting has been punished by the font!), but your solution doesn't look correct. Here it is rendered with LaTeX:

    $$\left( \begin{array}{cc} {27 - j4} & -3 \\ -3 & {27 - j4} \end{array} \right)
    \left( \begin{array}{c} I1 \\ I2 \end{array} \right) =
    \left( \begin{array}{c} 230 \\ 230 \end{array} \right)$$

    Can you check your determinant calculation? I see different magnitude and angle values.
     
  4. Sep 21, 2014 #3
    sorry about that :D
    and thank you for the matrix format and the reply

    the determinan is 704-j216?
    so it means 736.39 with -17.05 degree?

    so the matrix go like this?
    $$\left( \begin{array}{c} I1 \\ I2 \end{array} \right) = 230<0 / 736.39 < -17.05
    \left( \begin{array}{cc} 27-j4 & -3 \\ -3 & 27-j4 \end{array} \right)
    \left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$

    what is that $$\left( \begin{array}{c} 1 \\ 1 \end{array} \right)$$ mean?
    i get the formula from the ppt but i still dont understand why it is there
    and i dont understand how to calculate 230<0 / 736.39 < -17.05 manually
     
  5. Sep 21, 2014 #4

    gneill

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    Staff: Mentor

    Yes, that's the correct value for the determinant.
    I'm not sure. I haven't seen that particular method used before. The ##\left( \begin{array}{c} 1 \\ 1 \end{array} \right)## is a column vector. I presume that it's meant to be multiplied with the impedance matrix in order to yield a new column vector. That would make sense since there's a column vector to the left of the equals also. But I don't see how that equation is meant to solve for the currents.

    I would use Cramer's Rule to solve the problem (look it up). It uses determinants, and you've already got the value of one of them.
    That's just division of one complex number by another.
     
  6. Sep 22, 2014 #5
    OIC
    it is just an inverse
    moved the left 1st matrix to the right side
    then it make i1 = 7.59< 0.31
    and i2 = 7.59<0.31
    is this right?
    then IaA = I1
    InN = -(I1+I2)
    IbB = I2
     
  7. Sep 22, 2014 #6

    gneill

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    Staff: Mentor

    Please don't use text-speak abbreviations here on PF. They're not allowed.
    It doesn't work for me; I can't see how that math could be correct. If I wanted to "move" the impedance matrix from the left side to the right side of the equation, I'd pre-multiply both sides with the inverse of the matrix. But then what is moved to the right side would be the inverse, not the original matrix. Take a generic equation of this type; let I be the unknown current column vector, V the known voltage column vector, and Z the impedance matrix. The equation is written:

    ##Z \; I = V##
    ##Z^{-1}Z \; I = Z^{-1}V##
    ##I = Z^{-1}V##

    The problem I have with the equation that you've quoted is that I don't see the inverse of the impedance matrix coming about by simply dividing the matrix by its determinant. What you'd want is to use Cramer's rule which divides the adjoint matrix by the determinant, and solve for the I's individually.
    Yes, once you've solved for the mesh currents then then branch currents are obtained as you've shown.
     
  8. Sep 22, 2014 #7
    sorry about that
    and thanks for replying

    okay
    and the result is I1 = 9.44<9.45 and I1=I2 or I1=9.31+J1.55
    is this correct?
     
  9. Sep 22, 2014 #8

    gneill

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    Staff: Mentor

    Yes, that's correct.
     
  10. Sep 22, 2014 #9
    and for the phasor diagram just draw a line with 9.44 and 9.45 degrees right?
     
  11. Sep 22, 2014 #10

    gneill

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    Staff: Mentor

    Presumably you'll want to show the source voltage phasors along with the phasors for all the currents that you were asked to find.
     
  12. Sep 22, 2014 #11
    oohhh
    so there will be 3 line
    thx a lot
     
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