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Three phase system

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the current measured by Ammeter .
    zad2.png
    2. Relevant equations
    $$ U_f=\frac{U_l}{\sqrt{3}} $$, Millman's theorem ...


    3. The attempt at a solution
    I know I can't just find U_f, divide it with impedance, and add all three currents (I am not 100% sure why I can't do that, can someone explain when can I do just that, and when exactly is forbidden?)
    My first idea was to find U_f, find U_{0'0} (with Millman's theorem), subtract those two, and then find current for every phase, add them up and that's it.
    However, after trying something else, I found that finding U_{0'0} , and then evaulating
    $$ U_{0'0}\cdot (\frac{1}{Z_1}+\frac{1}{Z_2}+\frac{1}{Z_3}) \approx 60 $$

    gives correct solution.
    Can someone explain why , because it doesn't make any sense to me.
    Point 0' is where where 3 lines L1,L2,L3 meet
     
  2. jcsd
  3. Jan 27, 2016 #2

    gneill

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    The individual currents will not be in phase since each of the line voltages, while having the same magnitude, have different phases with respect to each other. So you can't simply add the current magnitudes.

    Now, thinking of Millman's Theorem was an interesting idea, but consider that you're not looking for the voltage at the end of a set of parallel branches here. The Neutral line effectively shorts your branches. Further, the voltage sources are not in phase so you can't just use their magnitudes. Here's your circuit with the implied voltage sources, including their return path via the neutral line:

    upload_2016-1-27_17-22-5.png

    I suppose you might use Millman's Theorem as a step in finding a Thevenin equivalent voltage with the neutral open, then find the short circuit current. But you'll still need to incorporate the phases of the voltage sources somehow.
     
  4. Jan 27, 2016 #3
    For the first one, I wasn't thinking of adding their effective values. I meant $$ \frac{U1 \angle 0}{Z1}+ \frac{U2 \angle -120}{Z2}+\frac{U3 \angle -240}{Z3}$$, and looking at your diagram, I would do just that. I do get around 63, which is the closest to correct solution,although the given answers are 40,50,60,70,80.
    I forgot to mention that U_{l} = 400V and that makes U_f= 231 V

    Should I use Thevenin theorem instead?

    EDIT: I do get the same with thevenin's theorem.
     
    Last edited: Jan 27, 2016
  5. Jan 27, 2016 #4

    gneill

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    I think your first solution is fine and that you need to choose the closest match from the solutions. The Thevenin method should give you the same results.
     
  6. Jan 27, 2016 #5

    haruspex

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    Not an area I know, but from a little reading I think you might be making this more complicated than it is.
    ##V=400e^{j\omega t}## (defining t such that the phase is 0), right?
    In those terms, ##I=\Sigma \frac V{Z_r}##. Isn't that easy to evaluate?
    I believe that gives one of the offered solutions exactly.
     
  7. Jan 27, 2016 #6

    gneill

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    I'm not sure what your V represents, or how t is being chosen.

    The 400V value that the OP provided should be the line-to-line voltage magnitude (So, |L2 - L1| = |L3 - L2| = |L3 - L1| = 400V). For the star configuration for the sources that I drew, that gives each voltage source the magnitude of 400V/√3, with relative phases of 120° between each.
     
  8. Jan 27, 2016 #7

    haruspex

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    Then I misunderstood what Voltage the 400 referred to, and I forgot to put back in the phases (whoops)
    So writing v for amplitude of the line-to-neutral voltage, whatever that is, the current should be ##\Sigma \frac{ve^{j(\omega t+r\alpha)}}{Z_r}##, where ##\alpha=2\pi/3##. With v=400/√3, that agrees with the 63.
     
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