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Find an interesting situation in a 3-phase circuit

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img210.imageshack.us/img210/2082/93284166.png [Broken]
    http://img210.imageshack.us/img210/2082/93284166.png [Broken]

    Find the correct reactances for the impedances so that the current in the neutral/grounding wire would be 0 A. (IN=0)
    Phase voltages are 231 V, and the voltage between phases is 400 V.

    The vector diagram is also given, and it changes as you change the reactive components.

    PS. The entered reactive components in the picture are random and incorrect. I inserted them to make the vector diagram better to read.


    3. The attempt at a solution

    Probably the best way to approach this is by using Kirchhoff's first law. Adding the current vectors Ia, Ib and Ic together should result in 0.
    The easiest is quite likely to set the reactive component of phase A to 0 and balance the currents changing the reactive components of B and C.

    At first, I set the reactive component of B to 0 as well, so the angle between currect vectors Ia and Ib was 60 degrees. Then solved the triangle with cosine theorem to find the current of C. Using the current of C, and the given phase voltage, it was easy to find the impedance and then the reactive component of phase C. This, however, resulted the current to be about 0.2 A in the grounding wire.

    As I thought about it, it turns out it really is an interesting situation, since finding the current/impedance this way doesnt really add up, because the angle between the current and voltage of phase C also has to somehow match the angle in that triangle formed of the currents.
    So the only way I see it, an isosceles triangle must form, which means the reactive components of phases B and C must be equal and opposite.

    Now where Im having problems in solving it is forming the actual equations binding the impedances (or currents) and the phase shift (the angle between voltages and currents). I only see ways of forming them which involve something like

    cos[arctan(x/40)]

    which seem way too complex and therefore force me to think this is the wrong way to approach it.

    PS. I would, again, like to point out that so far I've solved similar problems using the triangles in the vector diagram. If anyone knows a better way, I would love to know.


    Thanks in advance
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
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