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Calculating line currents if one leg of a star system is short circuited.

  1. May 16, 2012 #1
    Hey guys,

    Been struggling with this for ages any help would be awesome.

    In Electrical engineering;
    Suppose you have a balanced multiphase star system with a given line voltage. The impedance of each phase is also know.

    If one of the coils is short circuited, calculate the line currents.

    I'm assuming the two remaining phases are 90degrees apart so that the line voltage remains constant between all 3 remaining nodes. Also that the phase and line voltage of the remaining nodes are equal. Is this correct?

    I cannot understand why there are two different line currents? Shouldn't these both be equal to the line voltage divided by the impedance?
     
  2. jcsd
  3. May 16, 2012 #2

    NascentOxygen

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    Hi No36 ... http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    "Star" connected, so are you talking about 3ɸ https://www.physicsforums.com/images/icons/icon5.gif [Broken]

    "Impedance of each phase"? Are you talking about shorting the output of a generator? Or shorting one of the 3 load impedances of a star-connected load without a neutral connection?
    Coil? What coil? :confused:
    Assuming 3ɸ, why would the relative angle change from 120° to 90°? :confused:
     
    Last edited by a moderator: May 6, 2017
  4. May 17, 2012 #3
    Thanks for the reply. Just typed out a very long reply which got lost when I tried posting.

    In short:

    It is a three-phase star connected system.

    You are right the angle would remain 120degrees as the circuit cannot have an influence on the 3-phase source. (Am I correct in this assumption?).

    Let:
    There be 4 nodes R,B,Y and N.
    Let R,B and Y be connected to N such that RN, BN and YN exist as 3 similar "legs".
    Between R and N a coil is connected with a resistance and inductance.
    Vrnb = Vbny = Vynr = Line Voltage.
    Vnr = Vnb = Vny = Positive direction of phase voltages.

    Now let B be shorted to N.

    This now means Vrnb = Vrn, Vbny = Vny. These are now equal to the line voltage.
    In my opinion Vrn = Vny = V(Line Voltage) = Vrny.
    Yet it can be shown Vrny = Vrn + Vny ;
    Vrny^2 = Vrn^2 + Vny^2 - 2VrnVyncos(120)
    = 2V^2-2V^2cos120
    = 2V^2(1-cos(120)).

    I actually managed to solve my problem from the statements above whilst typing my reply :P Thank you so much for helping me find the right questions to ask.

    I still however can't understand why Vrny assumes this value. Shouldn't it be the same as the line voltage? Or does the potential difference just always asume the greater p.d. applied across it? Doesn't the line voltage that is still applied from R to Y "drag" down this voltage?

    Again, thank you for your time.
     
  5. May 17, 2012 #4

    NascentOxygen

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    N is the star point? (I wouldn't label it N when there is no neutral line.)

    So with one line, say B, connected to the star point, you have the line voltage applied across two loads. So R carries the phase current, Y carries the same phase current, and B carries the vector sum of those two.
     
  6. May 18, 2012 #5
    What do you mean by no neutral line? I guess there wont be a neutral line since its shorted to the B point, but this is originally a 4 wire system with a neutral point.
     
  7. May 19, 2012 #6

    NascentOxygen

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    Staff: Mentor

    You are shorting the line to neutral? Won't that trip a circuit breaker?
     
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