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Finding the value of a neutral earthing resistor

  1. May 5, 2016 #1
    1. The problem statement, all variables and given/known data

    An 11 kV motor is fed by cables from a transformer via switchgear,
    having a phase impedance of 0.3 + j0.3 ohm. The earth return path to
    the transformer neutral has a resistance of 0.42 ohm. Determine a
    suitable value of neutral resistance if the voltage rise at the motor in the
    event of an earth fault at the motor is not to exceed 430 V. Neglect
    transformer winding resistance.

    2. Relevant equations

    3. The attempt at a solution

    Hi there everyone. This is a question that I'm working on at the moment. My attempt so far is as follows. Maybe someone can tell me if I'm on the right path?

    Well using the information given the fault current would be 6350/.42 = 15121 amps

    The 6350 here is the phase voltage.

    To meet the requirement in the question the current would need to be reduced to
    430/.42 = 1023.8 amps

    Using an earth path impedance of 6350/1023.8 would give 6.20 ohms

    What I'm thinking of doing next is R = the square root of Z² - X²

    This gives me 6.19 ohms and we already have .3 ohms in the circuit so we would need to add a resistor of value 6.16 ohms.

    Would this be correct?

    Any help would be greatly appreciated!

    Last edited: May 5, 2016
  2. jcsd
  3. May 6, 2016 #2
    Hello again everyone!

    I'm actually just looking at the question there again and think that I made a mistake. I thought that the .42 ohms was the impedance but I think that the impedance from the transformer to the motor is Z = 0.3 + j0.3 which is .42 but I think that the earth return path to the transformer neutral is also .42 ohms which would give me an earth fault loop impedance of Z = 0.72 + j0.3.

    Does that make more sense I wonder?
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