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.
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!