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MrBolt
July 9, 2021, 08:37 AM
A given 4160V 3-phase wye connected system is grounded by a 12 ohm resistor, what would the ground fault current be limited to? the Answer is 200A but I don't understand how?
SecondGen
July 9, 2021, 03:03 PM
A given 4160V 3-phase wye connected system is grounded by a 12 ohm resistor, what would the ground fault current be limited to? the Answer is 200A but I don't understand how?

What was your answer and why?
rofo42
July 9, 2021, 04:53 PM
A given 4160V 3-phase wye connected system is grounded by a 12 ohm resistor, what would the ground fault current be limited to? the Answer is 200A but I don't understand how?

The key takeaway here is that it is a wye connected system.
MrBolt
July 12, 2021, 07:30 AM
What was your answer and why?

4160/12=346.66
rofo42
July 12, 2021, 02:02 PM
4160/12=346.66

That would be correct if it was a delta system. In a wye, you must divide by the square root of three and then your resistor.
SecondGen
July 12, 2021, 06:13 PM
4160/12=346.66

As others have stated, the key is to use phase to ground voltage.
Izeldeen
November 6, 2021, 06:54 PM
A given 4160V 3-phase wye connected system is grounded by a 12 ohm resistor, what would the ground fault current be limited to? the Answer is 200A but I don't understand how?

4160/1.732=2401
2401/12=200
karlap
August 26, 2022, 09:03 AM
The key takeaway here is that it is a wye connected system.


(4160/(root scuare 3))/12=(4160/root scuare 3)/12 = 200.1480 Amp
i used Ohms law V=IR...... I=V/R..... this V is the voltage divided by scuare 3 because is a wye