Jump to latest postA 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?
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Junior Member
Posted July 9, 2021
Pro Subscriber
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?
NETA Level III
Posted July 9, 2021
Pro Subscriber
What was your answer and why?Originally Posted by MrBolt
Seasoned Member
Posted July 9, 2021
The key takeaway here is that it is a wye connected system.
Junior Member
Posted July 12, 2021
Pro Subscriber
4160/12=346.66
Seasoned Member
Posted July 12, 2021
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.
NETA Level III
Posted July 12, 2021
Pro Subscriber
As others have stated, the key is to use phase to ground voltage.
Member
Posted November 6, 2021
4160/1.732=2401
2401/12=200
Junior Member
Posted August 26, 2022
(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
