Originally Posted by

**hello21**
A line to ground fault on the secondary side of the transformer, the fault current should make its way on the equipment ground back to the XO of the transformer then go on the phase conductor and hopefully trip out a breaker (usually the closest breaker to the fault if appropriately coordinated).

If current could make its way from the transformer XO to a phase conductor, that itself would constitute a fault. That is not the way this event would be terminated.

Originally Posted by

**hello21**
I assumed that the fault current would not be induced in the primary, and so regardless of the fault type and magnitude, the primary side of the transformer would not see a fault. Am I way off here?

Sorry, yes. The current on the primary side of a transformer is equal to the current on the secondary side multiplied by the ratio of secondary voltage to primary voltage. For example, on a 480 – 120/208 step down transformer, if the secondary current (all three phases – let’s talk about balanced loading for now) is 100 amps, then the primary current will be 100 x (208/480), or 43.3 amps. During a (again, balanced three phase) secondary fault that has, for example, a fault current of 10,000 amps, the current on the primary side would be 4,330 amps. That should trip the primary side OCPD.

For a line-to-ground or line-to-line secondary fault, it would be harder to describe the math. But the notion of the primary current becoming elevated would remain.

Charles E. Beck, P.E., Seattle

Comments based on 2017 NEC unless otherwise noted.

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