If a transformer has a secondary of 208/120 volts and 2% impedance, what is the maximum expected available fault current for a 300 kVA transformer?

To determine the maximum expected available fault current for a 300 kVA transformer with a secondary voltage of 208/120 volts and 2% impedance, it is essential to use the formula for fault current:

[

I_{fault} = \frac{V}{Z}

]

where ( I_{fault} ) is the fault current, ( V ) is the voltage, and ( Z ) is the impedance in per unit.

First, we need to convert the impedance percentage into a decimal for calculations. Since the impedance is given as 2%, it translates to:

[ Z = \frac{2}{100} = 0.02 ]

Next, calculate the full-load current for the transformer. The full load current (I) can be calculated using the formula:

[

I = \frac{P}{\sqrt{3} \times V}

]

In this case, the transformer's power rating (P) is 300 kVA, and for a three-phase system, the secondary voltage is usually considered as line-to-line (208 V):

[

I_{full load} = \frac{300,000}{\sqrt{3} \times 208} \approx