For a 500 kVA, 3-phase, 480-volt transformer with 5% impedance, what is the availability of fault current assuming infinite primary fault current?

To determine the fault current available from a transformer, one must use the transformer's rated capacity and impedance. The fault current can be calculated using the formula:

[

I_{fault} = \frac{(V_{line})}{(Z_{pu} \times Z_{base})}

]

Where:

• ( V_{line} ) is the line-to-line voltage (in this case, 480 volts).

• ( Z_{pu} ) is the per-unit impedance of the transformer (5%, which equals 0.05 in decimal form).

• ( Z_{base} ) is the base impedance, which can be calculated using the formula:

[

Z_{base} = \frac{(V_{line})^2}{S_{rated}}

]

In this case, the transformer's power rating is 500 kVA.

First, calculate ( Z_{base} ):

[

Z_{base} = \frac{(480)^2}{500,000} = \frac{230400}{500000} = 0.4608 , \Omega

]

Now, substituting ( Z_{base} ) into the fault current formula gives:

[

I_{fault