A transformer has a 120/208-volt wye secondary and is delivering 75 amps of line current to a 3Ø load. For safety, this transformer must be rated for at least how many KVA?

To determine the minimum transformer rating in KVA for a 120/208-volt wye secondary delivering 75 amps of line current to a three-phase load, the formula used is:

[

\text{KVA} = \frac{\sqrt{3} \times V \times I}{1000}

]

Here, ( V ) is the line-to-line voltage and ( I ) is the line current. In a 120/208-volt wye system, the line-to-line voltage is 208 volts. Plugging the values into the formula gives:

[

\text{KVA} = \frac{\sqrt{3} \times 208 \times 75}{1000}

]

Calculating this step-by-step, first, calculate ( \sqrt{3} ), which is approximately 1.732. Then:

[

1.732 \times 208 = 360.256

]

Next, multiply this result by 75 amps:

[

360.256 \times 75 = 27019.2

]

Finally, to convert to KVA, divide by 1000:

[

\frac{27019.2}{1000} =