What is the minimum interrupting rating required for an OCP on the secondary of a 150 KVA transformer with a 2.07% impedance?

To determine the minimum interrupting rating required for an overcurrent protective device (OCPD) on the secondary side of a transformer, we can use the formula that incorporates the transformer's KVA rating and its impedance. The formula used for calculating the available fault current on the secondary side is:

[ \text{Available Fault Current (A)} = \frac{KVA \times 1000}{\text{Voltage}} \div \text{Impedance (in decimal)} ]

For a 150 KVA transformer with 2.07% impedance, we first convert the impedance percentage to decimal by dividing by 100, resulting in 0.0207.

If we typically consider a secondary voltage of 480V (a common commercial and industrial voltage), we can calculate the available fault current:

1. Calculate the available fault current:

[ \text{Available Fault Current} = \frac{150 \times 1000}{480} \div 0.0207 ]

[ = \frac{150000}{480 \times 0.0207} ]

[ = \frac{150000}{9.936} \approx 15,089 \text{ amps