Protection Engineering Basics and Schemes Protection Engineering | Page 232

I FLL = 1600 √3 ∗ 415

I FLL = 2226A

I SCL = I FLL % Z

I SCL = 2226 6 %

I SCL = 37099.9A

If CT we have of 2500 / 5 rating, then CT output of Short circuit will be

I SCLS = 37099. 9

2500 5

I SCLS = 74.1998A

Voltage for Relay setting = V R = I SCLS ∗( R CT + 2R leads)

= 74.1998 ∗( 0.52 + 2 ∗ 0.021) V R = 41.7volts

Knee point voltage of CT must be 2 times the relay terminal voltage

pg. 232