The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 Ω. If the fault current is 20 times the rated primary current of the CT

Q. The total impedance of the secondary winding, leads, and burden of a 5 A CT is 0.01 Ω. If the fault current is 20 times the rated primary current of the CT, the VA output of the CT is

Solution:

Given that, impedance of secondary winding = 0.01 Ω

Rated current = 5 A

fault current = 20 times of rated current

= 20 × 5 = 100 A

VA output = I²z = 100² × 0.01 = 100 VA