Protection Engineering Basics and Schemes Protection Engineering | Page 15

Now let suppose we have 200 / 1 A CT, but CT secondary current is 0.99A but not 1 A, it means

I excitation = 200( 1 200) − 0. 99

I Excitation = 0.01A

It simply means 0.01A which is 0. 01

∗ 100 = 1 % of

secondary current is utilized for magetically coupling of Current Transformer. Let take one more example, take one CT of 200 / xA, Turn ratio is 5 / 200,5VA, Class 0.5 with ratio error of 0.5, Now we need to find I secondary and I excitation

0.5 = 200( 5

200) − x x x = 3.333A I Secondary = 3.333A

I excitation = I Primary( N Primary

N Secondary

) − I Secondary

Means 1. 667 5

= 200( 5 200) − 3. 333

I excitation = 1.667A

1

∗ 100 = 33.34 % of secondary current is utilized to magnetize the core for induction purpose.

pg. 15