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 .