Phase Angle

Huh?? By KCL(Kirchoff's Current Law), if current is flowing in to one end of a wire, then it is flowing out of the other end, assuming of course there are no other connections to that piece of wire. The difference occurs because of the orientation of the polarity marks on the CTs relative to each other.

There's a difference because of the orientation of the CT polarity marks relative to each other. For example, if we assume that phase current is flowing from left to right through the wire, then CT on the left side has its polarity marks facing the left end, and the CT on the right side has its polarity marks facing the right end.

The result of adding the CT currents together is to see if there is a difference in the currents. If 5A enter then 5A better leave so that the addition is (5A) + (-5A) = 0A; i.e., no current imbalance, what goes in comes out, no faults. If the polarity marks had the same orientation relative to each other, then the currents would be (5A) + (5A) = 10, and the only time there would be a zero differential current would be when no current was flowing.

That's not to say that the same result couldn't be achieved by swapping the secondary leads, but in power engineering consistency via standardization is everything.

dear sir

thank you for your replay.but if 5A with angle 0⁰ in neutral side sum with 5A with angle

180⁰ in line side .how the result become zero.please can you explain that,are we

convert these polar form to rectangular form then sum.

thanks

I already stated the simple version, but if you want to get complex:

5∠0° = 5 + j0

5∠180° = -5 -j0

5+(-5) = 0

There are plenty of online references that explain how a differential relaying circuit works. Use your favorite search engine to find them.

This is nothing more than Teslacles' Deviant to Fudd's first law.

I thought it was the BS Theorem: BS+BS=BS²

And its Corollary: (BS²)^-2 = BS⁴

Fudd's Corollary:

If it does not fall over, then you are not pushing hard enough, so then, push harder...