Fault Current Measuring in CT's

It won't. The circuit protective device(s) will operate before the CT reading will be noticed, disconnecting the circuit.

Have you ever wondered how protective relays sense and act on short-circuits ? After all, the relay is the 'brain' of the circuit breaker.

This subject of CTs, Rogowski sensors, and Hall-effect sensors is too big to cover here. You need to study some documents available on the internet.. here is one good site : http://www.kappaelectricals.com/technical.html

Another good document is ect170.pdf by Schneider Electric

You need a detailed course on current transformers, I fear! Depending upon the class of the CT and the Instrument Safety Factor (ISF) or the Accuracy Limiting Factor (ALF), a CT remians accurate up to a certain multiple of its rated primary current and after this limit, the CT saturates.

I suggest you attend to our training programme on Protective relaying.

It is not the CT that measures current. CT simply transforms high current to low current based on turns ratio, saturation effect etc.

The secondary current is used for measurement or protection.

The fault current can exist as long as the fault exists. This may be few cycles to few seconds depending on the type of protection and fault clearance provided.

Numerical Digital relays, record shortest possible duration fault current within few cycles. They work collecting samples of AC waves with an AC cycle, estimating the fundamental , harmonic components through fourier transforms, comparing the values to set values and triggering the protective mechanism if needed.

Hi Mohsinhamid

I see that the guys are giving you a lot of fancy answers on your question asked...I think they do confise you a lot....

You must just re-calculate your fault current ...it is far to high...you can not get fault current from a 2.5MVA transformer of that magnatude that you calculated...you must get the correct formula for the calculation....

To come back to your question:

If you get a CT 4000/5 then it is the nominal current where it can operate continuesly

On a CT name plate you find: STC=40kA/3s (Short time current)

That means that your CT can withstand 40000Amp for 3 seconds....then you must select suitable protection relays to clear the fault asap

I think, if others have given fancy answers, you have given an erratic answer and an unfair comment on others' answers.

And the OP is not wrong in calculating the fault current from a 2.5MVA Transformer to be 58kA. So, to say, it will be very close to that value.

I"k for a three-phase fault at the outgoing terminals of a transformer is given by:

I"k = Trf. FLC on the 433V side / % impedance of the Trf.

And, going by IS 2026, the typical impedance for a 2.5 MVA Trf. is: 6.25%.

FLC of Trf. is : (2500 x 1000) / (1.732 x 433) = 3333.52A;

I"k = 3333.52 / 0.0625 = 53.34kA.

And assuming that the short circuit happens after the location of the CT, the CT will still have to carry this current for the co-ordinated fault clearing time.

And, going by IS 2705, the duration of STC for CTs is not 3 sec. but 1 sec. For, 3 sec. STC duration would increase the cost & size of the CT.

What is the source impedance considered?

Source impedance cannot be zero and it will reduce the fault current.

Apart from the source impedance, pre-fault load flow conditions and thevenin's voltage also influence actual fault current.

Hi EE65

You are correct in saying the CT STC duration is for 1sec...I got confused with something else that is for a duration of 3sec...we work a lot with CT's and it is actually on the name plate

To make life simpler there is a nice little App called Transcalc, it's available on both Apple & Android, it calculates FLC and 3 Phase Short based on transformer ratings

Good one. First good reason for me to own an android phone ! GA ...