Calculate CT Burden

A CT will measure Current not frequency.

The CTR is the amount of current from source to device (The CT.)

CT is a current transformer and a CT ratio (CTR.) is a Current Transfer Ratio.

A CT from a 33KV bar will probably be wound for 50 or 60 HZ.

You can check the frequency response by testing the Q of the CT.

Thanks

Of course CT can't be used for measuring frequency.All devices has its own frequency response.Frequency response of CT depends mainly the core material's initial permeability. My requirement is How to get the CT burden from V-I characteristics and CT ratio.

Cheers

Faisal

There will be a current transfer ratio.

Pass 1 amp into the source measure the output of the CT and then pass 10 amps through the source and measure the output CT's I think have a fairly linear output and you can plot the expected graph.

Dear all,

The CT is for protection and 200/5A, excitation curve is available.

Let me say, half of knee point voltage (V) x 5 (A) is your practical and usable CT burden.

My assumption is that at the half of the knee point voltage, the CT is not saturated yet, and this voltage is induced when the above VA load is connected to the CT and when the secondary current 5A.

Regards

A agree with Akihito Shigeno. But there is one major concern here. Is this CT to be used for the purpose of measuring or protection? If it is measuring, half of knee point voltage can be considered but if it is protection CT, at best 10% of knee point voltage (for nP10 class or 5% for nP20 class etc., where n is the accuracy level) shall be considered. However, without knowing detail data for the CT, I wouldn't recommend to use this CT for the protection purpose. You can use this CT if it is a course measuring purpose (where the accuracy is not so important).

- MS

Hi Msamad

Many Thanks

I understand from your reply that , under normal operating condition ,protection CT is allowed to operate 10 % of Knee point region in the excitation characteristics. Which means when the fault current is 10 times more than the normal operating current , CT won't saturate. Is that what do you mean?

Regards

Faisal

Yes, you understood correctly. However, the % value depends on CT class. As I mentioned earlier it is 10% of the knee point voltage for class nP10, 5% of the knee point voltage for class nP20. See http://cr4.globalspec.com/thread/43440 for more detail about the CT class.

- MS

I agree with msamad.

If a half of knee point voltage is selected to calculate CT secondary load, as is assumed above, and a fault current is 20 times the CT rating, CT will be saturated very much.

So, I take the 5% of knee point voltage x 5A as practical burden.

Thank you for pointing out my error.

Regards