CT Cable Sizing

timm333

Senior Member
Location
Minneapolis, MN
I have a question about the sizing of the cable on CT secondary. Which IEC standard gives the criteria to calculate the wire size on the secondary of CT?

For example if the CT ratio is 400:1, CT rated burden is 10 VA, the distance from the CT to relay is 100 m, relay burden is 0.1 VA: then will a 4 mm2 cable be enough to connect the CT secondary to the relay? Thanks.
 

topgone

Senior Member
I have a question about the sizing of the cable on CT secondary. Which IEC standard gives the criteria to calculate the wire size on the secondary of CT?

For example if the CT ratio is 400:1, CT rated burden is 10 VA, the distance from the CT to relay is 100 m, relay burden is 0.1 VA: then will a 4 mm2 cable be enough to connect the CT secondary to the relay? Thanks.
Use IEC 60044 calculations. IMHO, you need to know what is the maximum short-circuit your CT will possibly be in that CT location!
 

Hv&Lv

Senior Member
Location
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Occupation
Engineer/Technician

paulengr

Senior Member
Use IEC 60044 calculations. IMHO, you need to know what is the maximum short-circuit your CT will possibly be in that CT location!
That makes no sense. A CT is normally shorted or it burns up. They have a maximum rated burden which is maximum allowed impedance. The rated output is listed on the CT and is on the side so 100:1 is a maximum nominal 1 A output continuous rating. It might get as high as say 20x the rating for short periods but continuous rating is listed on the CT. Since even North American CTs go up to 5 A minimum power conductor rules apply in all cases.
 

topgone

Senior Member
That makes no sense. A CT is normally shorted or it burns up. They have a maximum rated burden which is maximum allowed impedance. The rated output is listed on the CT and is on the side so 100:1 is a maximum nominal 1 A output continuous rating. It might get as high as say 20x the rating for short periods but continuous rating is listed on the CT. Since even North American CTs go up to 5 A minimum power conductor rules apply in all cases.
You don't want a CT to saturate and read wrong just because you missed to consider that possibility of the CT having to experience a short-circuit equivalent current! The CT will give a linear response according to it's given ALF. That's why you have 5P10, 10P20 CT ratings (10X and 20X ALF).
 

timm333

Senior Member
Location
Minneapolis, MN
For 5P10 CT, the total burden on CT is given by:

St = I^2 (Rct + Rw + Rr)

Where St=total burden in VA, Rct = internal resistance of CT secondary in ohm, Rw = wire (lead) resistance in ohm, Rr = relay resistance in ohm.

If St is 25%-100% of rated burden of the CT, then the CT accuracy would be guaranteed, so in this way we can show (without calculating ALF) that the CT is Ok for the application. The main issue is that sometimes they include Rct in the total burden calculations, and sometimes they don’t include it.

I think Rct should always be included in the calculation of total burden, Is it correct?
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
If St is 25%-100% of rated burden of the CT, then the CT accuracy would be guaranteed, so in this way we can show (without calculating ALF) that the CT is Ok for the application.
Yes, as long as the nominal ALF of a 5P10 is adequate for your application.

The main issue is that sometimes they include Rct in the total burden calculations, and sometimes they don’t include it.

I think Rct should always be included in the calculation of total burden, Is it correct?
They may not include it sometimes because Rct typically small compared to the other resistances.
If the external burden is intentionally made lower because you want to increase the ALF above the nominal value, then Rct becomes a more significant factor and must be included. But if you have the Rct available then it should be included in any case.
 

topgone

Senior Member
For 5P10 CT, the total burden on CT is given by:

St = I^2 (Rct + Rw + Rr)

Where St=total burden in VA, Rct = internal resistance of CT secondary in ohm, Rw = wire (lead) resistance in ohm, Rr = relay resistance in ohm.

If St is 25%-100% of rated burden of the CT, then the CT accuracy would be guaranteed, so in this way we can show (without calculating ALF) that the CT is Ok for the application. The main issue is that sometimes they include Rct in the total burden calculations, and sometimes they don’t include it.

I think Rct should always be included in the calculation of total burden, Is it correct?
Just KISS it! Make sure the required source voltage Vs is lesser than the knee-voltage Vk of your CT!
Formulas:
Required source voltage, Vs =KTD X (Ifsc/CTR) X (Rct +Rr +Rl),
where KTD is the transient dimensioning factor (take it as = 1 if you don't know); Ifsc is the the fault available;​


Knee-voltage of CT, Vk = ALF X ([VA/In] + [Rct X In]),
where ALF = accuracy limit factor, VA = CT burden; In = CT rated secondary current​
 
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