# NCEES #139



## Rei (Feb 22, 2010)

This question should be really easy that I'm afraid to ask, but I guess I miss understood the question.

An overcurrent relay is used to protect a circuit with a maximum 3-phase fault current of 8,000 A primary. A 400:5 current transformer ratio and a 5-A relay tap setting are selected. If the total secondary burden on the current transformer for the maximum fault condition is 1.10 ohm, the excitation voltage that must be developed by the current transformer is most nearly___.

Is/Ip = a = 400/5

which means Is = a(Ip), but the answer actually used 8000(5/400)=100A

Am I miss understanding the question?


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## Flyer_PE (Feb 22, 2010)

The concern is that the CT not saturate under fault conditions. In order to determine this, the secondary current under fault conditions (8000 amps) is the value of interest.


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## Rei (Feb 22, 2010)

Flyer_PE said:


> The concern is that the CT not saturate under fault conditions. In order to determine this, the secondary current under fault conditions (8000 amps) is the value of interest.


I still don't understand your explanation. You still want to find the secondary current, right? My question is why not 8000(400/5) instead of 8000(5/400) since a = 400/5? Thanks.


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## Flyer_PE (Feb 22, 2010)

Oh. You're looking for the secondary current on the CT. 8000 amps on the primary of a 400:5 CT. For current, you divide by the turns ratio. The secondary current with 8000 amps on the primary is 8000/(400/5) = 8000*(5/400) = 100A


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## Bluekayak (Feb 22, 2010)

Flyer_PE said:


> Oh. You're looking for the secondary current on the CT. 8000 amps on the primary of a 400:5 CT. For current, you divide by the turns ratio. The secondary current with 8000 amps on the primary is 8000/(400/5) = 8000*(5/400) = 100A


CT's are expressed as a ratio of the rated primary current to the rated secondary current where the primary is actually one (1) turn connected in series between the source and load. Therefore, using the turns ratio may not work as neatly as it does for a VT which is connected in parallel with both the source and load. Please correct me if I'm wrong Flyer.


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## Flyer_PE (Feb 22, 2010)

^For every CT I've ever encountered, the turns ratio is expressed as the primary circuit current required to generate either 1 amp or 5 amps on the relay/meter side. There's no difference in dealing with VTs and CTs as far as the turns ratio goes. Physically, there is technically only one turn on the primary side since the primary is actually the current carrying conductor. The CT itself is just a coil surrounding the current carrying conductor. Works just like a clamp-on ammeter.

In the real world, the difference is that you never want to open the secondary side of a CT (results in a pretty blue arc best viewed from a safe distance) and you never want to short a VT.


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## Bluekayak (Feb 22, 2010)

Flyer_PE said:


> ^For every CT I've ever encountered, the turns ratio is expressed as the primary circuit current required to generate either 1 amp or 5 amps on the relay/meter side. There's no difference in dealing with VTs and CTs as far as the turns ratio goes. Physically, there is technically only one turn on the primary side since the primary is actually the current carrying conductor. The CT itself is just a coil surrounding the current carrying conductor. Works just like a clamp-on ammeter.
> In the real world, the difference is that you never want to open the secondary side of a CT (results in a pretty blue arc best viewed from a safe distance) and you never want to short a VT.


That's why shorting blocks are so important when dealing with CTs! So for problem #139 one could actually solve the problem using the relationship Is/Ip=Np/Ns and Is=Ip(1/80)=8000/80=100A, correct?


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## Flyer_PE (Feb 22, 2010)

^Yes. The value for "n" in this problem is actually 80. It's just customary to see it expressed as a factor of the rated secondary current (typically 5 amps).


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## Rei (Mar 1, 2010)

Since current is Ip/Is=a instead of Is/Ip for the current transformer, does the power ratio reverse as well of it's still Vp/Vs=a?


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## Flyer_PE (Mar 1, 2010)

For any ideal transformer (VT or CT), Power Out = Power In. The result is that voltage is _directly_ proportional to the turns ratio and current is _inversely_ proportional to the turns ratio.


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## Rei (Mar 1, 2010)

Flyer_PE said:


> For any ideal transformer (VT or CT), Power Out = Power In. The result is that voltage is _directly_ proportional to the turns ratio and current is _inversely_ proportional to the turns ratio.


wait...from the question above, the current is not inverse. Given primary current is 8000A, CT ratio is 400:5, and we are looking for the secondary current.

According to what you said: Is/Ip=a=400/5 which means Is=(80)(8000)

but the answer is actually, Ip/Is=a=400/5 which gives Is=Ip/a=8000/80

right?


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## Flyer_PE (Mar 1, 2010)

Let me try this a different way:

Power in = Power out

VP*IP = VS*IS

VP/VS = IS/IP = a

In the given problem, the transformer ratio for the CT is 400:5. This tells you that 400 amps on the primary will give you 5 amps on the secondary.

(8000/400) * 5 = 100 Amps


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## Rei (Mar 2, 2010)

Flyer_PE said:


> Let me try this a different way:
> 
> Power in = Power out
> 
> ...


ok...you even use Is/Ip=a, then what is a? a = 400/5=80

we are giving Ip value, and so Is=Ip*a = 80*8000


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## Flyer_PE (Mar 2, 2010)

You have to divide by 80 rather than multiply by it.

CT Ration 400:5

IP=400

IS=5

ISC=8000 Amps

ISC/IRelay = IP/IS

Solve for IRelay

IRelay=ISC*(IS/IP) = 8000 * (5/400) = 100A


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## NJStilian (Aug 12, 2022)

It's asking for the excitation voltage, not the secondary current. Hence, Multiply by Resistance


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