# ncees sample exam afternoon power prob 539



## cabby (Sep 11, 2008)

Can anyone out there explain the solution to prob 539 in the NCEES sample Exam Power Module? I do not understand the approach. I thought I had a fault on the secondary side of a transformer and I need to find the voltage on the primary side. I am sure I am viewing the problem wrong. If anyone could explain this I would appreciate it.

thanks,

cabby


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## Flyer_PE (Sep 11, 2008)

It's a CT problem. You have 8000 amps on the primary of a 400:5 CT. The relay side of the CT will see:

8000 * (5/400) = 100 amps

The secondary burden is 1.1 ohm.

The voltage you are looking for is 1.1 ohm * 100 amps = 110 V

Hope this helps.

Jim


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## zhu (Oct 14, 2008)

Flyer_PE said:


> It's a CT problem. You have 8000 amps on the primary of a 400:5 CT. The relay side of the CT will see:
> 8000 * (5/400) = 100 amps
> 
> The secondary burden is 1.1 ohm.
> ...


What is the typical full scale output of a 400:5 CT? I got stuck on this problem because I thought the 100A will be way out of the scale of the CT.


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## mudpuppy (Oct 14, 2008)

The rule of thumb is for a fault the secondary current should not exceed 20 times the rating of the CT. In the U.S. CTs are typically rated for 5 A secondary, so this would typically be 100 A. I think this rule of thumb follows from the ANSI standard for CTs (which I believe is C57.13), so that you do not exceed the knee point of the CT's excitation curve and risk saturating the CT. It doesn't mean you can never exceed 100 A, but you need to be careful when you do.

To be clear, this is only for fault current, which will only last a short period of time. For continuous load, you usually do not want to exceed the secondary rating of the CT, i.e. 5 A. However, most CTs have a built in termal overload factor, often 2 to 3 times the secondary rating--for emergency loading.


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