# New NCEES Power problem 539



## jcreit (Mar 30, 2009)

Can anyone please explain the solution for 539 in mathematical terms.

Thanks


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## niurou (Mar 30, 2009)

jcreit said:


> Can anyone please explain the solution for 539 in mathematical terms.
> Thanks



actually very simple:

with normal load, the Z=1.0 pu.

short circuit means without external load, impedance will be exclusively transformer impeadance=0.1 pu

remember

I=U/Z; U=IZ

if your short the secondary terminal, you have only 1/10 of the Z of the normal condition. and you want same I, the voltage must be 10 times less than before.

say if you have 1A current, Z is 1 ohm. U= 1A x 1ohm = 1V

now you have 0.1 ohm, still same current 1A, U= 1A x 0.1ohm = 0.1 V


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## swooda2 (Mar 31, 2009)

Here is how I approached this problem. We are told that the s.c. test was done at rated current. I assumed a value of 1pu for the current, and we were given Zpu = 0.1. From there it was just V=IZ, and then turning the pu voltage into actual voltage.


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## Flyer_PE (Mar 31, 2009)

A little factoid to keep in mind to solve problems like this one VERY quickly:

The way the short circuit test is performed, the secondary terminals are shorted and the primary voltage is increased until rated current is flowing in the secondary.

By definition, the percentage value of rated voltage required to generate rated secondary current is, for all practical purposes, the percent impedance of the transformer.

i.e. If a 11% of rated voltage applied to the primary terminals generates rated current in the secondary with the secondary terminals shorted, the transformer impedance is 11%.


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