# NCEES PE Power Practice Exam 539 - Rated voltage for transformer short circuit test



## akyip (Mar 5, 2020)

Hi guys,

Working on the NCEES PE power practice exam, question 539 is one I don't get. I have trouble posting photos here, so I'll have to type it out.

"A one-phase transformer rated 100 KV / 7.2 KV has an impedance of 10.0%. During a factory short circuit test done at rated current, the voltage (KV) applied to the HV terminals is nearly?"

The solution states the answer is 10 KV, and the explanation is that under short circuit conditions the voltage required to produce rated current is only 10% of rated voltage. I'm guessing this has to do with the 10% impedance, but I don't fully understand how.

Can anyone explain? Thank you very much!


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## Orchid PE (Mar 5, 2020)

The % impedance of a transformer is the ratio of voltage applied for full load current during the short-circuit test to the rated voltage of the terminals.

In other-words,

Z% = V_sc / V_r

Where

V_sc is the voltage required on the primary winding to induce rated current on the secondary (shorted) winding
V_r is the rated voltage of the transformer's primary winding

So for the problem you gave:

10% = V_sc / Vr

Rearranged

V_sc = 10% * V_t = 0.1 * 100kV = 10kV

So 10kV is required on the primary winding to produce rated current on the shorted secondary.


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## Zach Stone P.E. (Mar 5, 2020)

akyip said:


> Hi guys,
> 
> Working on the NCEES PE power practice exam, question 539 is one I don't get. I have trouble posting photos here, so I'll have to type it out.
> 
> ...


Most of the NCEES questions each have their own thread on engineerboards. You can find them pretty quickly using the search function or just searching in google. Here are two threads on this question that I was able to find that you may find useful:


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## Dude99 (Mar 6, 2020)

First you must know what the sc test is: basically short the sec, increase v prim until you get rated i on sec (and prim)

current sec and prim = 1 pu, Z = 0.10 pu

v pu = i x Z = 1 x 0.1 = 0.1 pu

v = 0.1 x vbase = 0.1 x 100 = 10 kv


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