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# NCEES PE electrical and Computer practice exam 528

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A test of a 10 KVA, 2300/230V, single phase transformer gives an open-circuit test power consumption of 69W, and a short circuit test reveals a resistance referred to the high voltage side of 10.9 ohm. The efficiency of this transformer, operated at full load and unity power factor is most nearly:

I am confused with the steps in the solution.  Do we have any other methods?

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hey @leggo PE can you move this into the Power PE forum?

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2 hours ago, LyceeFruit PE said:

hey @leggo PE can you move this into the Power PE forum?

Done!

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On 10/15/2020 at 1:14 AM, Rajan said:

A test of a 10 KVA, 2300/230V, single phase transformer gives an open-circuit test power consumption of 69W, and a short circuit test reveals a resistance referred to the high voltage side of 10.9 ohm. The efficiency of this transformer, operated at full load and unity power factor is most nearly:

I am confused with the steps in the solution.  Do we have any other methods?

Which part of the solution you don't understand, though? The formula for efficiency is efficiency=Pout/(Pout+Core loss+Copper loss).

Pout=KVA output(pf)=10kVA(1.0)=10kW

Core loss=No load loss=Open ckt power consumption=69W

Copper loss=(I^2)R  where I is the current at the high-voltage side. I=Spri/Vpri=10,000kVA/2,300V =4.35A

Copper Loss=(4.35)^2(10.9)=206.05W

So,

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