# Power 524



## z06dustin (Oct 20, 2009)

> A power transformer has the following test data at rated voltage:% of Nameplate kVA || Total losses (W)
> 
> 0 _____________________________ 460
> 
> ...





> B, Poc=460W, P50%=2,370-460 = 1,910W, P100% = (4)(1,910) = 7,640W. PT = 7,640 + 460 = 8,100W.


Somebody (Flyer???) want to tell me where that equation came from / what it is? I mean is it as simple as "oh I assume that losses square" (hence the factor of 4?) or what?

Thanks


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## maxxpower71 (Oct 20, 2009)

I am giong to take a shot at it.

At % = 0 its only the core losses = 460

at 50% its core + copper losses (I^2 * R + Pcoreloss) therefore to only find out what the load is at 50% subtract 460. 2,370 - 460 = 1,910 is the copper losses at 50%

At 100% (2 * I ) ^2 * R + Pcoreloss = 4I^2 * R + Pcoreloss

I^2 * R = 1,910

At 100% , 4 * 1910 + 460 = 8,100W

So yes you were right about the square.


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## Flyer_PE (Oct 20, 2009)

^Very good sir! :appl:


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