# Transformer Open/Short Circuit Tests vs. Losses



## chicago (Sep 24, 2007)

Suppose a single-phase transformer is subjected to open-circuit and short-circuit tests. Additionally, the core loss (given in some arbitrary Watts) is at unity power factor. It just so happens that for this transformer,

*Copper losses (I^2 * R) = calculated short-circuit power*

Core losses = calculated open-circuit power

Strictly qualitatively speaking, can you imagine why/what constitutes both losses being equal to their respective short-circuit and open-circuit power ratings?

P.S. This is stated in the solution for Problem #4.29, p. 143, of the Kaplan Power Afternoon Sample Test

I owe you guys big time for all your efforts in helping me out...


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## Flyer_PE (Sep 24, 2007)

Here is my take on that one:

For the short circuit test, the output of the transformer is shorted. Since, theoretically, the output voltage is 0, the output power is zero. Therefore all of the power going into the transformer will be the copper losses (I2R).

Likewise, there will be some power going into the transformer when the secondary is not loaded. Again, the power output is 0. So any power going in will be a loss.

I'm not sure if I'm looking at the problem correctly though. I always have trouble with problems that start off with "Imagine if you will....".

Jim


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## chicago (Sep 24, 2007)

Jim, makes perfect sense. I think I was trying to over analyze things by taking the unity power factor into account. But once again, you've broken it down to laymen terms. Thanks


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## jdd18vm (Sep 24, 2007)

I just did that tonight, I got that part at the end by reading the EERM. WTG Jim as always.

What didn't get is how did they get 10 Ohms in the last problem, for the I Squared R which was stated as 5 Squared 10 for 250 watts? What did i miss about the Ohmic Value and Turns Ratio?

John


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## chicago (Sep 24, 2007)

John,

The 10 ohms comes from R_sc_total. If you recall, in Problem 4.25, they calculated it to be 10 ohms. So, the equation should have really read I Squared * R_total for Problem 4.29.

What I didn't get is why why split the 10 ohms up into 5 ohms = R1 = R2 in Problem 4.25. And isn't another typo the equation for R_sc = R_1 + a Squared * R__2_?


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## jdd18vm (Sep 26, 2007)

chicago said:


> John,
> The 10 ohms comes from R_sc_total. If you recall, in Problem 4.25, they calculated it to be 10 ohms. So, the equation should have really read I Squared * R_total for Problem 4.29.
> 
> What I didn't get is why why split the 10 ohms up into 5 ohms = R1 = R2 in Problem 4.25. And isn't another typo the equation for R_sc = R_1 + a Squared * R__2_?


Right the 10 Ohms was right in my face.

Regarding your question

I've been such a critic of Kaplan all do respect but is this another typo? In their statement "Clearly the resistance is Rsc=R1+a^2 R1where....." ??? From EERM 36.21, R=Rp+a^2Rs or their way should read R=R1+a^2R2 not R1? Shouldn't the second R1's be R2's? From there some Algebraic rearranging (which I am BLANKING on) to solve for R2?

BTW, I used the formulas in the EERM 36.22 to solve for R1 (Rp) direct which is Rp=Psc/2Isc^2 =250/(2)(5^2)=5, avoided all that...lol

my 2 cents on this one, I can be WAY off

John

In the EERM


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## Art (Sep 26, 2007)

short circuit

current approaches infinity limited only by the applied voltage and internal impedence (R in this case since pf=1)

so power loss = R I^2

open circuit

0 current in the secondary (it's open)

but with an applied voltage on the primary you must have some current flow (it's not infinite imped)

this must all be attributed to the excitation power, ie, core losses...


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