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I was wondering why exactly you have to use the nominal voltage of the transformer for the open-circuit or no-load test. Is this also the voltage over the parallel elements, even in a situation where the transormer is loaded? Or in other words, do the parallel elements produce as much losses in a loaded situation with a source applied and in a no-load situation when the nominal voltage is applied?

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  • \$\begingroup\$ The nominal output voltage is, by very definition, the voltage at the secondary under no load condition, assuming the input voltage is also nominal. \$\endgroup\$ – Bart Aug 25 '17 at 10:45
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I think I understand your question so first here's a representation of the equivalent circuit of a transformer: -

enter image description here

Under no load conditions the only current flowing into the primary is the current taken by the "parallel components" Xm and Rc. For a normal power transformer that current will be small compared to the current normally taken by the primary when the secondary is driving a load. For that reason you can ignore (short circuit) Xp and Rp and of course the secondary is only producing an open circuit voltage so Rs and Xs are of no consequence.

The "thing" in the middle that looks like a transformer is a perfect power transformer and because no current is being delivered to the secondary that perfect transformer takes no current.

So, it boils down to Rc and Xm being connected to the incoming power and no further components need to be analysed.

I was wondering why exactly you have to use the nominal voltage of the transformer for the open-circuit or no-load test.

There is one very important reason for this and that is core saturation - if you don't use the normal applied voltage you'll either have too much saturation or too little and you won't have a representive measurement. Saturation of the core is non-linear with voltage so it's important to use the right applied voltage. Look at the BH curve to see why: -

enter image description here

You can see that it is very non-linear once you start approaching saturation and, most transformers will be designed to run at a magnetic field strength (H) in the early to mid areas of saturation. This of course means a smaller transformer size and less iron. Commercial reasons prevail.

So, to do the test justice you need to run at nominal levels.

If you were doing a test to find out the values of the series components then you'd run the primary from a variac and short circuit the output. The sort of voltage that is now applied is a fraction of the nominal voltage so core losses are low (very linear BH curve) and eddy current losses (Rc) are also quite small.

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Is this also the voltage over the parallel elements, even in a situation where the transormer is loaded? Or in other words, do the parallel elements produce as much losses in a loaded situation with a source applied and in a no-load situation when the nominal voltage is applied?

The voltage across the magnetizing (parallel) elements is a little less when the transformer is loaded, and the core losses are a little less also. The combined voltage drop across the primary and secondary series elements also determines the transformer regulation. That is the percentage secondary voltage decrease as the load increases from zero to 100%. You could estimate that half of the regulation is due to the primary impedance and half is due to the secondary impedance. In a small transformer, say 5 to 50 VA, the regulation might be 10 to 15 percent and the change in magnetizing voltage might be 5 to 8 percent. For transformers above 1000 VA, the regulation is likely to be 2% or less and the effect of the magnetizing voltage 1% or less.

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