Transformer's Percentage Impedance

[hisham1986]

Dear All,

I have found two unseperable definitions whenever i tried to know what percentage impedance means for a transformer.

First Definition:

Percentage impedance is the voltage drop on the secondary winding at full load in a transformer due to the winding resistance and leakage reactance and is expressed as a percentage of the rated voltage on the secondary winding.

Second Definition:

It is also the percentage of the normal terminal voltage at the primary side that is required to circulate full load current at the seondary side while the secondary side is short circuited.


The problem is that i am facing a trouble understanding whether these two definitions have to do something with each other,are they kind of related or do they share a common sense between each other and maybe am not seeing it? what is the relation between these two definitions?

Could anyone please help???

Thank you.

 

[GoldDigger]

Dear All,

I have found two unseperable definitions whenever i tried to know what percentage impedance means for a transformer.

First Definition:

Percentage impedance is the voltage drop on the secondary winding at full load in a transformer due to the winding resistance and leakage reactance and is expressed as a percentage of the rated voltage on the secondary winding.

Second Definition:

It is also the percentage of the normal terminal voltage at the primary side that is required to circulate full load current at the seondary side while the secondary side is short circuited.


The problem is that i am facing a trouble understanding whether these two definitions have to do something with each other,are they kind of related or do they share a common sense between each other and maybe am not seeing it? what is the relation between these two definitions?

Could anyone please help???

Thank you.

If you look at the load and the transformer at full load current, you should see that the voltage drop is that voltage which is dropped within the primary and secondary of the transformer at nominal applied voltage. You can model that as an equivalent circuit in which there is a voltage source of the no-load input voltage in series with a resistor corresponding to the transformer internal impedance. So when you draw full load current, the voltage drop is the full load current times that resistor value. So the percentage impedance measures the voltage drop compared to the input voltage.
But if we replace the load with a short circuit, and slowly ramp up the voltage to the point where the current is the full load current, the applied voltage must have the same ratio to the no-load voltage that the voltage dropped across the resistor at full load had to the unloaded secondary voltage.

If this does not seem completely obvious (and it probably won't) then you need to look at both circuit conditions and actually work out the equations, given that the series resistance of the equivalent circuit will have the same value in both cases and the unloaded voltage ratio of the transformer will be the same in both cases.

 

[Sandman1110]

This is my first trip to the forums here, so here we go....

I understand it as a function of the second definition you listed. A given xfmr secondary is shorted and a voltage is applied to the primary until full load current is reached in the secondary. Whatever that voltage is (eg. 15 volts) is then divided by the rated voltage of the primary. 15/480 = .03 or 3% impedance. Like anything else in this trade, there's more than one way to skin a cat, but this is the way I was taught.