I was reading through this document

http://www.jensen-transformers.com/wp-content/uploads/2014/08/an002.pdf

the final section has examples of how transformers transforms impedance. I am a little confused.

How to does shorting the primary show 80 ohms on the secondary?

Later, they show more realistic examples, but they say that

If the load were 20k ohms, it would "look like" slightly less than 20k ohms because the open circuit transformer impedance is effectively in parallel with it.

So, if the resistance was closer, the are in series?

I guess it is kind of a bold statement to make with out math to back it up....

Also, since they are 1:1 transformers, wouldn't that mean that it would just see what is there since it is not transforming any impedance?

 

That PDF is hard to follow!
What they are doing is mixing DC ohms with AC impedance.

How to does shorting the primary show 80 ohms on the secondary?

The DC resistance of a winding is 80 ohms and the 1KHz impedance of any one winding is 150K because of inductance caused by winding turns on an iron core.

If the load were 20k ohms, it would "look like" slightly less than 20k ohms because the open circuit transformer impedance is effectively in parallel with it.

The 20K is in parallel with 150K of AC impedance at 1 KHz.

So, if the resistance was closer, the are in series?

Closer and farther don't mean anything in this case. The main point of the PDF is to educate you as to when an impedance is parallel or in series.

since they are 1:1 transformers, wouldn't that mean that it would just see what is there since it is not transforming any impedance?

Transformers always transform impedance, as soon as they are connected to anything. I think of them as like an optical lens. They can look like a telescope or a microscope, or the nearly insignificant glasses I am wearing, depending on which way the electricity is "looking", but they are always a lens.

 

I see, he says that the DC resistance is 40 Ohms in the paragraph, but the 80 ohms is indeed impedance. Bad reading on my part. I am still not sure how it changes to 80 ohms just because he shorted one side.

OK, so what is the driving factor that lets you says that the 20K ohm is in parallel, but the 680 ohm is not? That is what I do not get...

 

Bump...

I am still not sure what that driving factor is....is it an assumption about the current size?