Randy said:

I'm wanting to set up a transformer to deliver 50% of its' power by

applying a tap at the 7th turn of a 14 turn secondary. There are two in the

circuit but, only one will be modified. The only data I have on the

transformers is from my own calculations: (ignoring losses)

Power: .7kVA (each) Pri.: 120 turns Sec.: 14 turns Ratio: 116

I don't know what ratio refers to, here. If it was turns ratio, it

would be 120/14=8.57

The formula I used to get I is:

Isec = Ipri x Tpri / Tsec

What is Ipri?

Using this formula, I get 14 volts @ 49.97 amps. Again, using this same

formula for the tap, I get 7 volts @ 99.94 amps. Something has to be

wrong with my approach. I'm tapping for less but, according to this, I'm

getting more. What did I miss or, am I thinking in reverse?

The power the primary may be able to supply is not necessarily the

same as the secondary can deliver. If the transformer was designed to

supply 700 watts at 14 volts, that implies 700/14=50 amps. But this

also implies that the gauge of the secondary was chosen to handle up

to that much current, and not much more. If you use only half of the

turns, the remaining turns can handle a bit more than rated current,

but not twice as much. A better estimate would be about 1.4 times the

full winding current. This causes the half secondary to produce the

same heat as the full secondary at rated current (I^2*R, with R being

half of the full winding resistance). This concentration of heat is

offset by the fact that the primary will be running derated at only

71% of rated current, and producing only half the heat it was designed

for.

All that said, there is no reason to take this much current from the

transformer. That will be controlled by the load resistance. If you

have a load that draws 25 amps from whatever voltage the half

secondary delivers, that is all the transformer will deliver.