transformer measuring

[autir]

I have a transformer rated as 12Vac / 800mAh.
I measured the voltage rating of the transformer with no load as Vmax=14.3 Ohm.
If Rint is the internal resistor of the transformer, Rint= (14.3 - 12)/0.8 Ohm <=> Rint=2.875 Ohm. (I)

When connecting an R=4.9 Ohm resistor directly to the transformer, the voltage at the ends of the resistor was Vr = 8 Volt.
I=Vr/R, but I=Vrint/Rint. So Vr/R=Vrint/rint (where Vrint the voltage drop at the Rint, and Vrint=Vmax-Vr). So Rint=((Vmax-Vr)*R)/Vr <=> Rint=((14.3-8)*4.9)/8 <=> Rint=3.85875 Ohm. (II)

The cases (I) and (II) differ by far.
Why?

note: R was not a power resistor and it was connected for less than a second because it started to heat up really fast.

 

[audioguru2]

Hi Autir,
The answer is one or more of:
1) The value of the overloaded 4.9 ohm resistor changed by its very high temperature.
2) The internal resistance of the overloaded transformer changed by its very high temperature.
3) The core saturated in the overloaded transformer.
You don't have accuracy with overloaded parts. ;D

 

[1st Mil-Tech]

Measuring the transformer for resistance is of no value because it operates on AC, not DC.  It is only a indicator that there is no short or open in the coils.

The Resistance doesn't take into account the actual operation of the transformer when hooked up to A.C.
Your trying to measure resistance rather than the impedance,

impedance measurement is only applicable when you use use a impedance bridge. It provides the AC needed for measurement.
I don't understand why you are doing this method of calculation, but your transformer indicates by your data that you should use a 15 ohm resistor to carry the current.
P= I*E 
.8 amps times 12 volts = 9.6 watts.  That would be a very big resistor
So you would need a  I = E/R 
12v/15 ohm resistor. = .800milliamps