hello Gyula,

You probably are right by the LC meter and inacurracy to measure this type of inductances. Perhaps has to do with its probe measuring wave form or just aything.
i would like to try another small experiment to clarify this further .. but i am in vocations.. so it will wait a bit
Carry on.   

Quote from: gyulasun on July 12, 2012, 04:02:38 PM
Hi baroutologos,

... Regarding your measurements on transformers, it is indeed strange and I also found big differences in measured no load input current and the calculated input current which came from measured transformer parameters. 

The problem may come partly from the digital LC meter: it does not use 50 or 60 Hz test frequency for L measurements, my own LC meter (Maxwell DMM MX-25 304 old type) uses about 200 Hz in the some hundred milliHenry and Henry ranges and it is doubtful how the different mains transformer cores perform at such a "high" frequency, most cores may lose permeability to some percent but some other cores may lose even half of their '50 Hz' permeability.  This means that a 50 Hz test circuit should be used  for measuring the transformer coils.  I repeat this frequency difference does not fully explain the situation...

@ Microcontroller,
excellent. This is the measuring minimum standard that should be applied from very start.

@all

Well here is my last trials using a secondary configuration as well. Mind again that these transformers are step down and hence not a true isolation transformer.

MAINS: REGULAR WALL PLUG
VARIAC AT FULL: SUPERIOR ELECTRIC MODEL BP5715
INPUT A: 121.9 VAC @ 0.24A = 29.259 WATTS
OUTPUT B: 57.7 VAC @ O.5A = 28.85 WATTS
OUTPUT C: 43.3VDC @ 0.25A = 10.825 WATTS
TOTAL OUTPUT: 39.675 WATTS
HE HE HE that does not include the heat dissipation of the hand burning bulb.
This is crazy.

This will be up soon;
http://youtu.be/HGVWXMmMs1o

Enjoy.

wattsup

Hello
because I  am interested in the outcome of this experiment, I must give my 50c .
@microcontroller
You can add up output watts in this case , cause the bulbs consume reals watts (not like caps or coils).


You are not measuring real power at the input side. (The value is to high)
But this measuring error (Input watts) results in a benefit for the cop!!
Because the real input watts are a lower value the cop will be higher.


Nevertheless measure the right 'real watt' input walues.
But I think wattsup , you are on the right track.

At the input side amps and volts are not in phase. I would try 2 things:
1)Put a 'kill a watt meter' directly after the variac . It will show reals watts. If the input voltage at these meter is not too low the measuring electronic will do it's job. For example my energy power meter works in a input volt range between 100V to 240 V .as shown on the label. If your meter's voltrange is similar to this, this is the preferred input watt measuring method.


Method 2 : Use a shunt in series with the input and measure the phase difference between the voltage and the current with a scope. Multiply the volt and amp values that you read now as input values with the cosinus of this phase difference. Than you get the real power.
For the output : there is not phase difference therefor cos(Phi) = 1.

Wattsup:

The AC voltage that you are measuring is an RMS voltage.  It looks like the clamp ammeter is measuring RMS current.  Assuming that is true then you can't multiply the two together, you will get an exaggerated wattage calculation.  My suggestion would be to investigate this issue and make corrections if necessary.

For your DC current measurement you are at nearly the rock-bottom output from the digital clamp meter.  Hence your accuracy is very poor.  I would suggest that you just use a regular digital multimeter or an analog ammeter with the proper maximum scale to get a better DC power calculation.

MileHigh

You can multiply RMS values of U and I to get the average power.The meaning of the RMS Value, be it voltage or current is to get a value that is equal to the case the whole device would be operated with DC voltage / amps.
If you got Urms you can calculate the average power by P=(Urms ^ 2) / R
If you can't measure current , you can now calculate Irms = sqrt(P/R) . And I is now also the RMS - value, because these are DC - like values.
And it does not matter , if he is using meters , that measure rms or not, as long the device is operating with sinewave.
His device is operating at mains frequency and with sine wave.
And for this case meters can be uses that measure mean average U or I , but do internally a multiplication correction for to show the rms (root mean square) , because it's the necessity for power calculations to deal with rms values (meaning volts and amps). If the waveforms in the device were other than sine , [/size]
then he MUST use real RMS meters, for measuring the correct value. (Not necessary here)
(And U and I must relate proportional to each other by ohms law U=R*I no nonlinearities ,
or you can forget your meter with RMS measuring capabilities)
Only the input power measurements are false.