STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[Omnibus]

@LarryC,

You are using average V and I to calculate P and are not integrating the product of the  momentary V and I values, correct? If so, you're making a major mistake and until it is corrected any further discussion of your results is senseless.

[Omnibus]

@All,

Finally I was able to get my current probe working. How? Well, I brought the scope and the probe to one of the RadioShack locations and established that one of their 12V power supplies does the job. Finding a power supply caused a great deal of frustration but now it's behind and I was able to carry out some real measurements finally. You'll have to bear with me until I am able to post results. However, I'll mention at once that the preliminary tests I made confirm unequivocally the OU I reported in the region between 200 and 700 (and slightly beyond) kHz. The Hall effect based current probe will also resolve in a natural way and will disperse the confusion in some regarding the issue of inductive reactance and why only the real part of the impedance is to be used in calculating the losses (which is on top of it a conservative calculation since it doesn't take into account the hysteresis, eddy current etc. losses in the core). I'll emphasize again that the above applies to an unloaded transformer which means that we don't even need to have a transformer fro the OU effect to be observed. Things are now getting to be more exciting than ever.

[IceStorm]

@LarryC,

You are using average V and I to calculate P and are not integrating the product of the  momentary V and I values, correct? If so, you're making a major mistake and until it is corrected any further discussion of your results is senseless.

LarryC is Right, for a perfect Sine wave you just need to convert it to DC equivalent , so Veff =  (Vp * 0.707) give you the DC equivalent , since you only use resistive load, the current is in phase with the voltage so you get Ieff = (Ip * 0.707) , the end result is P = Veff * Ieffl.

AC source + 2 resistor = OU ? 

Best Regards,
IceStorm

Edit : LarryC specified to you that he was using RMS value so you dont do a integral to get the right value, RMS = ROOT MEAN SQUARE , the DC equivalent

[Omnibus]

LarryC is Right, for a perfect Sine wave you just need to convert it to DC equivalent , so Veff =  (Vp * 0.707) give you the DC equivalent , since you only use resistive load, the current is in phase with the voltage so you get Ieff = (Ip * 0.707) , the end result is P = Veff * Ieffl.

AC source + 2 resistor = OU ? 

Best Regards,
IceStorm

No, @LarryC is not right and he should know better when trying to carry out power measurements, especially when the reaction of the core inevitably changes the sine wave from being perfect, let alone if there is an offset.

Also, as I have shown OU effect isn't observed at all frequencies (this I'm confirming with the irrefutable Hall effect based current probe). So, there's more to be desired when one sets to debunk one's claims (the least that one should know is what part of the impedance is responsible for the losses).

[IceStorm]

I guess I can't do more at this point but to show you some more data similar to what I've already posted. Here are data taken at 0.1Hz and a simple resistor of about 10Ohm replacing the transformer. As you can see, the OU effect shows more than ever even at such a low frequency (it can be even lower).

I was refering to this post , you removed the trafo and used a resistor, so Vrms * Irms = P but look at what you got as result lol