Reactive power - Reactive Generator research from GotoLuc - discussion thread

[poynt99]

I must clarify something, because I was working on the notion of a pure LRC circuit. In this case, yes I would say the circuit is highly inductive and a signal inversion is taking place somewhere.

However, Luc is using a transformer as an inductor, and this changes things when the secondary is loaded or shorted.

What I found is this:

When the secondary is shorted and therefore left with the secondary's DC winding resistance as a load, that load is reflected back to the primary, effectively shorting it to a degree as well. As an example, let's say the primary inductance is 280mH and the secondary is 5H, with a DC winding resistance of 100 Ohms. When the secondary is shorted, the 100 Ohm winding resistance is reflected back across the primary as a ratio of the secondary to primary inductance, in this case 5H/.28H = 17.86. So this would result in the equivalent of a 5.6 Ohm resistor being placed across the primary winding.

When Luc says that the "magic" is in shorting the MOT secondary, what he is really doing is essentially eliminating the MOT inductance, making the circuit mostly capacitive, hence the "i leading v" result we see in the scope shot. The voltage swing across the primary will change from 350V to a few hundred mV when the secondary is shorted.

Now, there's still a problem; the capacitive reactance of the 25uF cap at 60 Hz is not enough to cause a 90º phase shift across the network, and we are back to what I said before about the circuit having to be purely capacitive (or mostly) with a capacitor value of about 1uF, not 25uF. I would like to see Luc measure his capacitor stack to see how close it really is to 25uF.

I'd also like to see Luc remove the MOT and test the circuit again. He should get a similar 90º result.

[poynt99]

Luc just emailed me this:

what is relevant is the zero effect to the generator and delivering over 20 Watts to the load!... this is what needs to be understood.

Actually, you've not proven that the resistor is dissipating 20W. Why? Because that 1k resistor is going to be inductive. The only way you are going to be 100% certain is to either measure its temperature and try to match it with a DC supply, or you use the scope to perform AVG[v(t) x i(t)] in order to get the average power of the resistor.

Before you get too excited about any measurement, especially a power measurement, you must confirm that measurement by at least one other means. This mantra has been stated endless times in this forum but it seems to always fall on deaf ears. And to date, no one here has made a single watt of OU power, which proves that they've been wrong 100% of the time they make a claim based on their botched power measurements.

[tim123]

In order to prove there's real power being developed, Luc could presumably replace the resistive load with a FWBR. Then put the same resistive load across the DC terminals (+ smoothing cap) and measure the DC voltage...

The trouble with the AC measurements is they have to include the power factor. I really don't think you can calculate AC power in the way that Luc was doing, but then he hasn't set out exactly how he's calculating these values (such as 10-20w out and zero in) in a post, so I'm not sure.

I have tried again - this time with the circuit from this post (i.e. with the load on the secondary):
http://www.overunity.com/14013/reactive-generator-research-for-everyone-to-share/msg377250/#msg377250

I found that if I calculated the power across the load *assuming* a power factor of 1 - then it appears to be OU.

All the calcs are complicated, and so I did what I suggested above - and used a FWBR. This time the output was sensibly UU...

[poynt99]

You can in fact calculate the real power in the resistor by measuring the rms voltage across it as Luc has done, but ONLY if the resistor itself is non-inductive, AND the two meter leads are right tight against the body of the resistor. But in this case since we are dealing with 60Hz, the latter requirement is not necessary, within reason.

Sure, Luc could use a FWBR on the output of the secondary as you suggested and tested yourself. This is yet another possible check that he has most likely NOT performed.

Are you using a resistor similar to his?

[tim123]

I was using a standard low-power resistor - 150ohms - probably wire wound - not sure...
As the instruction is to use only 1-2 watts - i thought it'd be ok. It was.
I didn't use the variac this time.