Please, note that the results from the tests are preliminary. I would not take these results too seriously.


I just do not like them because the step driver source is not a reliable power source for this particular application. I wanted to share them for argument sake and keep our discussion. Shortly, I hope to start the tests as intended by Mr. Figuera in his patent. Only then, we can discuss the results with more authority.


I have to say that I have gained a lot of knowledge from this setup. In this learning process, I have made few changes and witness the output current grow from 0.1 Amp to about 1.0 Amp. I am sharing my findings in this thread as a way to help you avoid my mistakes.


I cannot wait to build another one with an iron core having larger cross-sectional area.

There seems to be some confusion with the application of the Thevenin circuit. Well, the Thevenin circuit is the same for standard transformers also. The circuit has a voltage source (Thevenin) voltage in series with a resistors (Thevenin resistance). At open circuit, there is no current flowing and the output voltage is equal to the Thevenin voltage. As you load the circuit, the current will increase and the output voltage will decrease as per the voltage division formed by the load and the Thevenin resistance (power source internal resistor).
If you continue changing the load, there will be a point at which the load impedance is equal to the internal resistance of the power source (Thevenin resistance). This point is known as the maximum power transfer condition. If you continue changing the load to a zero impendance, the condition is known as short circuit. At short circuit, the power transferred to the load is zero. But the Thevenin circuit (secondary of the transformer) is heavily load showing the higher current value. The secondary of the transformer is producing power at a rate of Voc x Isc which is dissipated in the internal (Thevenin) resistance as heat. That is why the short circuit test shall be as quick as possible. Otherwise, you endup burning the transformer. For standard transfomers, the short circuit test is performed by reducing the applied voltage in order to limit the power being dissipated.
Notice that the maximum power transfer theorem is not used in power systems because 50% of the power being transferred is consumed by the load and 50% is consumed by the transformer. The maximum power condition is only used for communication and audio systems that deal with relatively low amount of power.
I kind of surprise that this basic concept has caused such a ruckus.

But coils on a separate cores is not a normal transformer. When I do the experiment with AC a normal coil on a normal core and place another coil on another identical core so the cores are very close together the supply voltage does not drop nor does the supply power increase (if it does it is not much, I cannot notice it, if anything it seems to reduce), only the induced voltage drops in the separate or induced coil and some current flows. If I use the theorem you used I would get similar results with that as well. I will test this second arrangement. The coils in this experiment will have less than 1 Ohm resistance.

Was the AC voltage RMS for all values in your equations ?

Cheers

I guess my point is that if there is extra energy with an arrangement as described then the same would happen with any such arrangement, the excitement method would make little to no difference. If there is extra energy I'll use it, if not I cannot use it. That is my angle.

Why use a special driver, why not power it with an isolation transformer and a sine wave ?

..

Hi Bajac,

On the Barbosa and Leal thread, SolarLab posted in reply #375, an interesting link to a website about Dr. Harold Aspden. In report #1, Dr. Aspden writes in length about "air gaps". I don't know if this will help or not, just thought it interesting.

Bob

Quote from: SolarLab on January 24, 2014, 04:55:07 PM
Hi Fellows,

Dr. Harold Aspden's two latest patents [UK Patent # 2,432,463 May 23, 2007 and #2,390,941 January 21, 2004] both relating to "Electrical power generating apparatus."

Here are several related links, not only to the patent information but Aether Electric theory in general.

http://peswiki.com/index.php/Harold_Aspden
Scroll down to the PATENTS heading.

http://haroldaspden.com/
http://haroldaspden.com/reports/index.htm     
Aspden's "Reports," especially No. 1 and No. 6; you may find provide a fresh prospective ().

Regards...

Bajac, Can I respectfully ask what the DC resistance of your 800 turn secondary coil is ?

My Tutor indicates that the output impedance is what the Thevenin's Theorem tells us. eg. an open circuit voltage of 50.5 volts and a short circuit current of 0.463 A tell me the output impedance is about 109 Ohms.

Handy online calculator
http://www.electronics2000.co.uk/calc/power-calculator.php

QuoteRESISTANCE, REACTANCE AND IMPEDANCE

Resistance causes the loss of (i.e. dissipates) power,
reactance does not. Pure (ideal) reactance returns all energy that it stores in its field.

http://www.magazines007.com/pdf/Brooks-RCI-2.pdf


And here.

http://www.allaboutcircuits.com/vol_2/chpt_11/1.html

QuoteAt a frequency of 60 Hz, the 160 millihenrys of inductance gives us 60.319 Ω of inductive reactance. This reactance combines with the 60 Ω of resistance to form a total load impedance of 60 + j60.319 Ω, or 85.078 Ω ∠ 45.152o. If we're not concerned with phase angles (which we're not at this point), we may calculate current in the circuit by taking the polar magnitude of the voltage source (120 volts) and dividing it by the polar magnitude of the impedance (85.078 Ω). With a power supply voltage of 120 volts RMS, our load current is 1.410 amps. This is the figure an RMS ammeter would indicate if connected in series with the resistor and inductor.

We already know that reactive components dissipate zero power, as they equally absorb power from, and return power to, the rest of the circuit. Therefore, any inductive reactance in this load will likewise dissipate zero power. The only thing left to dissipate power here is the resistive portion of the load impedance. If we look at the waveform plot of voltage, current, and total power for this circuit, we see how this combination works in Figure below.

Cheers