ENERGY AMPLIFICATION

[gyulasun]

Well, I cannot tell you from a distance either where the error is in your measurements.  If you understand what I am saying in my above post, you will surely clarify it. I have no intention to pull your legs.

All your instruments are run from the mains and perhaps the inputs of the instruments are not ground independents when two separate ones like the scope and the HP multimeter is connected to the common output of the transformer, I am just guessing.

Gyula

[gyulasun]

Maybe using a voltage doubler full wave rectifier at the output and then load the DC voltage output with a known resistor could better help in evaluating the output power because in that case you could use a simple handheld digital DC voltmeter (totally ground independent if the HP multimeters are not) to see the voltage across the known resistor. Here is a voltage doubling full wave rectifier schematic I think of:
http://www.play-hookey.com/ac_theory/power_supply/images/rectifier_doubler_expandable.gif  from this link:
www.play-hookey.com/ac_theory/power_supply/ps_v_multipliers.html   and no real need for grounding one leg of your output coils of course, as is suggested in the schematic.  The voltage doubling helps reducing the relative loss occuring in the actual output power, caused by the two diodes forward voltage drops. I suggest using at least two 1000 uF (microfarad) 25V or 35V DC rated electrolytic capacitors for C1 and C2.  I would use at least a good 10 Ohm resistor, not 1 Ohm to load the DC output of the voltage doubler so the 10 Ohm would be connected in parallel with capacitor C2.

I know the diodes will still cause some loss and to nearly compensate for this, the measured output DC voltage could be increased by say 1.2V (2 x 0.6V forward voltage drop) when calculating output power, this depends on your diodes.

[madddann]

@Farmhand
Since you are not a moderator, that post will stay right where it is  .

...and according to my calculations, the COP of this sistem is about 363%, which is about right for a Flynn paralell path system...

[Farmhand]

@Farmhand
Since you are not a moderator, that post will stay right where it is  .

...and according to my calculations, the COP of this sistem is about 363%, which is about right for a Flynn paralell path system...

If I was a moderator it would be already gone, I was just giving my opinion and explaining why I posted. Take it as you wish.

..

[Farmhand]

Gyulasun:  So if I use the same one OHM resistor and insert in series with the input, and measure the voltage across it (6.5 mv) and use the same calculation used on the output, (0.006.5² / 1= 0.00004225) than
the input current seems to be 0.042 ma, making the COP = .003/0.00004225 = 71 which doesn't seem right to me.

Just my 2 cents br549 but possibly the efficiency is low because there is no appropriate load, a 0.1 Ohm resistor is not really a good load. Maybe if you were to use a load at the output say maybe a 10 or 100 or even 1K resistor and use the current meter in series with the load and the secondary the efficiency would look better. Then it can also be checked by using the scope to measure the current across one of the resistors to get the voltage across the resistor then calculate the current from that using Ohms law.
EDIT: With AC currents the phase angle or power factor needs to be considered as well. Which usually works out in favor of a higher efficiency for the experimenters device, I thought I should add.

Also if the input is not affected by the output then the input will be always the same and always maximum so the way to get the best efficiency in such a setup is to load it appropriately so that the output better matches the input. It depends on the coupling and such things, if there is a path for energy to return to the supply or not ect.

Cheers