ENERGY AMPLIFICATION

[forest]

There is no end to this...even if somebody would show a closed loop system running like Tariel Kapanadze did, it ill be immediatelly attacked and treated as a hoax. Farmhand, how do you propose to PROVE overunity ?

[gyulasun]

Hi forest,

For instance, for me a self-looped setup would be a perfect proof (without mistereus wires of course). But I just found the service manual for the HP 3478A multimeter and it turns out that it uses an inside 0.1 Ohm resistor when measuring current...  this means that measurement error is possible because if you measure 1.51V rms across a 0.1 Ohm resistor then the 50 mA rms current measured is not correct or something else is not correct, sorry.
Link to service and operation manual: http://www.ko4bb.com/manuals/index.php?dir=HP_Agilent/HP_3478A_Multimeter
I am not saying anything other but to double or triple check the measurements, right?

Gyula

[profitis]

@forest it wont be treated as a joke by corporations if it works believe me,if they want to shelve it let them shelve it as long as you make a fastbuck or two in the process.

[br549]

Hi Br549,

Thanks for clarifying the load: it is the inner resistance (impedance) of the HP3478A multimeter when it is in AC amper mode.

Very interesting setup, have you considered to use a tuning capacitorI looked up the manual on the web for this multimeter but the inner resistance (impedance for AC) was not specified for the two current ranges neither for DC nor for AC,  lower range is 300mA, higher is 3A (fused).  If I accept the 1.51V rms voltage drop across the ammeter input and I accept the current is 50mA rms then the inner resistance (impedance for AC) is R=1.51/0.05=30 Ohm   So this means your load, willy-nilly, was a 30 Ohm resistor, right?  This 30 Ohm sounds a bit high to me for a current meter. Agree with this?  I also assume that such a current meter (HP) does not have an inductive inner impedance for current measurements, so the AC current and voltage more or less are in phase, right? 
Are you not curious if the 30 Ohm inner resistance could be measured from the outside by a normal handheld digital Ohm meter? Well I am...      Just for a simple cross check, probably the DC ammeter input is the same two terminal as the AC ammeter input, so by choosing the AC current mode from the DC current mode on the HP the input reistance will not change I suppose. You could see this resistance on the hand held DMM hooked to the ammeter input.

One more thing: would you mind checking the output voltage across the ammeter with the other HP multimeter, switched to AC voltage mode?  Also a simple cross check.  Another thing: if you have a 100 or 200 or 300 Ohm (non wire wound) resistor (wattage is 0.25 or 0.5W)  and used it as the load across the transformer output and also measure the AC voltage across this resistor, you could see the rms voltage either on the scope or the HP multimeter (no need to check the current in this resistor load case). at the output like at the input? This way the output would be more sinusoid too.

rgds, Gyula

gyulasun:  The Internal resistance of the HP 3478A (when set on the AC Current Scale and measured with the other HP 3478A) is one OHM. I replaced the meter on the transformers output with a 1 W (one ohm resistor, and measured the voltage across it. The measurment was (55 millivolts AC). I then removed the one OHM resistor and replaced it with the HP 3478A (set on the AC current scale) and it read 56 milliamps. The other 3478A was in series with the input, and is reading 6.373 milliamps.
The transformer does seem to become more efficient as the output load increases. I will do some more testing (load variation and graph) to see.

[gyulasun]

Hi Br549,

Very good, thanks for the efforts. 

So the measurements with the 1 Ohm resistor gives output power of about 3mW (0.055² / 1= 0.003)  If I assume that the input voltage was about 3.9 Vrms again at the now measured input current of 6.373mA then the input power was about 24.8mW,  (COP=3/24.8 i.e. under 1).

You may also wish to consider to tune out the output coils inductive reactance with a similar amount of series capacitive reactance.  And how the input current draw changes when the load changes at the output, that would be also a good info.


rgds,  Gyula