> VARs =3D Volt-Amps reactive and is the product of AC voltage The thread at metabunk has unceremoniously pulled QEG apart line by line, claim by claim, statement by statement An exchange between two posters about QEG figures as published by the "inventors" https://www.metabunk.org/threads/debunked-quantum-energy-generator-qeg-10kw= -out-for-1kw-in.3572/page-2 User2718218 - 1900 volts peak-to-peak, divided by two equals 950 volts. 950 volts times 0.7071 gives you 671.7 volts RMS 0.95 amps divided by two gives you 0.475 amps. 0.475 amps times 0.7071 gives you 0.3359 amps RMS. 671.7 Volts RMS x 0.3359 Amps RMS =3D 226 watts The electric motor is drawing 655 watts of power. Therefore the efficiency of the QEG is 226/655 x 100 =3D 34.5%. EsotericScience - Your calculations for the RMS power are incorrect. P =3D V x I =3D 1900 x 1 (as stated on video) =3D 1900 VA. Divide by 2 and times .7071 =3D 671W. This would give overunity, not by much but overunity nonetheless. At the very least it's close to unity depending on the exact values, but certainly much greater than 35% efficiency. User2718218 - No, my calculations are correct. P =3D V x I if V and I are constants. If you have a resistive load so that the current is always in phase with the voltage, and V is now V(t), some kind of arbitrary repeating waveform, then you have to take the RMS value for the V. The power can be Vrms^2/R or Irms^2 x R or Vrms x Irms, they are all the same thing. In addition, in looking at the clip, the light bulbs look like there is about 226 watts of power dissipation, not 671 watts. This is subjective from watching a video but if you had to pick the most likely fit, 226 watts fits what is seen in the clip. If you are in contact with "Allegedly Known as Dave" and James and HopeGirl, this serious mistake needs to be corrected. EsotericScience - They are not the same. Think about it this way. Lets consider a short time interval for which there is little change in the voltage and current, say around the peak. Lets assume for this example that V=3D2000V, I=3D1A then Power dissipated for that time interval is 500 VA. If you use your method the power for the same time interval is 2000x.353 =3D 707 times 1 x .353 giving 250 VA which is wrong. It's the average of the squares versus the square of the average type issue. User2718218 - At the peak like you are describing, the instantaneous power is 2000 watts. I don't know where you come up with "500 VA. "VA" is normally used for transformers. "RMS" means the square-root of the average of the squares. Please do a Google search on "measuring AC power across a resistor." Click on the first link and you will be taken to a web page called "ElectronicsTutorials" and you will land in the section called "Resistors in AC Circuits." You will see that they explain exactly what I explained in a previous posting. Dave and James really got it wrong and it's a serious serious problem that needs to be corrected. > So, they are not claiming an over unity machine per se They do - "The QEG was drawing 500 watts, the secondary was still powering its 600 watt load of light bulbs and yet on the primary side we were reading nearly 2000 volts at 1 amp. If I had gone by Jamie's furrowed expression alone I might have missed the fact that he had just achieved serious over-unity 500 Watts in and 2000 Watts out... I had to rouse him from his deep contemplation of how to reach his target of 10Kw to remind him to celebrate the fact that he had actually achieved 2.6kW... I managed to get a high five out of him before he resumed his tinkering. .... Breaking News: Hot off the press!!! I just spoke to Jamie and he mentioned that he has now achieved 3kW output, that'ssix times over unity!" --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .