COP 20.00 (2000%) Times, Reactive Power Energy Source Generator,

[wayne49s]

[@TinselKoala
Ya, what you have is VAR out/power in.

[wayne49s]

All,

Apologies for the crude schematic. This scheme is in alignment with slide 41 of the  just released Babcock Murray presentation. This doesnt need the caps to be be series linked for discharge.


Barry

I see, this is their circuit.. They discharge the positive cap charge in the negative half cycle, and vice-versa.
Worth trying to experiment.


Comment about classical analysis: OU cannot be computed with classical analysis? since it has been distorted to not have OU. No?


/Wayne

[wayne49s]

I see, this is their circuit.. They discharge the positive cap charge in the negative half cycle, and vice-versa.
Worth trying to experiment.


Comment about classical analysis: OU cannot be computed with classical analysis? since it has been distorted to not have OU. No?


/Wayne

Sorry, must be too early in the morning here (6am). I just read Garret's comment about the problem with the cap voltage in series..then something doesn't add up with what the authors are saying?

[Farmhand]

If you go to this page we can see the crazy claims made by these guys, using Tesla and a misuse of Tesla's words to try to make
claims for Tesla he did not make. http://teslaspowermagnification.com/

Power can be taken at a small wattage and then the energy from that power can be delivered at much higher powers, that has
nothing to do with OU and the energy is still the same and no claim of more "energy" is ever made by Tesla. Power is not energy.

From link.

In layman's terms, energy is supplied by a power transformer to run the lights and then is stored in a capacitor. The capacitor is then discharged back through the lights again to the power supply in a way that neutralizes the load seen by the power supply.

The consequences of this energy oscillation is that the load (bulbs) can be powered twice while the net energy supplied by the power supply is reduced to a very small value. The implications of this technology for energy use in the future is absolutely astonishing!

Here is an example of the input compared to the output as measured by a Tektronix scope... What this graph shows is that the SERPS device is drawing 1.1 WATTS net from the power supply but the light bulbs are actually burning 52.7 WATTS. 52.7 watts divided by 1.1 watts = a COP or coefficient of performance of 47.90, which is 4790% more energy than is required to run the machine.

The input represents the difference between the energy provided and the energy returned. The output represents the work accomplished as this energy oscillates in and out of the system.

They are misrepresenting Tesla's words.

..

I'll say it again .

1) The energy dissipated by the bulbs as light and heat and the power consumed on the charge phase does not charge the
capacitors.

2) The energy dissipated by the bulbs as light and heat and the power consumed on the discharge phase does not get returned
to the supply.

This leaves only one possibility, the power is not being measured correctly.

..

[G4RR3ττ]

@Farmhand,

I agree with what you're saying, the energy transformed into heat at the resistive load is 100% lost. So when you charge the capacitor through an RC circuit the total energy taken from the power supply is the energy transformed by the resistor PLUS the energy stored in the capacitor: E_taken(net) = E_c(stored) + E_r(loss). When the capacitor discharges, the same transformation of energy at the resistor happens again, so the net energy back to the source is: E_cap(stored) - E_r(loss) = E_returned(net). Which means the resistive load never disappears.

@All,

The only way the SERPS circuit can work--as claimed--is if the magnitudes of the current taken and returned are nearly equal and the times at which these exchanges of charge take place cause it to look 100% reactive to the source supply. The switched capacitor circuit does most of this, except for having unequal charge and discharge currents. The other circuit posted by Listener doesn't force current backwards so it can't force reactive currents to exist. If a transformer were used to reflect the cap discharge in the right direction then it may work, but not as shown.

So the question becomes, how can you--despite loss of energy at the resistor--make the circuit look reactive? Is this even possible? When you push all the extraneous details to the side, and just examine AC circuits, the phase of the current to the voltage is how you determine the magnitude of resistive and reactive components. This effectively means, if you were to somehow shift the phase relation to some arbitrary angle, you could make a resistive load "look" reactive, despite it transforming energy into heat. Now that's theory. But can you really do it? Murray and Babcock seem to think so...