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

[listener191]

Hi G4RR3TT,

Slide 42 of the Bbcock/Murray presentation shows capacitor charged in the positive half sine is discharged in the negative half sine period. The video of the Japanese SERPS demo shows fixed links between the two sets of caps that rule out parallel charging, series discharging.. so what arrangement do you suggest that will deliver equal magnitude charging and discharging current?

Barry

[listener191]

Hi G4RR3TT,

Perhaps what I should have said is.. other than a center tapped transformer, what arrangement do you suggest?

A center tapped transformer cannot be ruled out as, the three variants of SERPS that I have seen, all have transformers.

Personally I would like to see the current variants run on a small stand alone generator, as it would be easy to hear if a given load on the generator is lightened, when  the SERPS circuit is switched in.

The absence of line input waveforms from the wall, showing the I & V phase, is disturbing. Not much point of the device if the 1W real draw on the secondary side of the transformer for 50W in the load, is supported by 500VA from the line!

Barry

[Farmhand]

Thread is difficult to read due to oversized images, if people paragraph the text it can be read without scrolling sideways.
Nothing will put people off reading a thread like having to scroll sideways for every line of text.

Now my point to the drawing I posted was to see if anyone can determine the net power drawn from the supply and the power
dissipated by the resistor, not how efficient cap charging and discharging is.

If no one can determine the power dissipated by the resistor then how can we even know what will be OU and what won't ?

Seems to me that when the caps are charged through the resistor the energy that charges the capacitor does not heat the
resistor and on discharge any energy returned to the supply does not heat the resistor either.

So where does the OU come into it ? I say it is in the incorrect measurement of power. To put it simply the power dissipated in
the resistive load cannot charge the capacitors and the power returned to the supply cannot heat the resistors, the
circuit behavior I think is normal for what it is, the thing that is not normal is the way the power would need to be measured.
I don't think regular measurement methods will show accurate values.

If anyone can determine the powers in the circuit below we might be able to see where the problem lies.

..

[Farmhand]

Surely we can agree on those simple points.

1) On the charge phase the power dissipated by the resistors cannot charge the capacitors.

2) On the discharge phase the power dissipated by the resistors cannot be returned to the supply.

If we can not all agree on those two points, we have strange things to discuss. 

If anyone disagrees with the two points above please say so and explain why and how it can be different.

..

P.S. Basically these people are claiming that they can draw power from the supply and dissipate almost all of that power in
the light bulbs and then return that same power to the supply.

Our job as experimenters is to determine what is actually happening because what I just described cannot be what is going on.
Can't have your cake and eat it too.

..

[SeaMonkey]

Agreed, in the process of "recycling" the electrical
power there must be some "losses" in the Load.

Is the return of some fraction of the power at
critical points in the waveform sufficient to "fool"
the prime mover into thinking that it is working
into a very small load?

What is going on seems to defy common sense.

The proclaimed "COP" is simply an illusion?