Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[Zeitmaschine]

looking at the wiring for the capacitors on the bottom-right of the picture (you provided), it appears that there's 2 capacitors wired in series (labelled I and II).

But not connected as it seems because one wire is cut. And that indicates that a precise capacitance is needed here (to get resonance).

2) The current oscillating back and forth, via one wire, is stored to a internal capacitance.

Why not to an internal inductance? A resonant LC circuit needs a capacitance and an inductance. Therefore the energy can be stored in the magnetic field. How do we calculate that?

Regards

[Grumage]

Interesting reference article.

It indicates that L and C should be in series.

Dear Haan.

Also note that the same formula for calculating resonance can be used for "SIMPLE" series or parallel circuits!!

Cheers Grum.

[Haan]

But not connected as it seems because one wire is cut. And that indicates that a precise capacitance is needed here (to get resonance).

It only appears to be cut because of the low image resolution resulting in pixelation.

I thought you would comment on the 6 sets of capacitors observed ...

[Zeitmaschine]

I see this as a clear cut. These two capacitors are connected in series and not 3-phase. So they are apparently not for the same purpose like the others (if they where connected at all). The white wire on the right side goes to a different direction (upwards) than the other white wires. Maybe they are for the second resonance with higher frequency thus the series connection which lowers the capacitance.

Question: When the device is a 3-phase one then it works most likely with a 3-phase transformer. But does each phase then need a separated high voltage input or just one high voltage input (and therefore just one high voltage resonance) for all 3 phases together?

[baroutologos]

Why not to an internal inductance? A resonant LC circuit needs a capacitance and an inductance. Therefore the energy can be stored in the magnetic field. How do we calculate that?

Regards

SImply because internal inductance poses no problem. Its easily made. Internal capacitance of huge figures for a single terminal oscillator is an impossibility!