Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[verpies]

Question for verpies. What happens in below circuit at resonance ?

This.

At resonance impedance of the whole system is zero.

At resonance the input impedance is infinite (open circuit), because the primary winding forms a parallel LC tank with the cap.
The secondary winding forms a series circuit - an important difference.

When load is applied, <input> impedance is still zero ?

No, as it was never zero to begin with.
When a resistive load is connected to the secondary winding, the apparent inductance of the primary winding decreases due to reflected secondary impedance caused by mutual inductance and the Lenz law.
As a consequence, the primary LC tank's resonance frequency increases under load.
In other words, the primary LC tank goes off-resonance under load.  This video illustrates this behavior.

Of course, the secondary load in your circuit and in the circuit from the 1^st video is not purely resistive (R), but the principle remains.
In those scenarios, when the magnetic coupling is complete (1), the load is a series RC load and when the magnetic coupling is incomplete (<1) the load is a series RLC load.
BTW: Magnetic coupling coefficient (0..1) is related to the mutual inductance.

When power is taken <from the secondary>, the more power you take, the less power will be consumed in the source ?

No


P.S.
It shall be emphasized that the brightness of the input light bulb does not indicate the average power drawn from the power source.  It only indicates the average input current.
However, the brightness of the output bulb, does indicate the power dissipated in the output bulb.
Measuring input power and output power with light bulbs is very different due to MPTT.

[x_name41]

to all: http://realstrannik.ru/forum/39-kapanadze/134730-generatory-kapanadze-obshhaya-tema-2.html?start=180#193261
          http://realstrannik.ru/forum/39-kapanadze/134730-generatory-kapanadze-obshhaya-tema-2.html?start=180#193101

[Zeitmaschine]

Question for verpies. What happens in below circuit at resonance ?

The question should be, what happens at PARAMETRIC RESONANCE. Because if we rule out high voltage and high frequency as well as a spark gap, then all what's left is parametric resonance that could do anything useful with a coil or a transformer.

Going on with wrong connections: When both anodes of the two thyristors are connected to a transformer's coil and the two capacitors are also connected to the anodes then we have a resonant LC circuit made out of two capacitors in series and one coil in parallel. What if these two capacitors are shorted in alternating mode by the thyristors? (provided that SR193 is not fake) How to connect a three-pin two thyristor construction to a four-pin transformer in an intelligent way?

Given the size of the two Stepanov capacitors I think they are too large to just connect the gates of the thyristors to something.

[Zeitmaschine]

One more guesswork: Either it is the poor image quality or it is another strange coincidence. But as it seems one wire (of the cable) coming from the two capacitors is connected to the thyristor's anode or cathode (depends on the construction type of the thyristor) but not to the thyristor's gate. That's the same unusual connection method like the odd SR193 wiring.

Furthermore that whole capacitor switching technique seems to be not very far from that idea of mine a year ago. Interesting ...

[Zeitmaschine]

Two capacitors in series = low overall capacity = high resonant frequency
One capacitor shorted = high overall capacity = low resonant frequency

The curious thing is, shorting one of the capacitors (in order to get parametric C resonance) on the peak of a half-wave means destroying the energy stored in that capacitor because on voltage peak all energy of the LC circuit is stored in the capacitors as an electric field. Or I'm wrong? On the other hand, shorting a charged capacitor 50 times a second would explain why the thyristors have to be on big heat sinks cooled by a fan in addition.

Thinking twice: In case of the pendulum, does down shifting of the weight (go to low frequency) on its end positions not also destroy the elevation energy of that weight? Nevertheless the energy budget of the parametric excited pendulum is positive.