Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[Zeitmaschine]

Interested in more conceptual errors ... err ... theories? Here it comes:

Let's say I have two capacitors each 1µF charged with 10 volts each, so the overall stored energy equals 100µ Joules. Then I discharge each capacitor into a coil, so I have a magnetic field in two coils accordingly. Next I connect these two coils in series so that the back voltage adds up to 20 volts. Now I take these 20 volts and charge a 1µF capacitor with it. Then as a result I should have a capacitor of 1µF at 20 volts, so the stored energy should equal 200µ Joules.

Therefore: 2µF/10V becomes 1µF/20V which equals an energy gain of one hundred percent (in an ideal circuit).

But let me guess: There has to be something wrong with this conception, hasn't it?

[verpies]

...I have a magnetic field in two coils accordingly. Next I connect these two coils in series so that the back voltage adds up to 20 volts.

The coils do not store energy as voltage. Ideal inductors store energy as electric current/magnetic field ( W= ½LI² ) and they don't lose that energy only while they are shorted.
What do you mean by "back voltage" ?

[Zeitmaschine]

Two magnetic fields in two coils are generated by two charged capacitors simultaneously. As long as there are no changes made in the circuits then each magnetic field of the coils will create a back voltage (Back EMF) and will recharge its own capacitor in an oscillation (with slight losses). That would be the behavior of an ordinary LC circuit. So far, so good.

But as soon as the voltage of the capacitors is zero (zero crossing) the coils are disconnected from these capacitors and connected in series to a single capacitor of the same capacity as one of the initial two capacitors. Hence the series back voltage (Back EMF) of the coils should have enough power to charge this single capacitor to the combined voltage (with slight losses) previously received from two capacitors each of the same capacity. Since the energy (Joules) stored in a capacitor is a linear function of the capacity but a square function of the voltage, half the capacity but double the voltage means twice the stored energy (as shown by means of the Capacitor Charge Calculator).

In other words: Nothing is gained by connecting two parallel charged capacitors in series. The voltage doubles but the capacity is divide by four so the stored energy stays the same. But transferring the energy of two capacitors to the magnetic fields of two coils and then connecting these coils in series while receiving the Back EMF the outcome could be somewhat different.

The remaining question is: How to do this practically?

[elementSix]

Has anyone ever gotten these kind of sparks before.  I took these off of the video from the homepage of our site here.  I don't know if its the metal he is using, but they are very much like the energy sparks that Tesla observed from his HF Pulsed DC experiments with Aether Energy in his magnifying transmitter.  It was Cold electricity that was totally different from regular hot currents.  I don't think this is, but they are nice sparks between his gap and grab pretty far.  Of course he is using like 6 amps..  He doesn't say why he made this device or what use it has...

[Farmhand]

Interested in more conceptual errors ... err ... theories? Here it comes:

Let's say I have two capacitors each 1µF charged with 10 volts each, so the overall stored energy equals 100µ Joules. Then I discharge each capacitor into a coil, so I have a magnetic field in two coils accordingly. Next I connect these two coils in series so that the back voltage adds up to 20 volts. Now I take these 20 volts and charge a 1µF capacitor with it. Then as a result I should have a capacitor of 1µF at 20 volts, so the stored energy should equal 200µ Joules.

Therefore: 2µF/10V becomes 1µF/20V which equals an energy gain of one hundred percent (in an ideal circuit).

But let me guess: There has to be something wrong with this conception, hasn't it?

Capacitors are non linear devices ie. if we take a 1000uF capacitor and charge it to 100 volts we have 5 Joules of potential energy.
If we then charge the 1000uF capacitor to 200 volts it will have 20 Joules of potential energy, which is 15 joules more for the same voltage increase.
Then to charge the 1000uF to 300 volts it will have 45 Joules of potential energy which is 25 Joules for that 100 volt increase.

One 2uF capacitor at 10 volts is 100 microjoules so two capacitors equals 200 microjoules and a 1uF capacitor at at 20 volts is 200 microjoules so that's unity.

Cheers