Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[sparks]

http://www.youtube.com/watch?v=qtI1CPBSm-o&feature=related


Hmmm...

Tesla would never have wasted the current flow charging the capacitor.  He would have used the current to charge the capacitor to drive a load by installaing a current transformer on each leg of the capacitor leads.  Then he would have flipped the capacitor over and shorted it back into the battery again using the reverse current to drive a load.  He regularly used an electrical capacitor as a spring and built highspeed mechanical oscillators to build "Earthquake machines"

[bass]

1V on 1â,,¦ = 1A, 1V × 1A = 1W, 220V on 220â,,¦ = 1A, 1A × 220V = 220W, 5000W ÷ 220W = 22.73, 220â,,¦ ÷ 22.73 = 9.68â,,¦

Thus Kapanadze's five 1KW lamps should have a combined parallel resistance of 9.68â,,¦, that means the voltage source (in the tin can) should have no higher resistance than that. If the voltage source is a coil then this is not the DC resistance but the apparent resistance (impedance) at 50Hz.

So, what's next?

Grate work!!!

Calculate the active resistance of the secondary winding of the transformer Kapanadze in the bank.
The formula to calculate the following

where S - square of wire
L - length of the winding wires
0.0175 - resistivity of copper
For simplicity, in your case, we assume that the primary of the transformer consists of 344 turns. one volt/one turn  . Then the secondary winding consists of 220 turns on it that would be 220 ​​volts.
Suppose that the winding is wound on the transformer core with a radius equal to 25 mm. Then the length of the one turn of secondary winding will be equal to 2 * 3.14 * 0.025 = 0.05785 m
Then, the total length of the wire of the secondary winding of the transformer will be 220 * 0.05785 = 12.727 m
We assume that the secondary winding of the transformer is wound with wire 2.5 mm2 that would pass current 25 A.
Then the active resistance of the secondary winding is equal to R = 0.0175 * 12.727/2.5 = 0.09 ohms.

much less than the load active resistance
Although I may be wrong

More about what you should consider: the higher the frequency the greater the Q

Ie higher the voltage on the primary winding of the transformer, means more reactive power.

[bass]

.

[xenomorphlabs]

Grate work!!!

Calculate the active resistance of the secondary winding of the transformer Kapanadze in the bank.
The formula to calculate the following

where S - square of wire
L - length of the winding wires
0.0175 - resistivity of copper
For simplicity, in your case, we assume that the primary of the transformer consists of 344 turns. one volt/one turn  . Then the secondary winding consists of 220 turns on it that would be 220 ​​volts.
Suppose that the winding is wound on the transformer core with a radius equal to 25 mm. Then the length of the one turn of secondary winding will be equal to 2 * 3.14 * 0.025 = 0.05785 m
Then, the total length of the wire of the secondary winding of the transformer will be 220 * 0.05785 = 12.727 m
We assume that the secondary winding of the transformer is wound with wire 2.5 mm2 that would pass current 25 A.
Then the active resistance of the secondary winding is equal to R = 0.0175 * 12.727/2.5 = 0.09 ohms.

much less than the load active resistance
Although I may be wrong

More about what you should consider: the higher the frequency the greater the Q

Ie higher the voltage on the primary winding of the transformer, means more reactive power.

That would be the ohmic part, in the case of AC in the secondary the impedance has to be considered as well ...

[jbignes5]

Tesla would never have wasted the current flow charging the capacitor.  He would have used the current to charge the capacitor to drive a load by installaing a current transformer on each leg of the capacitor leads.  Then he would have flipped the capacitor over and shorted it back into the battery again using the reverse current to drive a load.  He regularly used an electrical capacitor as a spring and built highspeed mechanical oscillators to build "Earthquake machines"

What are you talking about? I showed a bifilar coil being energized with zero current driving a loaded coil/ferrite in the core.


If you were talking about The other Hmmm... Then I would have to agree. Anyways a battery doesn't put out negative or positive electrons. Think of the wire as a static surface. The positive gets sucked to one end when in circuit. This is exactly how a cap works to. The positive electrons in the metal plates get moved to one side and the lack of electrons on the other side of that plate.


http://upload.wikimedia.org/wikipedia/commons/thumb/d/d9/Electrostatic_induction.svg/440px-Electrostatic_induction.svg.png

Current is the charges contained in the material. The electrostatic voltage must be present for current or charges to move. Charges that are moving is called current and can and will have a magnetic or sympathetic movement of charges outside the body(magnetic).

This is exactly what is happening and why we are missing part of the puzzle. we have ignored the electric and only concentrated on the magnetic.

In most of our traditional circuits we have to complete the path or close loop the device to get it to work. But my example of what is going on completely explains this young mans experiment with the bifilar coil and load.

again replicate this yourself and see.

http://www.youtube.com/watch?v=qtI1CPBSm-o&feature=related

Now even in his description he mistakenly says they are magnetically coupled which they are not because there is zero current flow in the bifilar coil. This is an electrostatic capacitance coupling. A very different beast all together. But as you have seen it can and will drive a magnetic load (wire closed looped plus magnetic ferrite core closed looped to a load). If there is no current in the bifilar coil then there is zero magnetic field.

If we take this one step further we could put the open coils into a toroid and drive this to infinity if one wanted. No current would be used from the source and yet I believe it can power larger loads. I will be working on this concept since it already shows OU type action and a perfect separation of the source to load.