Free energy from capacitor displacement current

[dgreen264]

Hi all

Just wanted to put this idea out there... its regarding the displacement current of a capacitor.
 A capacitor stores energy in the electrostatic field between the plates, but little or no attention is normally given to the fact that a magnetic field is created between the plates when current flows in the wires.

This is called displacement current and produces a (magnetic) field within the capacitor that is equivalent to that which would be created if a wire were to connect the two plates together.

A coil placed between the two plates of the capacitor can receive this oscillating magnetic energy and create an ac voltage on its terminals. A load can be placed on the coil's output and real power can be drawn from it. When current (and hence power) is taken from this coil, the coil generates its own magnetic field which opposes the original field generated by the displacement current (Lenz's law).

The net field will be reduced due to this back flux from the coil (superposition theorem applies). But as there is no 'primary' coil generating the magnetic flux, there is no mechanism for the back flux to reflect electrically back onto the input circuit thus demanding real power input as in a normal transformer.

In the capacitor-generated flux case, all that happens is that the flux is diminished due to the back flux, and this limits the power out because the net flux reduces and hence the output voltage - which is proportional to net flux - reduces.

But the way I see it, there is now way for this to change the input impedance. The input still looks totally reactive i.e.no real power, and can be resonated away to practically zero, allowing for losses etc.
There is no way for this reduction in magnetic field (due to load current) to electrically get back onto the input circuit.

So this would be free energy then?
If anyone can make suggestions or comments as to why this wouldn't work and where I am missing something in the theory, please do tell! Guests cannot see images in the messages. Please register at the forum by clicking here to see images. Guests cannot see images in the messages. Please register at the forum by clicking here to see images.


thanks

[Jim Mac]

Hi all

Just wanted to put this idea out there... its regarding the displacement current of a capacitor.
 A capacitor stores energy in the electrostatic field between the plates, but little or no attention is normally given to the fact that a magnetic field is created between the plates when current flows in the wires.

This is called displacement current and produces a (magnetic) field within the capacitor that is equivalent to that which would be created if a wire were to connect the two plates together.

A coil placed between the two plates of the capacitor can receive this oscillating magnetic energy and create an ac voltage on its terminals. A load can be placed on the coil's output and real power can be drawn from it. When current (and hence power) is taken from this coil, the coil generates its own magnetic field which opposes the original field generated by the displacement current (Lenz's law).

The net field will be reduced due to this back flux from the coil (superposition theorem applies). But as there is no 'primary' coil generating the magnetic flux, there is no mechanism for the back flux to reflect electrically back onto the input circuit thus demanding real power input as in a normal transformer.

In the capacitor-generated flux case, all that happens is that the flux is diminished due to the back flux, and this limits the power out because the net flux reduces and hence the output voltage - which is proportional to net flux - reduces.

But the way I see it, there is now way for this to change the input impedance. The input still looks totally reactive i.e.no real power, and can be resonated away to practically zero, allowing for losses etc.
There is no way for this reduction in magnetic field (due to load current) to electrically get back onto the input circuit.

So this would be free energy then?
If anyone can make suggestions or comments as to why this wouldn't work and where I am missing something in the theory, please do tell! Guests cannot see images in the messages. Please register at the forum by clicking here to see images. Guests cannot see images in the messages. Please register at the forum by clicking here to see images.


thanks

This is interesting.  Let me rephrase and ask a question to see if I understand what was proposed..

1. Discharging a Cap produces a magnetic field between the terminals.
2. Placing a coil between the terminals of a capacitor while discharging it should induce a current that can power some load.
3. Since capacitors can not be charged / discharged with magnetism, the output shall be decoupled from the input.

If all is true, this suggests one could fill capacitors to use on loads, while tapping the extra magnetic field for a little extra bonus.

Did I get that all right?  If so, should be an easy test to see if discharging caps induces coils if placed between terminals.

[dgreen264]

Yes that's right . You've got it!
The cap produces a magnetic field between the plates (i.e. in fresh air!) whenever it takes current, so this happens when charging or discharging.

So you just need to drive the cap with an ac voltage source and supply some current so that you get a constantly changing magnetic field. The phase angle between volts and current is always 90 degrees with a capacitor (or at least near as), which means it doesn't dissipate any real power. It just consumes reactive power which can be made negligible by resonating it with an inductor. (Not the one inside the cap, that is separate!, and the two should not be allowed to couple magnetically).


One of the practical issues with this idea is that because the capacitor is going to have reasonable distance between the plates in order to insert a coil of wire, its capacitance is going to be very small.
With very low capacitance there is going to be very little current flow, hence very little internal magnetic field.

The current in the capacitor is given by I=V/Xc where Xc = 1/ 2PifC
From the above formulas you can see that to maximize current you need to increase V and also increase f (the frequency).

So essentially what this device does is to turn reactive power into real power.

[andy]

So you are talking about Mislavskij Transfomer?
https://www.physicsforums.com/attachment...pg.106612/


  Mislavskij.pdf (Size: 152.35 KB / Downloads: 28)

[dgreen264]

Thanks for that Andy. Yes it's the same.

[dgreen264]

The power output of the device is a function of frequency cubed. This is because:

i) As the frequency increases for a given flux level, you get more volts per turn at the output coil, which translates into proportionally more power. The back emf flux is proportional to output turns x output current. This is what generates the back flux that diminishes the input flux. So output current would still be limited producing the same back emf for a given output current. And the same magnitude of flux at the input (the input flux has not increased). But output volts do not increase back emf flux provided that no extra turns are added in order to achieve the increase in volts.

ii) The capacitor current increases. Thus the capacitor magnetic flux increases and this increases the coil volts proportionally. The available current increases also because there is now more input flux. Thus power increases by frequency squared due to this factor.

[andy]

I am more interested now (after some thinking about it).
What do you suggest to solve problem with capacitor plates to not be so far apart?
In Mislavskij pdf is shown on the last picture something like A4 (I guess) plates. But I am not sure if this can be enough to create some usable capacity (which would create enough magn. flux)

   

[phoneboy]

Hi all

Just wanted to put this idea out there... its regarding the displacement current of a capacitor.
 A capacitor stores energy in the electrostatic field between the plates, but little or no attention is normally given to the fact that a magnetic field is created between the plates when current flows in the wires.

This is called displacement current and produces a (magnetic) field within the capacitor that is equivalent to that which would be created if a wire were to connect the two plates together.

A coil placed between the two plates of the capacitor can receive this oscillating magnetic energy and create an ac voltage on its terminals. A load can be placed on the coil's output and real power can be drawn from it. When current (and hence power) is taken from this coil, the coil generates its own magnetic field which opposes the original field generated by the displacement current (Lenz's law).

The net field will be reduced due to this back flux from the coil (superposition theorem applies). But as there is no 'primary' coil generating the magnetic flux, there is no mechanism for the back flux to reflect electrically back onto the input circuit thus demanding real power input as in a normal transformer.

The problem is that in the case of a parallel plate capacitor there is no magnetic field generated between the plates.  If the electric field is uniform (even if its changing) there is no vorticity (curl) so no magnetic field. The concept is still valid just not as presented.  

In the capacitor-generated flux case, all that happens is that the flux is diminished due to the back flux, and this limits the power out because the net flux reduces and hence the output voltage - which is proportional to net flux - reduces.

But the way I see it, there is now way for this to change the input impedance. The input still looks totally reactive i.e.no real power, and can be resonated away to practically zero, allowing for losses etc.
There is no way for this reduction in magnetic field (due to load current) to electrically get back onto the input circuit.

So this would be free energy then?
If anyone can make suggestions or comments as to why this wouldn't work and where I am missing something in the theory, please do tell! Guests cannot see images in the messages. Please register at the forum by clicking here to see images. Guests cannot see images in the messages. Please register at the forum by clicking here to see images.


thanks

Hello, all.  This was one I was always intrigued by (Mislavskij Transformer).  Unfortunately, in the case of a parallel plate capacitor (same size plates/area) if the E field is uniform (even when its changing) there is no vorticity (curl) so no magnetic field.  That said, the concept is still valid just not the way it was presented there.  I can show a design and simulation if you like.

[andy]

@phoneboy
I am interested in your setup. Can you show it please?
Andy.

[Sandy]

Hi all

Just wanted to put this idea out there... its regarding the displacement current of a capacitor.
 A capacitor stores energy in the electrostatic field between the plates, but little or no attention is normally given to the fact that a magnetic field is created between the plates when current flows in the wires.

This is called displacement current and produces a (magnetic) field within the capacitor that is equivalent to that which would be created if a wire were to connect the two plates together.

A coil placed between the two plates of the capacitor can receive this oscillating magnetic energy and create an ac voltage on its terminals. A load can be placed on the coil's output and real power can be drawn from it. When current (and hence power) is taken from this coil, the coil generates its own magnetic field which opposes the original field generated by the displacement current (Lenz's law).

The net field will be reduced due to this back flux from the coil (superposition theorem applies). But as there is no 'primary' coil generating the magnetic flux, there is no mechanism for the back flux to reflect electrically back onto the input circuit thus demanding real power input as in a normal transformer.

The problem is that in the case of a parallel plate capacitor there is no magnetic field generated between the plates.  If the electric field is uniform (even if its changing) there is no vorticity (curl) so no magnetic field. The concept is still valid just not as presented.  

In the capacitor-generated flux case, all that happens is that the flux is diminished due to the back flux, and this limits the power out because the net flux reduces and hence the output voltage - which is proportional to net flux - reduces.

But the way I see it, there is now way for this to change the input impedance. The input still looks totally reactive i.e.no real power, and can be resonated away to practically zero, allowing for losses etc.
There is no way for this reduction in magnetic field (due to load current) to electrically get back onto the input circuit.

So this would be free energy then?
If anyone can make suggestions or comments as to why this wouldn't work and where I am missing something in the theory, please do tell! Guests cannot see images in the messages. Please register at the forum by clicking here to see images. Guests cannot see images in the messages. Please register at the forum by clicking here to see images.


thanks

Hello, all.  This was one I was always intrigued by (Mislavskij Transformer).  Unfortunately, in the case of a parallel plate capacitor (same size plates/area) if the E field is uniform (even when its changing) there is no vorticity (curl) so no magnetic field.  That said, the concept is still valid just not the way it was presented there.  I can show a design and simulation if you like.

Hi phoneboy. This was my experience. This can be verified experimentally with a roll of twin and earth house wire. Charge two of the conductors as a capacitor with a high voltage supply and connect a spark gap across opposite ends of the conductors. The third conductor will have a high voltage induced across it when the gap fires, but only if the gap is a
connected across opposite ends of the wires. If gap connected at the same end, no voltage is induced in the third wire.
Kind regards, Sandy.