A long time ago, in an apartment far far away, I built a small test rig I titled the Inverse Transformer. I have thought about it ever since.
YouTube - Inverse Transformer.wmv
Before I can get to my point, I have to explain the device a bit.
1)The input into the device is a rotating disk with magnets of alternating polarity.
2)The "primary" of the transformer is a ferrite core with 1 turn.
3)The "core" is an electrical winding of copper, with 300 or so turns.
4) The "secondary" is a ferrite core with 1 turn.
When the secondary is left open, we see Lenz generator drag on the rotor. The magnetic primary induces a current within the copper core, which opposes the motion which created it.
When the secondary is closed, it creates an equal and opposite force within the electrical winding which opposes the first current. The rotor now rotates much more freely, with very little drag.
THINK ABOUT THIS....
The addition of a closed path on the secondary is the equivalent of shorting the secondary on a normal transformer. This act lets the rotor rotate freely, almost eliminating the impedance to the rotor. A dead short, or the largest possible LOAD on the secondary, constitutes a free spinning rotor with minimal drag. We are loading it to the max on the output, and the result is free passage of the rotor, or minimal consumption on the input.
This is very similar to a normal transformer, in that shorting the secondary, reduces the impedance in the primary, allowing more current to pass through the primary. This increase in current in the primary, NORMALLY constitutes increased energy consumption at the source, but that is only because we have chosen an INCORRECT SOURCE.
Imagine instead an inverse inverse transformer.
In place of magnets we could place electrically charged plates, in alternating predominance. Think the exact same setup as in the video above, except that in place of magnets we have electrostatic elements. In place of magnetic primary and secondary, we have electric conductors. In place of an electric core, we have a magnetic core.
Now what happens when we load this machine? questions to consider...
1) if we leave the secondary open, will we see a Lenz drag on the rotor? (the same as if we left the secondary open on the video above, where we saw a drag with the open magnetic secondary)
2)If we dead short the secondary of our inverse inverse transformer, will it induce an equal and opposite magnetic component within the magnetic core? Modern transformer theory says YES definitely we will. Will the result of this be a decrease in impedance of the primary winding? Modern transformer theory says YES definitely we will.
If there is very little impedance in the primary, does equate to a free spinning rotor? Are we allowing the free passage of electrical energy through the primary, which results in little or no drag on our electrostatic rotor?
In the inverse transformer, our secondary is magnetic, it is hard to use this in our electrically oriented circuits, and the demo shows a dead short equates to a free spinning rotor, which is the equivalent of maximum current draw at the secondary.
If we do the inverse of this, replace magnets with electrets, can we cause a situation where a shorting of the secondary creates a free passage of the electret rotor?
In a normal generator, the relationship is, increased current draw equates to increased mechanical load.
In an "inverse generator", or "inverse inverse generator" (or I like to call it the inverse^2 generator ) the relationship is increased draw at the secondary equates to decreased mechanical load on our rotor. Here the closer you get to a dead short, the better! But if there is no load on the secondary, we might just see a heavy drag on the rotor...what an interesting inverse relationship!!!
YouTube - Inverse Transformer.wmv
Before I can get to my point, I have to explain the device a bit.
1)The input into the device is a rotating disk with magnets of alternating polarity.
2)The "primary" of the transformer is a ferrite core with 1 turn.
3)The "core" is an electrical winding of copper, with 300 or so turns.
4) The "secondary" is a ferrite core with 1 turn.
When the secondary is left open, we see Lenz generator drag on the rotor. The magnetic primary induces a current within the copper core, which opposes the motion which created it.
When the secondary is closed, it creates an equal and opposite force within the electrical winding which opposes the first current. The rotor now rotates much more freely, with very little drag.
THINK ABOUT THIS....
The addition of a closed path on the secondary is the equivalent of shorting the secondary on a normal transformer. This act lets the rotor rotate freely, almost eliminating the impedance to the rotor. A dead short, or the largest possible LOAD on the secondary, constitutes a free spinning rotor with minimal drag. We are loading it to the max on the output, and the result is free passage of the rotor, or minimal consumption on the input.
This is very similar to a normal transformer, in that shorting the secondary, reduces the impedance in the primary, allowing more current to pass through the primary. This increase in current in the primary, NORMALLY constitutes increased energy consumption at the source, but that is only because we have chosen an INCORRECT SOURCE.
Imagine instead an inverse inverse transformer.
In place of magnets we could place electrically charged plates, in alternating predominance. Think the exact same setup as in the video above, except that in place of magnets we have electrostatic elements. In place of magnetic primary and secondary, we have electric conductors. In place of an electric core, we have a magnetic core.
Now what happens when we load this machine? questions to consider...
1) if we leave the secondary open, will we see a Lenz drag on the rotor? (the same as if we left the secondary open on the video above, where we saw a drag with the open magnetic secondary)
2)If we dead short the secondary of our inverse inverse transformer, will it induce an equal and opposite magnetic component within the magnetic core? Modern transformer theory says YES definitely we will. Will the result of this be a decrease in impedance of the primary winding? Modern transformer theory says YES definitely we will.
If there is very little impedance in the primary, does equate to a free spinning rotor? Are we allowing the free passage of electrical energy through the primary, which results in little or no drag on our electrostatic rotor?
In the inverse transformer, our secondary is magnetic, it is hard to use this in our electrically oriented circuits, and the demo shows a dead short equates to a free spinning rotor, which is the equivalent of maximum current draw at the secondary.
If we do the inverse of this, replace magnets with electrets, can we cause a situation where a shorting of the secondary creates a free passage of the electret rotor?
In a normal generator, the relationship is, increased current draw equates to increased mechanical load.
In an "inverse generator", or "inverse inverse generator" (or I like to call it the inverse^2 generator ) the relationship is increased draw at the secondary equates to decreased mechanical load on our rotor. Here the closer you get to a dead short, the better! But if there is no load on the secondary, we might just see a heavy drag on the rotor...what an interesting inverse relationship!!!
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