Food for thought.
The nature of a reactive circuit such as a tank circuit is that you can produce thousands of volts and thousands of amperes with just a tiny excitation, depending on the "Q" of the circuit. Theoretically, in an ideal L-C circuit the voltage and current would go to infinity, but in the real world this is not possible, as the dielectrics would break over long before that.
It does take many input impulses for the energy to accumulate in the reactive circuit, but once it does build to it's maximum value, it will stay at that high value, limited only by the resistive and radiative losses of the tank circuit. These tend to bleed off real power which must be resupplied by the source of excitation.
Thus far all "known" attempts to get at that reactive power and have it do real work by becoming real power, have AFAIK failed, but it is an area ripe for investigation.
I have postulated that the key to OU may be in using high Q circuits to advantage by not directly bleeding off the circulating energy, thus lowering the Q, but letting it build such that it can act as a catalyst in a reaction which releases energy from "some other potential source". A catalyst is typically "not used up" in the reaction, but acts as a trigger.
If you could capture the instant of connection of your LED lamp to the stepper output with your scope you would probably catch the first impulses from the stepper motor pumping up the tank circuit. It would look like an oscillation that builds in amplitude with each cycle.
So for each cycle a max of 76 mA is pushed into the tank until it accumulates it's maximum, which is limited by the stepper winding resistance and the power dissipated in the LED.
If you try tuning the output of the stepper motor with a capacitor of the right value, instead of the LED lamp, you should get even higher values of voltage and current circulating in the tank.
You asked:
But first a question-->and this may sound silly,but 
When measuring AC current and voltage,how do we know which way-or,from what device(source) the bulk of the power is coming from?.
To make my question more clear,lets say we have two black boxes,and we have a common or neutral wire from one box to another,and a live (or hot) wire with a 1 ohm CVR in series on the hot wire to measure the current flowing from one to the other black box. Lets assume that it's an AC current flow. So we place one channel of our scope across the neutral/common and live wires to measure voltage. We then place the second channel of the scope across the CVR to measure current. Without knowing which black box is the power source,and which is the sink,how do we know which way the power is flowing ?.
One way to do this is by measuring the voltage across the CVR with respect to ground, this should tell you which direction it is flowing. If it is flowing out of the source into the DUT, it will be higher on the left and lower on the right side of the CVR, and vice versa.
Also if you compare the phase of the current across the CVR WRT to the voltage input you can tell which way it is flowing if it is a pure resistve load. If it is reactive, it gets trickier.
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
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Topic: Eddy Current Heating (Read 15827 times)
