Confirming the Delayed Lenz Effect

[SchubertReijiMaigo]

Hello Gotoluc I have DSO 2090, just a tips to measure the power, use the maths function (CannelA*ChannelB) to display power, if you see a curve that have equal pulse above and bottom the zero line you have reactive power, maybe it will be much easier to read.

Channel -> Maths -> Activate -> SourceA(Channel1) -> SourceB(Channel2) ->
Operation -> A*B...

[CRANKYpants]

At everyone,

There could be a mistake somewhere so lets look at this over carefully. Maybe the power meter I'm using is being fooled in this kind of configuration?

A youtube user posted this: "Sorry Luc, but I have to correct you on the voltage leading the current. In a capacitive circuit the current leads the voltage. This is what your scope is showing. Scope traces go from left to right with time."

I thought it was the other way around?... but I may not have it correct. Can you look over the scope shot below and comment.

I think a pure sine wave output inverter would help confirm but I don't have one. Only modified sine wave I have. I'll give that a try and let you know.

Thanks for sharing

Luc

LUC,

THE DIAGRAM BELOW SHOWS A SERIES RLC CIRCUIT. RESONANCE OCCURS WHEN XL AND XC ARE EQUAL AND IS MINIMUM IMPEDANCE AND ZERO PHASE SHIFT.

WHEN XL AND XC ARE MADE UNEQUAL AS IN YOUR CASE... TOTAL IMPEDANCE INCREASES, CURRENT DROPS, AND MOVES OUT OF PHASE WITH THE VOLTAGE.

THE PROBLEM IS THIS DIAGRAM IS FOR AN INDUCTOR BUT YOU HAVE A TRANSFORMER THERE WHICH IS ACTUALLY TWO INDUCTORS IN PARALLEL AND YOUR INDUCTANCES ARE EQUAL AND THE MAGNETIC COUPLING BETWEEN THEM IS PERFECT (M = 1).

SO... THE EQUIVALENT INDUCTANCE WILL BE ZERO AS THE TWO EQUAL INDUCTORS CANCEL EACH OTHER OUT. BECAUSE THE COUNTER FLUX FROM L2 WILL CREATE A NET 0 FLUX IN L1. IN THE DIAGRAM BELOW B WILL = 0 BECAUSE BL1 AND BL2 WILL BE EQUAL AND OPPOSITE.

SO WHAT YOUR METER AND SCOPE ARE READING IS THE POWER DISSIPATED ACROSS A 1 OHM RESISTOR AND A SMALL CAP.

I SUGGEST THAT YOU 1) REMOVE THE SECONDARY LOAD AND SEE WHAT HAPPENS AND 2) REMOVE THE TRANSFORMER AS WELL...?

CHEERS
T

Mutually Coupled Inductors in Parallel

When inductors are connected together in parallel so that the magnetic field of one links with the other, the effect of mutual inductance either increases or decreases the total inductance depending upon the amount of magnetic coupling that exists between the coils. The effect of this mutual inductance depends upon the distance apart of the coils and their orientation to each other. Mutually connected inductors in parallel can be classed as either "aiding" or "opposing" the total inductance with parallel aiding connected coils increasing the total equivalent inductance and parallel opposing coils decreasing the total equivalent inductance compared to coils that have zero mutual inductance. http://www.electronics-tutorials.ws/inductor/parallel-inductors.html

[Jack Noskills]

Is it possible to select C for parallel resonance so that it blocks 50/60 Hz signal ? If so, what if you then put transformer that has one way induction (e.g. gabriel device), you could then take power from it but there would be no current flowing in the source because of parallel resonance.

Normal transformer would not work because of mutual inductance that would drive the parallel circuit out of resonance. Or there would need to be tunable capacitor and adjust it differently for every different kind of load.

[Overunityguide]

[...]

WHEN XL AND XC ARE MADE UNEQUAL AS IN YOUR CASE... TOTAL IMPEDANCE INCREASES, CURRENT DROPS, AND MOVES OUT OF PHASE WITH THE VOLTAGE.

THE PROBLEM IS THIS DIAGRAM IS FOR AN INDUCTOR BUT YOU HAVE A TRANSFORMER THERE WHICH IS ACTUALLY TWO INDUCTORS IN PARALLEL AND YOUR INDUCTANCES ARE EQUAL AND THE MAGNETIC COUPLING BETWEEN THEM IS PERFECT (M = 1).

SO... THE EQUIVALENT INDUCTANCE WILL BE ZERO AS THE TWO EQUAL INDUCTORS CANCEL EACH OTHER OUT. BECAUSE THE COUNTER FLUX FROM L2 WILL CREATE A NET 0 FLUX IN L1. IN THE DIAGRAM BELOW B WILL = 0. BECAUSE BL1 AND BL2 WILL BE EQUAL AND OPPOSITE.

[...]

Dear Thane,

This is exactly what I was thinking, it runs almost totally capacitive. (sure no resonance)
So power factor is indeed almost: 0 (or almost 90 degrees capacitive). And indeed a heavily loaded transformer will respond more like a much smaller (lower ohmic) normal resistive load. So in Luc's video it could be that he is indeed showing his first steps to overunity.

With Kind Regards, Overunityguide

[CRANKYpants]

LUC,

I HAVE A 12V DC - 120 VAC INVERTER IF YOU WANT TO BORROW IT.
LET ME KNOW AS I'LL BE IN OTTAWA TODAY.

CHEERS
T