skycollection - TRUTH ABOUT LENZ / ZERO LENZ

[skycollection 1]

Thanks for the info Skycollection1.

Is that how the capacitor stops the Lenz, by preventing the voltage from reversing?

I used many kinds of capacitors and only one Works, is a small capacitor from a high voltaje lamp (http://www.afinidadelectrica.com/articulo.php?IdArticulo=35) i really don´t know the reason
why this small capacitor reduce or delate the Lenz effect...! If you can investigate this effect you will find that the rotor do not stop with this capacitor. I hope it serves for you...!

[tomd]

A Capacitor provides low impedance for faster changes in voltage. This leads to large current, whenever voltage tends to change instantaneously.

When the voltage increases fast, capacitors draw large current from the voltage source. This current creates drop in internal resistance of the voltage source.

When the voltage reduces instantaneously, capacitors provides large current to the voltage source, which partially compensate the original reduction in the voltage. This current crates drop in internal resistance of the voltage source.

So can we say that capacitor opposes instantaneous changes in voltage? I think, this statement /question is little confusing. Better way is to say that "capacitor absorbs instantaneous changes in voltage"
https://www.quora.com/Why-does-a-capacitor-oppose-instantaneous-changes-in-voltage

"If you have 10v across a 10uF capacitor, and the voltage has been steady for long enough, then no current flows. If you now try to change the voltage to 20v, ramping it up at 10^6 volts per second, so it takes 10uS to change from 10v to 20v, the current will go to 10A for that 10uS, and back to zero when the voltage is steady again. If you try to change the voltage at 10^7 volts/second, the current pulse will be 100A.

That sort of current amounts to a fairly violent 'objection' to the voltage being changed. If the power supply cannot supply it, then the voltage will not change as fast as expected."
https://electronics.stackexchange.com/questions/361661/how-does-a-capacitor-resist-changes-in-voltage


Could it be that the capacitor is absorbing the increase in voltage across the primary due to cemf? In this case the source would be the primary coil. The actual circuit source would be unaffected and would only have to supply enough current to maintain the magnetizing current.

[tomd]

A Capacitor provides low impedance for faster changes in voltage. This leads to large current, whenever voltage tends to change instantaneously.

When the voltage increases fast, capacitors draw large current from the voltage source. This current creates drop in internal resistance of the voltage source.

When the voltage reduces instantaneously, capacitors provides large current to the voltage source, which partially compensate the original reduction in the voltage. This current crates drop in internal resistance of the voltage source.

So can we say that capacitor opposes instantaneous changes in voltage? I think, this statement /question is little confusing. Better way is to say that "capacitor absorbs instantaneous changes in voltage"
https://www.quora.com/Why-does-a-capacitor-oppose-instantaneous-changes-in-voltage

"If you have 10v across a 10uF capacitor, and the voltage has been steady for long enough, then no current flows. If you now try to change the voltage to 20v, ramping it up at 10^6 volts per second, so it takes 10uS to change from 10v to 20v, the current will go to 10A for that 10uS, and back to zero when the voltage is steady again. If you try to change the voltage at 10^7 volts/second, the current pulse will be 100A.

That sort of current amounts to a fairly violent 'objection' to the voltage being changed. If the power supply cannot supply it, then the voltage will not change as fast as expected."
https://electronics.stackexchange.com/questions/361661/how-does-a-capacitor-resist-changes-in-voltage


Could it be that the capacitor is absorbing the increase in voltage (by way of current draw) across the primary due to cemf? In this case the source would be the primary coil. The actual circuit source would be unaffected and would only have to supply enough current to maintain the magnetizing current.

The idea is to filter out any voltages higher than the voltage which maintains the magnetizing current.

[tomd]

Further on this. The voltage required to maintain the magnetizing current is the source voltage minus the self induced voltage from the primary. So the idea is to stop the source from "seeing" the cemf voltage from the secondary. As this voltage will bring the voltage accross the primary back up to the source voltage and will result in a larger current draw.

http://www.physics.usyd.edu.au/~khachan/PTF/Transformer%20explanation.pdf

[seychelles]

YO ALL THIS IS A CIRCUIT  THAT WILL REDUCE LENZ EFFECTS.