I'm not an expert on plasma, and the relativistic effects inside a black hole or a solar centre, but the continuous partial discharge of hydrocarbons with magnetic and dielectric materials, has led me to discover the root cause of MVA distribution transformers from contaminants in a transformer factory process cause these effects under high field strength in operation. Such that within the 1 year warranty period the DGA dissolved gas analysis of transformer oil, when it exceeds 4% or the LEL lower explosive level is exceeded by partial discharge, (PD) the transformer is taken out of service. ( It was a multi-million $ factory epidemic liability from very tiny defects.
PD is a result of plasma physics like an "avalanche electric switch" with almost zero discharge time of charges shorting out detonating molecules and breaking down HxCy oil in H2 and longer explosive gas molecules with higher activation energy. My DSO was slow so the current rise time was in picoseconds.
You might want to ask some questions from an expert like Briane Greene to validate your assumptions.
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The few Bedini 's experiments I have observed , were disappointing on the non-scalable demonstration of tiny scale electrostatic engines.
If you were applying 8V @ 500 mA, what was the initial condition Vc, of the Caps and for how long. Obviously the more time, the more energy either pulsed or continuous, but energy is lost in the ESR of both components when connected.The 9v battery I tickled the caps with is 8v at 500ma. That’s 4 watts. It made a pretty nice arc.
Now if we are using at 7000v pulse at 10ma, that’s 70 watts.
If we progressively increase the frequency of pulsing into the caps, are we going to get an increase of energy stored in the caps?
It’s a fairy simple question.
The caps are kept discharged.If you were applying 8V @ 500 mA, what was the initial condition Vc, of the Caps and for how long. Obviously the more time, the more energy either pulsed or continuous, but energy is lost in the ESR of both components when connected.
Where can I look up the symbols in your formula, as some of them are unfamiliar to me.Current thru a capacitor or a megacap like a tiny 9v battery is Ic= Cdv/dt+ΔV*ESR{C&Bat)
That makes perfect sense.Hi,
A pulse will give some charge into a capacitor.
If we ignore the increasing capacitor voltage, then every pulse will push the same amount of energy into the capacitor.
But to keep the voltage constant you need to draw the same amount of energy from the capacitor as you put in ... in the same time.
If you don't draw energy from the capacitor, it's voltage will increase. It starts with an (almost) contant rate, but the rate will decrease with increasing capacitor voltage.
Usually the higher the capacitor voltage, the less energy you can push into the capacitor at the same time....because the current is dropping more fast.
So if you increase the frequency ... there will be a limit.
In all my years I've experienced the laws of physics are true. But new inventions need new ideas.
If you think it's worth the effort ...
Klaus
That’s a good question.
ΔV=Vi−Vc
is the change in instantaneous voltage due to internal resistance with a step current = I*ESR
For measuring voltage you basically don't need a load. But your measurement equippment most uses some "non_infinite" resustance inner circuitry, like a voltage divider. So, although not needed, there will be some tiny input current to your voltmeter.In measuring the voltage and amperage output of the caps, would I need a load on the caps to get meaningful readings on my test equipment?
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