I'm trying to imagine how the voltage level would look if one were to compare to water pressure. One thing that I think all of us to date have confirmed that no matter how much we try to restrict amps, we simply do not get gas with high voltage without some current involved.
We also know there is a minimum amount of voltage required to perform normal electrolysis. 2.2V or somewhere there abouts.
Lets look at this from another perspective; I'm going to use an example where balloons are the electrodes and water the electricity.
Lets say that the balloons are full of tiny holes. Those holes will represent electron leakage in the water.
If we fill the balloon only to the point where the balloon resists any further input without any additional pressure, there is minimal water (or in electronic terms, electron) leakage. The more pressure we add the greater the leakage since the tiny holes in the balloon expand and allow more water to flow out.
Now, lets say we put an instantaneous high pressure of water into the balloon. What will happen? The balloon may instantly explode? It may instantly swell up briefly before releasing its load. But ultimately, the load is still expelled with a higher volume than it would with less pressure. UNLESS that pressure imposed is so short and pulls back so fast that it barely makes a ripple in the load and intact perhaps even creates a reverse load to balance the pressure within the balloon to keep the leakage constant.
Just really thinking out loud here...
Not done with this thought, but am posting what I have come up with so far as I work out my thinking on this.
TS
Thanks for sharing your idea's, TS.
Maybe able to help your thoughts:
What is happening at electrodes when you put voltage on them?
One will electrode will have more electrons then the other one. Thats the potential difference. Thats the 2.2V you spoke of.
As soon as you you put more volts on the electrodes, you create a bigger difference between them.
Then U=I*R comes in.
When you put 2.2V at your electrodes or 10KV doesnt matter. The reaction will be the same. Voltage = Amps times resistance
The higher volts, the same resistance, the more amps. Thats a law.
In case of water, higher volts create even less resistance when the molecules align, and even more amps will go....
The electrons give their charge to the ions in the waterbath.
So, if HV is the goal, then you must switch off the pulse before the water molecule are getting aligned. Thats Meyers system.
Ultrashort pulses.
But the issue is if that creates gas or not.
Sofar no results by anybody.
I know from first hand observers that the alternator setup was pumping high amps, because the observer had put his hand on the cables going to the cell of that setup.
I think that the high voltage pulses where used to get more amps in a smaller unit and to be able to use any kind of water.
Steve
