Matt or anyone, would you say the polarization process is used in Brut force electrolysis?
hows this for real testing :
https://www.youtube.com/watch?v=5kPcZ7fZTS8
~Russ
bloody hell and bloody hell again I have to sleep over that brain is frying
bloody hell can not be that simple
As you know the water molecule has two hydrogen and one oxygen so there would be two cancels on the neg plate compared to one cancel on the positive plates. Question again: if the battery is left on, would they be more charge accumulate on the neg side than on the positive side?
Everyone sleep on that tonight.
So by the logic in that video (Good find Ris), if we have a dry cell (K = 1) that measures 10pF, without even taking another measurement, we know immediately when the cell is filled with water (K = 80), the capacitance has just increased to 800pF.
That much is pretty easy and it completely defines the working range of our cell.
Yes, I agree, more charge would accumulate on the negative plate. The tougher question is what happens with the voltage in this unbalanced system? And where do you place a reference point? That's the "bloody hell and bloody hell again" spot for me. It's like an asymmetric capacitor where the two plates are of different dimensions. I think T. Townsend Brown was probably the only one that really comprehended such a scenario.
Guess I'll have to sleep on it, because I sure don't have an answer for it at the moment.
My intuitive notion is that there's equal amount of charge on both sides (anode and cathode) on the cell tubes, which leaves one of the hydrogen atoms with it's naturally net positive charge, until........the next step........?
Matt the tube cell is also asymmetric.
The charge MUST remain = on both plates
Like a thunderbolt.
wait.. whay arnt you in bed... :)
Here is where I'm am at.............
We have 12 numbers we need to know...
L1-L2-S1-P1-C-1Cell-1(C-1)
Most of us are not smart enough to do the math to get this correct
Well....How about 1 real number? With the twin core you made
Cell=?
L-1=?
L-2?
S-1?
P-1?
Dan there is more to it than just a capacitance value. Voltage plays a big role in this. If you look at this photo of Stan's, the first level is the polarization process. You want this process to take place with around 2 volts on the primary. If you tune to resonance at 2 volts and you see no gas being produced at all then you know there is something wrong. By leaving it at resonance at 2 volts, raise the voltage from 2 volts 4,6,8,10,12 somewhere in that voltage range you should see some gas being made. What ever that voltage is that you see the gas being made let's say 6 volts. That should tell you, the turn ratio is off on the secondary, because you want it to take place at 2 volts to the primary not 6 volts. This is where everything gets tricky to adjust. In order to keep the impedance match, what you have to do is take turns off the chokes and add them to the secondary to increase the voltage. By doing this you change the inductance which will change the resonate frequency. So it's a balancing act that your shooting for. Again if the polarization process takes place at 6 volts and it's suppose to start at 2 then you've lost 4 volts in the process that you can no longer do anything with.
So by the logic in that video (Good find Ris), if we have a dry cell (K = 1) that measures 10pF, without even taking another measurement, we know immediately when the cell is filled with water (K = 80), the capacitance has just increased to 800pF.
That much is pretty easy and it completely defines the working range of our cell.
Yes, I agree, more charge would accumulate on the negative plate. The tougher question is what happens with the voltage in this unbalanced system? And where do you place a reference point? That's the "bloody hell and bloody hell again" spot for me. It's like an asymmetric capacitor where the two plates are of different dimensions.