Evening Dan,
Thanks for twisting my arm into studying the Faraday Equation. <grin>
>>At 01:13 PM 2/19/2007, you wrote:
but look over the formula below. The voltage is not part of the
equation, although it will affect how fast the process progresses. Both
cells have the same current, being in series, so the amount of silver
released is the same.
I have studied the equation in detail, and read other explanations of it
as well.
It is easy enough to understand. I tried to understand each number and see
where they came from.
It seems that when an electron leaves the positive electrode, specific and
positive things happen, no matter what.
The equation originated about 1900. If no one has disproved it by now, it
must be right. Everyone accepts it as such.
I still find it hard to believe, understand, or accept that any electrical
process or action does not adhere to Ohms Law.
In trying to re analyze the CS process, and the two units in series,
I may have figured out how it does in fact adhere to ohms law and we have
no violation of old accepted theories.
Even if, we had an LED and Resister installed in one of the units, and not
the other, the current is the same at all electrodes of course.
The voltage drop across the units would be different due to this and the
slight difference in conductivity of the solution.
So, where is the voltage drop? Not at the Electrodes. This junction of
the electrodes and the solution must be so close to zero there is no
voltage drop.
According to the Faraday Equation, the spacing of the electrodes
does not even enter in. Only the electrons passing from the positive
electrode to the solution does any work.
Work without Wattage is a near miracle.
So, ....... If the voltage drop is different between unit one and unit two,
it must come from the varying conductivity ( or resistance ) of the
solution itself. That is, if we had two identical units without resistors
and LED's.
We understand that the lower the conductivity, the less voltage drop and
the higher the conductivity the greater the voltage drop.
I think we both agreed that making the units precision and alike would be
important. Not so, ...... when the electron leaving the positive electrode
is all that matters.
That is the only place any work is done.
I could build two units, break out all the volt and current meters I can
find, and see what happens. Unless the meters are very high quality, they
will disturb the circuit to the point the test is worthless.
We both realize, I trust, this is mostly academic as we agreed the parallel
units solve most problems and will operate more uniformly.
Understand please, I am not trying to be ornery or hard headed, I simply
want to better understand the process.
If the voltage drop is different on the units, for whatever reason, It must
be someplace. Yet this does not matter, only the electrons leaving the
positive electrode matters.
If you have different ideas and better explanation, please tell us.
I hope a few others are interested in the workings of the process.
If you think I should forget it, or Mike wants us to stop this thread, let
me know.
Wayne
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