There appear to be many ways to respond to this, and since I am just a simple man (heh, heh) and not a scientist, I reply in my simple way...
Well, of course Ohm's law applies. When you increase the voltage, you get more current. When you move the electrodes further apart, you have more resistance, so you need more voltage to sustain the current. I suspect the problem you are considering is the Watts, or the "work" that is being done, and if I say you don't have to consider the Voltage then you think, well, what happened to the "work" (watts)? The answer to this is the question, "what is the work that is being done?" It has been shown, apparently by experiment, that the amount of silver released is proportional to the amount of current per unit time that is going through the silver anode. One atom of silver is released for each electron passed (as far as I know, I am not a chemist...). If you move the electrodes close together you get your current with very little applied voltage. If you move the electrodes apart, you have to increase the voltage to maintain the current. So, Ohm's law applies, but what is the work being done? Most of the "work" is done to overcome the resistance of the cell which can be variable due to the electrode spacing (and other things). If the electrodes are closer together you have less work to do. If they are far apart, you have to work harder. The point of an equation of this sort is to simplify the process and remove as many variables as possible. You can change the applied voltage to affect the "time" in the equation. You could replace the "current" variable with "volts/resistance" etc, but I don't think the equation would be simpler and easier to deal with (and I think I am starting to confuse myself... ;-)) Also, it might be helpful to think of the two cells in series as a current controlled system (at any instant in time...). Dan -----Original Message----- From: CWFugitt [mailto:c_wa...@earthlink.net] Sent: Wednesday, February 21, 2007 9:12 PM To: silver-list@eskimo.com Subject: CS>Faraday Equation, CS Process 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 -- The Silver List is a moderated forum for discussing Colloidal Silver. Instructions for unsubscribing are posted at: http://silverlist.org To post, address your message to: silver-list@eskimo.com Address Off-Topic messages to: silver-off-topic-l...@eskimo.com The Silver List and Off Topic List archives are currently down... List maintainer: Mike Devour <mdev...@eskimo.com>
