George wrote:
>>> The above calculation is for a single elecrode.  Most use 2 so to
>>> get the actual current density of your setup you will need to
>>> divide the calculated current density by 2 

I wrote:
>> Both electrodes are passing the same current, so I don't think
>> you need this last step... I would think only the anode surface
>> area need be calculated.

George replies:
> The current is flowing through BOTH the anode and cathode and is
> equally distributed (assuming they are the same in all dimensions),
> therefore:
> 
> Total current / total surface area = (total) current density. Double
> the surface area, no change in current = 1/2 the current density. 
> Same result as if you had doubled the length of a single conductor. 

All the current that is flowing in the circuit passes through each 
electrode. Both electrode surfaces will offer close to the same 
resistance if they're both the same size and shape, and together with 
the bulk resistivity of the water/CS determine the *amount* of current 
that flows. In fact, the cathode could even have a different surface 
area or be made from another metal than silver, for all that matter. 

>From the standpoint of making CS, what goes on at the Cathode has 
no effect on the dynamics at the Anode, where silver atoms are leaving 
the surface and undergoing agglomeration and hydration -- except its 
role in determining the overall resistance of the circuit, which also 
becomes irrelevant if you're doing current control.

You may define "current density" differently than I do, but it seems 
that you need only calculate current density for the anode. We *are* 
both talking about a typical LVDC system, right?

Anybody else have an opinion?

Mike D.
[Mike Devour, Citizen, Patriot, Libertarian]
[[email protected]                        ]
[Speaking only for myself...               ]


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