On Wed, 10 Oct 2001 23:04:46 +0000, M. G. Devour 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 which would give a current density of =>> 500 uA per mm^2. => =>Err, George? 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.
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. Regards, George Martin => =>Other than that, it makes sense. <smile> => =>Be well, => =>Mike D. =>[Mike Devour, Citizen, Patriot, Libertarian] =>[mdev...@eskimo.com ] =>[Speaking only for myself... ] => => =>-- =>The silver-list is a moderated forum for discussion of colloidal silver. => =>To join or quit silver-list or silver-digest send an e-mail message to: =>silver-list-requ...@eskimo.com -or- silver-digest-requ...@eskimo.com =>with the word subscribe or unsubscribe in the SUBJECT line. => =>To post, address your message to: silver-list@eskimo.com =>Silver-list archive: http://escribe.com/health/thesilverlist/index.html =>List maintainer: Mike Devour <mdev...@eskimo.com> => =>
