Sean Breheny wrote:
> The voltage distribution in the water DOES have to do with its
> conductivity because it isn't being fed by a true voltage source.

Only if you are drawing so much current as to drop the feed voltage.  I
think for the purpose of the original question, the voltage at the
connection point between the two cords is fixed enough.  Even if the pool
was a significant enough load as to drop the feed voltage (and the breaker
somehow didn't trip), the distributed voltages would all still have the sam=
e
ratio to each other.

Think of it this way.  The water is a whole bunch of little resistors all
strung together in a network.  Put a fixed voltage between any two points o=
f
this network, and you can calculate the voltage of all other nodes.  Now
notice that everything is based on ratios of the resistors.  For example,
you could double the value of all resistors and the voltages would be the
same.

Think of a voltage divider, which is a very simple case of a resistor
network.  If the top resistance is R and the bottom 2R, then the output wil=
l
be 2/3 the input.  It doesn't matter if the absolute resistances are 100 an=
d
200 ohms, or 1500 and 3000 ohms, or any other combination as long as the to=
p
is half the bottom.

More complicated resistor networks work the same way.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.
--=20
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist
.