As a discharging battery (providing
power) we have...
emf*I = V*I + losses
emf is greater than V
If we treat emf, V, and I as
unsigned numbers, then when we reverse
the current we have
-emf*I = -V*I + losses
We don't change the sign in front of
losses. The cell does not go endothermic.
It is exothermic either way.
I rearranged this to
V*I = emf*I + losses
because we are now doing work on the
battery rather than the battery doing work
for us. I like to put the agent doing
the work on the left, and put where
that work is going on the right. It's
just a personal preference.
How does the voltage know what to do?
If you push in on a piston very slowly to
compress the gas in a cylinder; then you do
it again but push much faster; how does the
gas know to push back harder in the
non-reversible case?
When a spinning ice-skater pulls in his arms,
how do the various parts of his body know they
must rotate faster?
These things are the way the world
operates. We don't explain them, we
describe them.
Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton College
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu