Re: emf and batteries

[Bob LaMontagne <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

At this point a hand is going to go up in class asking why
we are 'adding' the losses to the EMF - after all, they're
losses! How does the terminal voltage know enough to
increase to provide the energy for the losses?

Bob at PC

*********** REPLY SEPARATOR  ***********

On 3/23/2004 at 5:26 PM Edmiston, Mike wrote:

> The work you are doing (actually rate of work)
> when you run the current backwards is
> V*I.  If you have a very small current such that
> you are doing a thermodynamically reversible
> process, then V*I equals emf*I.
>
> However, for larger currents you have not
> satisfied the conditions of a thermodynamic
> reversible process, so your work goes other
> places in addition to running the chemical
> reaction backwards.  In this case your
> work (power) is
>
> V*I equals emf*I + losses (heat)
>
> This means that V must be greater than emf.