The r is the "effective" internal resistance of the battery, so the (I^2)r losses end up as heat. The IE part goes into chemical energy to charge the battery, but you don't get all this energy back out as electrical energy because of the (i^2)r' losses within the battery as it discharges. Note that r may change a bit as the battery charges and discharges....
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Savinainen Antti
Sent: Monday, January 21, 2008 11:32 AM
To: phys-l@carnot.physics.buffalo.edu
Subject: [Phys-l] Energy balance in charging battery
Hi all,
my student asked today a question which I'd like to pass on to you.
When a battery is being charged the power at which energy is transported to the battery is given by the "standard" model:
P = (E + Ir) * I
where E = emf, r = internal resistance of the battery and I = current.
The question is: what part of this power goes into heating the battery and what part goes into chemical energy? Is the heating part simply the same as when the battery is used as a power supply (rI*I)?
Regards,
Antti Savinainen
Kuopio Lyseo High School
Finland
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