Olin's point is that the current flow through the internal "ideal
battery" is the same as that coming out of the physical battery,
regardless of whatever series resistance you place inside the battery.

Batteries are very complex devices to model. What's going on inside
are chemical reactions which depend on the ability of ions in solution
to move around to parts of the electrodes which have not yet reacted
with the electrolyte or other electrode. In addition, if the reaction
proceeds faster than the diffusion rates allow, then it will grind to
a halt quickly until you allow it time to recover.

So, in other words, no, the reduction in capacity is not just because
of internal resistance. The term "capacity" usually refers to the
current integrated over time, NOT the energy.

The effect of discharge rate on capacity is called the Peukert Effect
in Lead Acid batteries (I don't know to what extent other batteries
have this issue or what the effect is called in those cases). It is
due to what I was talking about above - the formation of a layer of
insulating material around grains of electrode material during fast
discharges.

With a Lead Acid battery rated for, say, 10AH at a 1 amp discharge
rate, if you try to discharge it at 10 amps, you will probably get
something like 63% of the 1amp rate capacity (so the battery will run
for only 0.63 hours instead of 1 hour. However, in order to get there,
you will also have to allow the battery voltage to drop a lot more
than usual because of the high current and high internal resistance,
which means that you get even LESS than 63% of the energy out.

HOWEVER, if you then allow this "dead" battery to sit open-circuit, it
will recover some of its capacity as electrolyte slowly seeps into the
electrode grains and then you can draw more AH out of it.

If you Google for Peukert Effect you will find some really interesting stuf=
f.

Sean


On Sat, Jul 5, 2008 at 12:25 PM, Tom=E1s =D3 h=C9ilidhe <toe@lavabit.com> w=
rote:
>
>
> Olin Lathrop wrote:
>> Tom=E1s =D3 h=C9ilidhe wrote:
>>
>>> Is the reason that it depends on the current you're drawing from it to
>>> do solely with the internal resistance of the battery?
>>>
>>
>> Think about it.  If a perfect battery with 0 internal resistance was rat=
ed
>> for 1AH and you put 10ohms in series with it, how many AH can the battery
>> and resistor deliver to the load?
>>
>
>
> Depends on the resistance/impedance of the load. But if I have my device
> and I know it has an "average resistance" of 100 ohms, then I can do the
> maths to work it out.
>
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