I have 2 Energizer brand AA batteries, rechargeable NiMH type. 2500
milliamp-hours. They are labeled as 1.2 volt as well.

Now, if put these two in series, I believe the voltage is additive.
Also I now have 5000 milliamp-hours (or 5 amp-hours) of total usable
energy, theoretically.

My question is, how do I determine how many amps these two batteries
can load at a given time? Is there a limit?

For instance if I have 5 amp-hours available, can I run a 5-amp motor
for an hour? Or a 10-amp motor for half an hour? Or is there some
lower limit of the current draw of the motor, and determined
by...what?

If someone knows the answer, I'm assuming it's a quite simple, one-
sentence answer. A link would be appreciated as well.

see at least two questions here:

1) Do capacity of the battery, and voltage add
   when you connect them in series?

    Yes. You will have  now 5 amp-hours, and so you can run
 
    .1 A  bulb for 50 h  or  .05 A bulb got 100h 

2) Can you run  10-amp motor on a dry cell battery?

   No.  Batteries have an inner resitance, which limits the current
        they can deliver. Those resistances add also, in series:
   One battery has upper limit  V/Ri on current it will supply at short,and 
  two baterries in series will have same max current limit 2*V / 2 * Ri.

        If you connect in parallel, max current will double.
       

capacity
http://en.wikipedia.org/wiki/Battery_(electricity)#Battery_concepts

inner resistance
http://www.dokidoki.ne.jp/home2/tohrus/battery/battest0E.html

In general, for a load (e.g. a bulb) with resistance R, the current
supplied by the battery is:

           I= V / (R + Ri)    (Ohms law for equivalent circuit)


Equivalent circuit is shown here ( our R.i is their r )
under heading: The emf of a battery: where it says:

The equivalent circuit is shown on the left, where r is the internal
resistance of the battery with emf E, connected to a circuit with
resistance R. The voltmeter measures the potential difference across R
and not the emf of the battery which is equivalent to the potential
difference across (R + r).

http://www.physchem.co.za/Current%20Electricity/Heating.htm



Hedgie - rating appreciated

---

Request for Answer Clarification by jjsonp-ga on 27 Mar 2006 11:45 PST

Thanks hedgie. That clears things up somewhat.

However, you did not provide a specific answer to my question: "...how
do I determine how many amps these two batteries can load at a given
time? Is there a limit?"

Do I need to use a multimeter to measure the internal resistance, and
then calculate the maximum current they can provide using the formula
you mention (i=v/ri)?

I still want to know exactly how to determine the maximum current 2 AA
batteries can provide.

---

Clarification of Answer by hedgie-ga on 28 Mar 2006 12:52 PST

jjsonp-ga

I do not recommend 
"using the formula you mention (i=v/ri)".

as it requires shorting a battery and that is not a good idea 
(for several resons) which I will not elaborate (since I know
the answer) and you are right, that:
  
" If someone knows the answer, I'm assuming it's a quite simple, one-
sentence answer. .."

I suggest that you use the formula for equivalent circuit with a reasonable load

I= V /(R + Ri) 

for two ( or several values of R ) (one, two, three, flashlight bulbs,
in series and parallel ..)

 For each mesurement your get pait of values (R and I) (same V for all)

You can then plot 

 V/I  vs  R + Ri

and get a line,  roughly

 Ri  is  a constant  (you get when you extrapolate  V/I-->0

and do not forget that R.i  will change somewhat over the life of a battery
and with temperature.

Rating appreciated

Hedgie

The internal resiatance o the battery consumes power as seen by
heating of the battery. so part of 5Ah is consumed inside especially
aat higher loads. But baaery capacity is again deFined to a cell end
voltage. There will be still some power after %Ah has been used. But
this may not be adequate to run the equipment.

"I have 2 Energizer brand AA batteries, rechargeable NiMH type. 2500
milliamp-hours. They are labeled as 1.2 volt as well."

If you place these two AA batteries in SERIES, the resulting voltage
adds to 2.4 volts but the current capacity = 2.5 amp-hours (NOT 5.0
amp-hours).

If you place these two AA batteries in PARALLEL, the resulting voltage
= 1.2 volts and the current capacity = 5.0 amp-hours.

This is a logical consequence of the fact that each AA battery has a
fixed ENERGY-storage capacity = 6.0 volt-amp-hours = 6.0 watt-hours.

larry is right.  Connecting batteries in series will not give you more
amp-hours.    Also, the maximum current that can be supplied is the
same with one battery or 2 in series.  Only the voltage is doubled. 
There is no minimum current.

also every battery has rating based on discharge rate.
so you may get 5Ah at say 5hour rate of 1A but at 1 hour rate you dont get 5A.

isnraju is very correct. I did lots of research on different battery
types (NiCad, NiMH, etc.) during an ASME design project. Discharge
rates can vary by battery type, temperature and how much stored energy
is left. Generally, for everyday batteries (A,AA,etc.) NiMH are the
best in every respect compared to NiCad or Alkaline--though all three
will vary between brand so be wary. Lithium Ion generally have crazy
energy density but not as great at power discharge.

I'm not sure if you still need help, but Energizer keeps data sheets
on their batteries, which might be of some use to you.

In reference to battery nh15-2500 (An Energizer 1.2v 2500mAh
rechargeable NiMH battery):
"The internal resistance of the cell varies with state of charge, as follows:
<u>Cell Charged:</u>
30 milliohms

<u>Cell 1/2 Discharged:</u>
40 milliohms
(Tolerance ±20% applies to the above values)"

(http://data.energizer.com/PDFs/nh15-2500.pdf , 1)

For other Energizer batteries, you can search on
http://data.energizer.com/ with the appropriate category.