Thank you everyone for the replies.
Here are a few more details if anyone is inclined to comment further.
The cells are AAA alkaline, not NiMH.
Starting voltage is around 6.25. Cut-off is 5V, being the lower Vss
limit of the amp which is a CA 3130.
Phil's product of 6.8 ohms seems very low to me. Can we confirm this?
Phil is giving you the right info. Explanation below.
I get 909 ohms for DC. 2V/(1.100AH/50Hrs). Is AC really that different
in practice?
This is the wrong result, but you didn't even do this division right. Looks to me like
1.1AH/50Hrs is .022 amps; then 2v/.022 amps gives 90.9 ohms. But besides that error, the
2VAC is a peak-to-peak value. The RMS value of the sine wave is .707 volts, and that's
the number you need to use. And there's more to it than that, anyway.
You have to do this on an energy basis. I am going to do the calculations with some
false assumptions. You say the cut-off voltage is 5V, but this is for 4 cells in series,
so the cut-off per cell is 1.25 volts. If you only take the cells down to 1.25 volts, you
won't get the full 1.1 Ah out of them.
See:
http://professional.duracell.com/product_data/datasheets/PC2400.pdf
But, let's assume that you do. Then let's use as the average cell voltage during
discharge as (6.25 + 5.00)/2 = 1.40625, or about 1.4 volts. (The real discharge curve is
non-linear, but for this first crack at it, we'll assume it's linear with time.) If we
assume that we'll get exactly 1.1 Ah during the 50 hour discharge, that means the current
delivered by the cells for 50 hours would be 1.1/50 = .022 amps if the output voltage per
cell were a constant 1.4 volts. So, if the 4 cells in series can put out .022 amps at 5.6
volts, that is a power of .1232 watts that the 4 cells can deliver for 50 hours.
This is the key to solving this problem; you have to calculate the *power* the cells can
deliver as DC and then figure out what resistive load will consume this power as AC (with
conversion efficiency taken into account).
The DC provided by the 4 cells has to get converted to AC somehow, and that process will
happen with less than 100% efficiency. Phil picked 70% efficiency, so let's go with that
number. Thus, you will be able to provide .7 * .1232 watts, or .08624 watts to your load
for 50 hours. The AC voltage is 2 volts, peak-to-peak, which is .707 volts RMS. Now
we're finally at a point where simple AC ohm's law can be used. Power =
volts^2/resistance (P = E^2/R), so .08624 = .707^2/R, or R = .5/.08624 = 5.798 ohms.
The reason I get a slightly different result than Phil is because you told us that the
cells are alkaline, which have a higher voltage during discharge than NiMH cells do.
But, I will say that you'll get substantially less than full capacity from these cells
if your endpoint voltage is 1.25 volts per cell. See the curves in the PDF linked above.