I'm doing something wrong mathematically. Can anyone spot my mistake?

I'm powering some laser sensors and an arduino with 8 AA batteries and a step-up box for the lasers. I calculated a theoretical battery life using some empirical Watt-hour findings for my batteries here: http://rightbattery.com/57-1-5v-aa-duracell-procell-alkaline-battery-tests/

Both I and the source website used a cutoff point of 1V per cell. My cutoff point was 8V.

I first used their Wh values and plotted them out as in the attached image. Then I found an exponential approximation, input a measured current draw, and got an answer for my theoretical Wh (see bottom right of image under "Dual laser test setup (no LCD, Xbee)".

Then I divided the Wh by Watts, which is the measured system voltage times the measured current and got 0.169 hours which is 10 minutes. I ran a test though, and it took 85 minutes to reach 8V, starting from 11.7 with 8 batteries. There seems to be some discrepancy.

The batteries lasted 8 times longer than predicted! But I was under the impression that for batteries in series, you don't add the mAh, since the current through each battery is the same, as described here: Adding mAh when wiring battery cells in series?

So what gives?

 See Screengrab Here

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    \$\begingroup\$ Did you account for the fact that you're calculating Wh per battery and using 8 batteries? \$\endgroup\$ – Brian Drummond Apr 27 '16 at 21:46
  • \$\begingroup\$ As I referenced, I was under the impression that for batteries in series, you don't add the mAh, since the current through each battery is the same. The link to that discussion is here: electronics.stackexchange.com/questions/20701/… \$\endgroup\$ – LegitimateWorkUser Apr 27 '16 at 21:49
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    \$\begingroup\$ Right but your "Watts" calculation incorporates the voltage from 8 cells not 1. You are then dividing the Wh rating for a single cell by this power to get time, and wondering why the time is 8* smaller than measured. \$\endgroup\$ – Brian Drummond Apr 27 '16 at 21:53

Brian Drummond (see comments) is correct. For each battery that you have:

$$Power = V_{batt}I_{batt}$$

If you put the batteries in parallel, your voltage doesn't change, but your potential current changes:

$$Power = V_{batt}(8I_{batt})$$

If you put the batteries in series, your potential current doesn't change, but your voltage does:

$$Power = (8V_{batt})I_{batt}$$

Either way, you get 8x the Power with 8 batteries:

$$Power = 8V_{batt}I_{batt}$$

Add in the time factor and you have Amp-hours or Watt-hours.

  • \$\begingroup\$ No problem, I worked through similar pain on a project a few months back (looking at your spreadsheet). As it happens, at low current draws, you can typically predict within about 10% the life of a product given its discharge curve. Enjoy! \$\endgroup\$ – slightlynybbled Apr 27 '16 at 22:53

Your discrepancy comes from the failure to account for the capacity of the 8 cells. Your presumption of not using the collective total is wrong when talking about watts or watt hours but correct when considering amp hours or total amps.


Two things, the manufactirers usually publishna guaranteed capacity (some will be higher), second, the capacity can be quite dependent on how fast you drain them, a clock might get 2-3Ah from a AA while a high power flashlight might only 1Ah. As a side note, Dave Jones from the eeVblog found you could get an extra 10% just by squashing them a bit so their capacities have quite wide margins


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