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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
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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.

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  • \$\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
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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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