Let´s say I have a 2 AA 1,5V 1000mAh Alkaline battery cells in series.

The circuit it´s connected to consumes 500mA, works at 3V and it´s powered on always, and taking into account other electrical specs as ideal, the time it would take to last would be:

1000mAh / 500mA = 2h

Is this correct?

I´ve checked this question and it confuses me.

  • 1
    \$\begingroup\$ Calculate the watt-hours (Wh). \$\endgroup\$ – Jeroen3 May 12 '20 at 12:28
  • 1
    \$\begingroup\$ Are you likely to get both in AA sizes? \$\endgroup\$ – Andy aka May 12 '20 at 12:35
  • \$\begingroup\$ Add the datasheet of the specific batteries concerned. As Andy is hinting, one is probably lying about its spec. To address the question : if both are honest about their capacity, and used according to their ratings, the answer would be No. \$\endgroup\$ – Brian Drummond May 12 '20 at 12:46
  • \$\begingroup\$ Two AA 1,5V 1000mAh Alkaline batteries connected in series Making that a 1,5 V + 1,5 V = 3 V battery. Connecting in series doesn't affect the current so it remains 1000 mAh. So basically you're asking if 1 liter of fuel lasts as long as two times halve a litre of the same fuel. \$\endgroup\$ – Bimpelrekkie May 12 '20 at 12:49
  • \$\begingroup\$ Thank you all for the comments. This is a theoretical question, not based in any datasheet. \$\endgroup\$ – LazyTurtle May 12 '20 at 12:50

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