I've searched for an answer for this question for a long time. Specifically I've found this thread which is closely related to my own question: Why are batteries measured in ampere-hours but electricity usage measured in kilowatt-hours?

Here is my question: I would expect that the electrical energy that a battery produces would be constant, but according to tyblu's answer to the thread above, the charge is the one that's approximately constants. It seems odd to me, because it seems to violate the law of conversation of energy, because if a battery's charge is constant, but its voltage output decreases over time, then the electrical energy must decrease too, because E=V*Q.

What am I missing?

  • \$\begingroup\$ The answer to the banner question is that Ah gives a simple, rough & ready guide to battery life to the non-technical (who buy 99%+) \$\endgroup\$
    – Chu
    Jul 4, 2015 at 16:41
  • \$\begingroup\$ A quick thought : Energy delivered depends on the power lost in the internal resistance. Amp-hours should be more constant across different load cases. \$\endgroup\$
    – tomnexus
    Jul 5, 2015 at 9:02

1 Answer 1


You may have missed tyblu's clarification on his answer. As stated, a battery can store a fairly constant amount of charge, this decreases as the battery is discharged, a batteries output energy is dependant on it's operating conditions, and partially the length of time the battery spends powering something.

This can be seen on the datasheet of many batteries. Such as this. Note that the scale is not linear on all of the graphs.

(I'm completely rusty, and just getting back into electronics, so please feel free to point out any corrections needed, or the possible non-sensical answer.)

  • \$\begingroup\$ A battery has it's total output energy reduced as the length of time taken for it to output energy is increased No. First approximation, energy is the same with time. Better approximation, internal resistance means less energy with less time. \$\endgroup\$
    – tomnexus
    Jul 5, 2015 at 9:01
  • \$\begingroup\$ Battery is an electrochemical "CAPACITOR" and the charge on a capacitor is measured as Q = I*t vtingole \$\endgroup\$ Jul 5, 2015 at 12:52
  • \$\begingroup\$ @tomnexus I'm afraid I do not follow, I'm not sure what I have missed, nor can I find anything which points to a battery being able to deliver more energy over a week, than it could in an hour. Or is that discrepancy due to factors other than internal resistance? \$\endgroup\$
    – Phizes
    Jul 6, 2015 at 3:12
  • \$\begingroup\$ @vijayingole could you please clarify the implications one might infer from that statement? \$\endgroup\$
    – Phizes
    Jul 6, 2015 at 3:15
  • \$\begingroup\$ It is due to internal resistance. In an extreme example, with Rload = Rinternal, only half the energy will be delivered. With Rload >> Rinternal, most of the energy will end up in Rload, over a longer time. In both cases, the Ah delivered will be the same, the difference in terminal voltage under load is what makes the difference. \$\endgroup\$
    – tomnexus
    Jul 6, 2015 at 5:05

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