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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?

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Brrch
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  • 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%+) – Chu Jul 04 '15 at 16:41
  • A quick thought : Energy delivered depends on the power lost in the internal resistance. Amp-hours should be more constant across different load cases. – tomnexus Jul 05 '15 at 09:02

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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.)

Phizes
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  • *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. – tomnexus Jul 05 '15 at 09:01
  • Battery is an electrochemical "CAPACITOR" and the charge on a capacitor is measured as Q = I*t vtingole – vijay ingole Jul 05 '15 at 12:52
  • @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? – Phizes Jul 06 '15 at 03:12
  • @vijayingole could you please clarify the implications one might infer from that statement? – Phizes Jul 06 '15 at 03:15
  • 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. – tomnexus Jul 06 '15 at 05:05
  • Q=I*t means Q is the charge being stored in the battery, I is the current in Ampere and t is the time duration during which current has flown in the battery. and the energy=Q*V where V is the voltage of battery so energy= V*I*t so for energy one has to multiply AMP Hr by battery voltage to get the energy stored therein. vtingole – vijay ingole Jul 07 '15 at 14:00
  • @tomnexus I have removed my incorrect statement, I'm still unsure as how to differentiate then between the energy a battery supplies into a load, excluding the "loss" due to internal resistance. Though, it's likely I'm still way off somewhere. – Phizes Jul 25 '15 at 10:40