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I've seen a Duracell alkaline AA battery on Amazon. It can supply 1.5 V, but I don't see any information about the current (in A) or the power (in W). Where can I find this information?

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    \$\begingroup\$ You should look in the datasheet of that AA battery and check the discharge curves. That gives you an indication. Note that the highest discharge current that is mentioned is 1000 mA = 1 A. That does not mean you cannot discharge with 2 A but realize that the battery's capacity will be less at such a high current. You will get less energy out of the battery compared to a more realistic discharge current of for example 100 mA. You might want to watch this EEVBlog video: youtube.com/watch?v=R8hTQXqURB4 \$\endgroup\$ – Bimpelrekkie Jun 15 at 14:44
  • \$\begingroup\$ Check the factory website, they probably post that information or ask the battery vendor. \$\endgroup\$ – Gil Jun 16 at 2:25
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Short-circuit current of a new alkaline AA battery is in the low amperes.

About 3A for a fresh Kirkland AA cell. 2.4A for a Panasonic Platinum power.

Source: actual measurements

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    \$\begingroup\$ FYI: Kirkland batteries are made by Duracell. finance.yahoo.com/news/… Used to be, most store brand batteries were made by Rayovac, but the private label market has changed since that company's assets were acquired by Energizer in 2018. \$\endgroup\$ – Solomon Slow Jun 15 at 16:52
  • \$\begingroup\$ @SolomonSlow Thanks. They definitely seem interchangeable in my experience in terms of capacity, shelf life and lack of propensity to leak (a common issue with no-name alkalines). \$\endgroup\$ – Spehro Pefhany Jun 15 at 17:04
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The datasheet for a Duracell AA alkaline cell shows that its voltage drops very quickly with a 1A load.

Beware: I bought some Duracell chargers that came with 4 AA Ni-MH cells and 4 AAA Ni-MH cells at Costco. The cost was lower than possible because the Duracell batteries were made in China with half the capacity of American or Japanese ones.

I do not know if Duracell or Amazon are selling cheap poor quality alkaline batteries now. Duracell

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In the EE world, almost every question of this type will be answered by the manufacturers datasheet of the part in question. Here is the one for the Duracell AA you mentioned:

enter image description here

From the impedance of the battery, you only need Ohm's law to calculate the peak current and power the battery can supply. I'll leave the calculations for you and your understanding.

Here is a datasheet from Energizer instead which is more useful for your purpose. Inner resistance is listed as 150-300 mohm. Shoutout to @Hearth and @ScottSeidman for pointing this out.

enter image description here

Duracell AA datasheet

Energizer AA datasheet

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    \$\begingroup\$ Technically, that's not the DC impedance, but it's likely that the DC impedance will be quite close to the 1 kHz impedance. \$\endgroup\$ – Hearth Jun 15 at 14:41
  • \$\begingroup\$ This doesn't tell how how much current can be provided for how long, which is really useful info to have. That impedance will also change as a function of time. The info is in the provided graphs, though I confess, those aren't the most useful discharge graphs I've seen. \$\endgroup\$ – Scott Seidman Jun 15 at 14:49
  • \$\begingroup\$ Is impedance of 120 m-ohm equals to R = 0.120 Ohm, and so U = 1.5/0.120 = 12.5 A and so the current of this battery is 12.5 A ? is that true ? It seems me too high 12.5 A for a battery like this... \$\endgroup\$ – tata stereo Jun 15 at 14:50
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    \$\begingroup\$ it may (possibly) be able to supply 12.5A for a little bit of time if you directly short the terminals together i.e. the output voltage will be effectively 0V. The maximum power that you can (maybe) extract from this battery is ((1.5^2)/(0.12*2))/2=4.7W with an output voltage of 0.75V, again for a short time. Battery chemistry and behavior is complicated, and will diverge significantly from the idealised model of voltage source + resistor at the extremes (even the not so extremes), i.e. full capacity, low capacity, high current, high pulse current, high temperature, low temperature, etc \$\endgroup\$ – BeB00 Jun 15 at 15:00
  • \$\begingroup\$ Only for a very short amount of time. You can calculate backwards from the drain curves how much the DC voltage is dropping and get far more accurate the steady state values. \$\endgroup\$ – winny Jun 15 at 15:30
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In response to your comment to @winny's answer:

Is impedance of 120 m-ohm equals to R = 0.120 ohm, and so U = 1.5/0.120 = 12.5 A and so the current of this battery is 12.5 A?

You can draw 12.5 A into a short-circuit but a short circuit will have zero voltage and since P = VI you'll get P = 0 x 12.5 = 0 W.

The Maximum Power Transfer Theorem says that you will get maximum power when RL = RS so that would be 0.12 Ω load. The current would be reduced to 1.5/0.24 = 6.25 A and the power into the load (and dissipated in the battery) would be P = VI = 0.75 × 6.25 = 4.7 W.

is that true? It seems me too high 12.5 A for a battery like this...

Try it! It will deteriorate quickly as the cell's internal resistance increases.

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  • \$\begingroup\$ In your last calculation for power you use 0.75 for the voltage? Why is that instead of 1.5? \$\endgroup\$ – TheValyreanGroup Jun 16 at 3:24
  • \$\begingroup\$ @TheValyreanGroup, because the cell's internal resistance and the load form a potential divider. The battery model is an "ideal" 1.5 V source in series with the internal resistance of 0.12 ohms. That means that with a 0.12 ohm external load half the voltage will be lost to the internal resistance and half to the external load. 1.5 / 2 = 0.75 V. \$\endgroup\$ – Transistor Jun 16 at 6:30

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