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I have a simple question. How many amperes can an AA battery draw? I'm studying electronics and looking at the Ohms law: V=IR and taking as an example a 1.5V AA alkaline battery, and considering the battery has an internal resistance in the order of 0.1 then it could be capable of delivering up to 15 Amperes?

Reading I see these can deliver up to 2 Amperes in real life so I'm a bit confused. What else is playing here that I'm missing?

Thank you,

Matias.

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  • \$\begingroup\$ Yes, I noticed my math wasn't right, I just corrected my question numbers. Now, it can truly deliver that much? I guess not for too long but I'm wondering if it can. \$\endgroup\$ – Matias Apr 27 '15 at 20:55
  • \$\begingroup\$ It probably could deliver 15A, but not safely. It may overheat, or damage it in some other way. But otherwise, yes, the internal resistance implies it could deliver 15A. The internal resistance is not always a simple ohmic linear resistance - you may find that when you push the current draw, the internal resistance increases. \$\endgroup\$ – CL22 Apr 27 '15 at 20:57
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As the battery begins source current, it heats up and the chemicals inside begin to polarize, causing the internal resistance to increase significantly reducing it from the 15A theoretical to the realistic DC output of a few Amps max.

Chemistries like Lithium Ion can pump out far more amps for longer, because the chemistry does not break down so easily as those in Alkaline and Nikel Metal Hydride battery chemistries.

Another issue is the voltage of the AA battery while under such a huge load will sag significantly. Total rated capacity also takes a massive hit for certain chemistries, meaning sustained loads that large will drain the battery 5+ times faster than expected.

Again, Lithium Ion chemistries are more resilient to voltage sag and capacity loss under load. Unfortunately though, Lithium Ion batteries tend to explode into fiery awesomeness so care must be taken - with great power [density] comes great responsibility.

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  • \$\begingroup\$ Great, so here's what I found on the battery spec: Temperature 21ºC (70ºF) RI=0.5 0ºC (32ºF) RI= 0.8 -21ºC (-4ºF) RI=5.0 Then it means the Internal resistance is 0.5 at 70F? That makes more sense and now my formula tells me the max amperage will be 3 amps \$\endgroup\$ – Matias Apr 27 '15 at 21:06
  • \$\begingroup\$ @Matias those values only give the initial internal resistance due only to temperature. The rated test current for those is probably very very low. When you actually draw useful current from the batteries at those various temperatures, it will be very different, very quickly. \$\endgroup\$ – KyranF Apr 27 '15 at 21:08
  • \$\begingroup\$ Thanks, so where you can find the internal resistance value in the Battery Specs? I'm just curious. \$\endgroup\$ – Matias Apr 27 '15 at 21:13
  • \$\begingroup\$ You cant, you can only get what the manufacturer gives, if they even bother (it can be very hard for some brands to get this sort of info) and even then it is for unrealistic test conditions, in order for their marketing numbers to look better. \$\endgroup\$ – KyranF Apr 27 '15 at 21:17
  • \$\begingroup\$ Take a look at this site powerstream.com/AA-tests.htm This will give you an idea of what voltage an AA cell can put out for various currents, and how this will change with time. \$\endgroup\$ – WhatRoughBeast Apr 27 '15 at 22:04

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