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If we assume a short circuit, with zero resistance (not even within the battery), would this 'infinite current' deplete all voltage within the battery? For example, if I "short" a 9V alkaline battery, it will run for about an hour (though the wire gets pretty hot, if not melts) until it depletes all its power. Is this only because the battery has internal resistance, or why does the battery 'keep running'? Additionally, does a battery ever give its Amp-hours or how long it can run?

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Because in the real world there is nothing such as a perfect wire or a perfect battery. There are parasitic resistances, inductances, and capacitances everywhere.

Plus there is the issue of the chemical reactions in the battery that is taking place to produce that electricity. That can only happen so fast even if you had a battery with zero parasitics.

Also, I find it pretty hard to believe that your shorted 9V runs for an hour. What is your definition of depleted? Batteries are considered dead long before they reach 0V in the same way you are too tired to do useful work long before you drop dead from exhaustion. I can only assume you are talking about reaching 0V across the terminals for purely academic purposes.

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  • \$\begingroup\$ yes, something like 0V. Also, I didn't let it go down all the way to zero, I stopped it a bit before to do a few measurements on it when it cooled down. \$\endgroup\$ – David542 Jan 21 at 23:49
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Just so this actually has an answer: yes, it is because the battery has internal resistance and because your wire is not an ideal short circuit.

Information about the capacity of a given battery can be found in the battery's data sheet, which should be available from the manufacturer. However, batteries are generally not designed to used with an extremely high load current like this so the actual capacity with a (near) short circuit is probably much less than what the manufacturer has specified for a more reasonable load scenario.

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The bigger the battery in a single cell, the lower it's internal resistance. Of course battery chemistry like electrolytic caps has a wide range in ESR (effective Series Resistance) So shorting its voltage across it's internal ESR will produce the short circuit current. This is dangerous on Lithium cells and lead acid cells due to the energy stored. Voc/ESR=Isc But for Alkaline which has almost the same energy of Li Ion cell it is safe since Alkaline has a higher ESR. We call the voltage drop with load resistance, Load Regulation Error, measured @ Amp with % V error also being a ratio of ESR to Load R.

For bigger battery short current pulsses Isc must be safely. It is not necessary to go to 0V to measure ESR, since you now understand where Load regulation comes from with ESR=ΔV/ΔI.

Although this image has slightly more energy than a Lithium cell both have a thermal runaway effect, once the activation energy threshold is reached. It accelerates to fuel a lower resistance with exothermic reactions of heat to ever increasing currents. In the case of a nuclear bomb, it takes a hydrogen bomb to Trigger it. Once triggered it has a negative incremental ESR , so the rise time is so short, it's equivalent bandwidth BW=0.35/Tr reaches gamma wave frequencies.

enter image description here

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  • \$\begingroup\$ Skill Testing question. How many cells are in a 9V alkaline battery. If the total resistance was 9 Ohms what is each cell's ESR? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 22 at 0:16

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