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I recently learned how how energy actually flows in batteries, the electric and magnetic fields create an energy vector pointing into the circuit at all points and out of the battery. This broke my preconception that electrons are converted into energy which produces light when powering something like a light bulb. I'm confused on how batteries lose charge then if all energy is conserved in the system and electrons are not directly converted into energy. (this is my first post, so sorry if I didn't follow the rules correctly)

How do batteries lose charge?

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    \$\begingroup\$ Where did you learn this? When a battery isn't providing any power to a load no current is flowing so there is no magnetic field. Electrons are not converted into energy. You appear to have a lot of misunderstandings. I suggest that you study some basics about electricity and after that how batteries work. Maybe this helps: youtube.com/… \$\endgroup\$ – Bimpelrekkie Dec 9 '20 at 19:47
  • \$\begingroup\$ Another good source could be Tony Kuphaldt's series. \$\endgroup\$ – rdtsc Dec 9 '20 at 20:11
  • \$\begingroup\$ @Bimpelrekkie why is there no current when there is no load? don't the electrons still want to move from high to low concentration? isn't there a lower concentration of electrons in the negative end of the terminal? \$\endgroup\$ – Micah Weiss Dec 9 '20 at 21:09
  • \$\begingroup\$ This "electric and magnetic fields create an energy vector pointing out of the battery and that's how energy flows" thing is interesting, but not useful. Don't learn it. Learn voltages and currents. \$\endgroup\$ – user253751 Dec 10 '20 at 20:58
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What depletes batteries

Batteries work on a chemical reaction between materials. The chemical reaction creates an excess of electrons on one connection (the - pole of the battery) and a lack of electrons at the other connection (the + pole of the battery).

When a load is connected to the battery, between the + and - poles of the battery, the electrons flow through the load and their movement provides the energy.

The chemical reaction "consumes" the materials. When one of the materials is used up, the chemical reaction can no longer continue and the battery is depleted.

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  • \$\begingroup\$ So I understand how the chemical reaction facilitates the flow of electrons that gives the system its energy and how it depletes as the reaction runs out of reactants. What I don't understand is how energy lost through heat when a lightbulb is attached affects the energy of the system. Suppose both ends of the terminal were just connected to each other and no energy was convered to heat. Are the two batteries, one connected to a light bulb and the other only to itself, different somehow when they are depleted? (my question was worded poorly) \$\endgroup\$ – Micah Weiss Dec 9 '20 at 20:16
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    \$\begingroup\$ @MicahWeiss "Suppose both ends of the terminal were just connected to each other and no energy was convered to heat." - energy is converted to heat - there is resistance in the wire you use to connect the terminals, there is resistance in the terminals themselves, and if you eliminate all of that, you end up with nothing more than a chemical reaction in a jar which generates heat (exothermic). \$\endgroup\$ – Tom Carpenter Dec 9 '20 at 20:20
  • \$\begingroup\$ Maybe I am mistaken, but it sounds more like you know that chemical reactions produce electrons and deplete reactants but it does not sound like you know howthis happens. See my answer here: electronics.stackexchange.com/questions/517893/… \$\endgroup\$ – DKNguyen Dec 10 '20 at 23:54
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Batteries lose charge because of an internal leakage current, this current is present in all batteries to some extent.

A battery is dependent on a chemical reaction with the electrolyte reacting with the anode or cathode. There are many reasons why a battery has leakage currents, but the main reason is the electrolyte reacts with impurities in the battery which creates a very small current (which represents the reactions taking place).

Over time these leakage reactions\current can totally deplete a battery and degrade it to a point that it is no longer useful.

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