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In my career I have mostly ignored this question and just taken it for granted that if a 9V battery says it produces 9V that it will produce 9V until it runs out of "battery" but curiosity finally caught up with me and I want more details.

I recognize that all things are made up of atoms. Atoms have electrons. Electrons moving is what we call electricity. We can quantify the number of electrons moving past a point into a measurement called current.

Voltage is a difference in the amount of electrons between 2 points. In a 9V battery the manufacturer might label the battery has 500 mAh (it can produce 500 mA for one hour.) I conclude there are enough electrons in this battery to create 500 mA for 1 hour.

So lets say I connect a light bulb that consumes 500 mA at 9V. In one hour the battery runs out and the light bulb turns off. If electrons are flowing from the negative terminal through the light bulb and back to the positive terminal then how is it possible to have a 9V voltage difference between terminals for the entire hour? After 30 minutes wouldn't I be left with 4.5V worth of electrons? Half of my electrons have flowed from the negative terminal to the positive terminal. I conclude the potential difference must be less than 9V significantly.

Obviously reality doesn't match my expectations but the saying "voltage is pressure; current is flow" doesn't really explain this situation either.

Hence, how does a 9V battery generate 9V for an extended period of time?

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  • \$\begingroup\$ Snarky answer: a 9v battery doesn't stay at 9v for any reasonable definition of "extended period of time". See the voltage vs capacity or time graph of any 9v battery datasheet. \$\endgroup\$
    – Passerby
    Commented Mar 29, 2022 at 0:26

4 Answers 4

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I'll just talk generally about the chemistry. And keep in mind that the usage of anode and cathode as a chemist applies them may at first seem confusing to someone in electronics. (I'll also talk about a rechargeable battery case, with reversible reactions -- although that's not critical to what I write below, it makes it a little easier for me to write.)

Suppose you fabricate a chemical battery, using any of a variety of practical arrangements. When all the materials are first assembled (for example, the diluted sulfuric acid is first poured into a lead-acid wet cell), the reversible chemical reactions inside the battery cell have a preference to start removing electrons from the cathode and adding electrons to the anode. This continues for a while and as it does an increasing field gradient (electric in nature) develops because of the accumulating charge differences. The increasing electric field gradient increasingly retards the chemical reaction rate, too.

Eventually, the early preference for one side of the balanced reversible chemical reaction for moving electrons from the cathode to the anode has diminished enough so that the other side of the balanced chemical reaction occurs at the same rate and the battery finds an equilibrium state. This sets the unloaded battery voltage that you'd measure using a high impedance device designed for that purpose.

When you place the battery in service and the accumulated charges disperse rapidly into a circuit to motivate a current to move (very rapid process that is only very rarely discussed in electronics), current in the circuit occurs and charges at the two battery terminals will recombine and this will disturb the field gradient inside the battery, lowering it, and thus the equilibrium state shifts out of that equilibrium with a renewed, slight preference for one side of the chemical equation, again, and electrons are moved through the inside of the battery until things are back in equilibrium.

The battery voltage is held up in this way. At least, it is so long as the current is low enough that the reaction rates can sustain it.

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Volts aren't a measure of how many electrons are left in a battery; a half discharged 9 V battery doesn't have "4.5 V worth of electrons [left]". Instead, volts are a measure of energy per electron. A 9 V battery is a device that can pump electrons around a circuit, providing up to 9 joules of energy for every 6.24 x 1018 electrons (or 9 joules per coulomb of electrons).

It does this by sustaining a series of chemical reactions which provide that much energy per electron. When the energetic chemicals inside the battery run out, the battery stops providing energy to the circuit.

When you discuss amp-hours, you actually get at this point - 500 mA hours is 1.12*10^22 electrons. A 500 mAh battery has enough energetic chemicals inside it to perform this reaction for approximately 1.12x10^22 electrons passed through it - this might be 1.12x10^22 molecules, each reacting to pump one electron, 0.6x10^22 molecules, each reacting to pump two electrons, etc...

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  • \$\begingroup\$ So you are saying the battery chemical properties adjust automatically to produce a 9V difference in potential energy until it can't (like a car running at 90 mph until it runs out of fuel)? \$\endgroup\$
    – user306837
    Commented Mar 28, 2022 at 18:03
  • \$\begingroup\$ @MechEE The chemical properties don't adjust automatically. The actual reaction has a fixed energy associated with the energies of the reactant and product chemicals. \$\endgroup\$
    – nanofarad
    Commented Mar 28, 2022 at 18:16
  • \$\begingroup\$ At the beginning a battery has 9V worth of electrons available. One minute under load passes. There are now less electrons available. The chemicals react supplying more electrons. Or am I way off base still? \$\endgroup\$
    – user306837
    Commented Mar 29, 2022 at 13:31
  • \$\begingroup\$ J = C * V so to keep V constant means that C has to adjust itself "automatically" as time under load passes. \$\endgroup\$
    – user306837
    Commented Mar 29, 2022 at 13:35
  • \$\begingroup\$ @MechEE J = C * V is not an adjustment. It's a law of physics and an interpretation of the units. The number of coulombs is a direct consequence of the current drawn by the load. If the load draws more current, the number of coulombs and the number of joules both increase: the battery does 9 joules of work for every coulomb of charge, and now it's doing work on more charge. \$\endgroup\$
    – nanofarad
    Commented Mar 29, 2022 at 13:40
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The chemicals in a disposable battery get used up as it is used when it gets weaker and weaker. The datasheet of the battery from a manufacturer's website shows it.

https://data.energizer.com/pdfs/522.pdf

9V

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Batteries operate on a chemical reaction. For common alkaline batteries, Wikipedia shows the reaction. According to the article, "The chemical energy is stored mostly in the zinc metal."

The voltage from batteries is not stable, but decreases over time as shown in the graph below from the Duracell MN1604 alkaline battery data sheet.

enter image description here

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  • \$\begingroup\$ That is very informative. I didn't realize you could look up voltage/current graphs for batteries. Makes sense in hindsight though. \$\endgroup\$
    – user306837
    Commented Mar 28, 2022 at 18:05
  • \$\begingroup\$ It makes so much more sense now on why you shouldn't mix batteries... thanks for the clarity! \$\endgroup\$
    – user306837
    Commented Mar 28, 2022 at 18:23
  • \$\begingroup\$ As a chemist, thanks for the statement about the source of the chemical energy. We get a fair number of cell and battery questions over at chemistry SE and it gets tiring pointing out the energy source in lemon cells, the Daniell cell, etc. It is surprising how many people think seawater or saltwater is the energy source in some simple cells. (+1). \$\endgroup\$
    – Ed V
    Commented Mar 28, 2022 at 20:36

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