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So I understand the basics of ohm's law and batteries. The electrons move from the cathode to the anode via the circuit and when the anode and the cathode have the same amount of electrons they stop flowing.

So when it comes to V=IR if the resistance in the circuit is increased the current decreases, so wouldn't this mean that the more resistance in a circuit the longer it would take the battery to become flat?

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Yes - if you draw a low current from a battery, it will last longer (provide useful voltage for longer) than if you draw a high current.

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    \$\begingroup\$ Not only that, but with most battery chemistries, if you draw half the current you get more than twice the time... \$\endgroup\$ – Ecnerwal Jun 24 '17 at 1:36
  • \$\begingroup\$ @Ecnerwal Look at a battery equivilent model. Think about ESR, ohms law and conservation of energy. \$\endgroup\$ – winny Jun 24 '17 at 6:28
  • \$\begingroup\$ @winny: Your model doesn't account for the fact that batteries rely on chemical reactions. Discharge at a high rate causes heat to be generated inside the battery, which changes the way the reactions occur in the battery. The end result is that you less of the available energy out of the battery if you put a heavy load on it. Then there are other effects like high discharge rates damaging the battery such that its internal resistance rises... \$\endgroup\$ – JRE Jun 24 '17 at 6:58
  • \$\begingroup\$ @JRE Sure thing, but even a first order model with just ESR is enough to explain the phenomenon. \$\endgroup\$ – winny Jun 24 '17 at 7:22
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In general it is true. Charging and discharging of batteries should be slow and gradual. Just like in many other aspects of physics , an adiabatic process which maintain equlibrium gives longeivity and stability to things... Resulting in less volatality.

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