Why do batteries drain so quickly if they are forced to draw more current than they normally do?

When a battery is pushed to use twice the current it normally does, it lasts for less than half as long before dying...

In fact, batteries often come with a 'C' rating that gives you an idea of how long they might last when run at a high current level...

What is going on inside the battery when this occurs? What is the chemistry, so to speak?

• The chemical reaction the drives a battery does not depend on the currant that is produced. It is only a question of kinetics. If the contact between the poles is good, the currant is high, and the battery is soon discharged. If there is an important resistance between the poles, the currant is low and the battery will be operational a long time. But the chemical composition of the electrodes does not depend on the intensity of the currant. Sep 6 '20 at 15:10
• High current leads to increased temperature, leading to increased parasitic internal discharge, which leads to further temperature increase.
– Karl
Sep 7 '20 at 8:14
• Batteries store chemical energy. They have a finite amount of it. if you use that energy faster (all other things being equal that is what "higher current" means) then the capacity will be reduced faster. That's the simple explanation. If you are asking why does the energy sometimes decrease faster than expected given that basic observation, you need to clarify the question. Sep 7 '20 at 9:16

The capacity of a cell is measured toward the terminal cut off voltage. If e.g.a double current is drained from a cell, the external cell voltage is decrease by the doubled voltage drop on the cell internal resistance: $$U_\mathrm{e}=U_0 - R \cdot I$$

That caused reaching the terminal voltage sooner then in the half time.

It is additionally leading to the ohmic internal energy wasting $$P=R \cdot I^2$$ But this energy loss affects the useful energy output, not the capacity ( directly).

Additionally, higher than optimal current causes lowering of the immediate cell open voltage. Limited rate of diffusion of electroactive components and/or of electrode reactions are not able to maintain electrode equilibrium potentials.

In result, even the $$U_0$$ (non linearly) decreases with the higher current and the terminal $$U_e$$ is reached after providing smaller total charge.

• Also, energy is wasted in the internal heating of a cell. At low currents, the cell is hardly warmed, but at high currents, the voltage dropped across the cell's internal resistance is wasted as heat. This is just a "common sense" explanation of the answer. Sep 7 '20 at 4:22
• @DrMoishe Pippin This is involved in RI, as P=RI^2. Sep 7 '20 at 5:34
• Yes, I was just making the point that the cell gets noticeably hot at high drain. Sep 7 '20 at 19:51