My question essentially is concerning what (if anything) prevents really high current flow (amperes) in a battery if the terminals were to be connected directly with a low resistance conductor?

For Example this website has the following chart pertaining to the internal resistance of various batteries:

enter image description here

So lets assume I have an AA NiMH battery and use Ohms Law, we will also account for a minimal ammount of additional resistance (.003 ohms) from the conductor wire itself just to further illustrate the point:

V = IR

V = 1.5

R (total) = .023

I = V/R = 1.5/.023 = 65.22

enter image description here

Is this what really happens when battery terminals are directly connected, and if not what is the other limiting factor that prevents a super fast current from occuring besides the natural resistance of the battery and the wire?

  • $\begingroup$ The internal resistance can only be used for smaller currents, as long as the current-voltage curve is linear. Not for a short-circuit. $\endgroup$
    – user137289
    Dec 13, 2016 at 21:00
  • $\begingroup$ And, ultimately, there is a chemical reaction going on that will have rate limiting steps associated with it. $\endgroup$
    – Jon Custer
    Dec 13, 2016 at 21:07
  • $\begingroup$ @Pieter okay, but if we then ignore the internal resistance for the short circuit, the current is even higher. So what are the missing factors to the equation, because most AA batteries max out at 2 amps and that's pushing it. Is it simply that once the short circuit occurs the voltage will drop and no longer be 1.5 which will reduce the current? $\endgroup$
    – person0
    Dec 13, 2016 at 21:12
  • $\begingroup$ No, oh no, the short-circuit will be smaller than your estimate. As @JonCuster says, there is chemistry involved: diffusion, gradients, bubbles may form, heat is developed, etc. $\endgroup$
    – user137289
    Dec 13, 2016 at 21:25

1 Answer 1


Batteries are decently complicated devices. They're certainly more complicated than an ideal voltage source. As you've noticed, we often model them as a voltage source with a series "internal resistance." This is a better model which works well in a large range of conditions, but still not a true model of how batteries work.

When you short a battery out, initially you do get the very high amperages you calculated. However, as the short continues, chemistry gets involved. Inside the battery you have two materials reacting with each other to provide the electrical energy. In the case of your example battery, NiMH, the reactions are:

Negative terminal: $H_2O + M + e^− \Leftrightarrow OH^− + MH$
Positive terminal: $Ni(OH)_2 + OH^− \Leftrightarrow NiO(OH) + H_2O + e^−$

Noe that the negative terminal produces hydroxide ions and the positive terminal consumes them. Also note the double arrows on both equations. The equilibrium point depends strongly on the concentrations of the compounds.

In normal operation, the hydroxide ions have time to work their way from the negative terminal to the positive terminal, keeping the reactions going smoothly. However, in a short, there will be a build up of hydroxide near the negative terminal because it takes time to diffuse away from the terminal. This will reduce the rate at which the reaction occurs, reducing the maximum current the battery produces in this short example to below what you would expect from modeling the internal resistance alone.

Also, depending on your battery, there may be 3rd order effects. You're going to generate lots of heat in the battery, and that causes reactions to change rates. Depending on the battery, this may accelerate the discharge. For an example of what these effects might be like, I point to this video of a lithium-ion battery. In this case the battery was *ahem* "encouraged" with a knife, but runaways like this for lithium ion batteries are not unheard of.

  • $\begingroup$ That actually clears it up very well for me. Thanks for the answer! $\endgroup$
    – person0
    Dec 13, 2016 at 22:58

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