How do you calculate the internal resistance of a battery?

1 Answer
Jan 24, 2015

A real-life battery can be described as an ideal voltage source with an internal resistance.

If you measure the voltage of a battery with a Volt-meter, which has a very high resistance, you'll get the raw voltage.

If you add a smaller resistance between the poles of the battery, you will see that Ohm's law is not followed, but the amperage is lower than expected.

An example
Raw voltage: 9V (no current)
Measuring the current with a 9Omega external resistor between the poles, we would expect a current:

I=U_"raw"/R_"ext"=(9V)/(9Omega)=1A current.

In reality we measure 0.8A.
Therefore the total resistance R_"tot" =(9V)/(0.8A)=11.25Omega

So the internal resistance of the battery in this case:

R_"int" =R_"tot"-R_"ext"=11.25Omega-9Omega=2.25Omega

Extra
You can also work with the voltage loss on the poles of your battery. I'll leave that to you.