# How do you calculate the internal resistance of a battery?

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: $9 V$ (no current)
Measuring the current with a $9 \Omega$ external resistor between the poles, we would expect a current:

$I = {U}_{\text{raw"/R_"ext}} = \frac{9 V}{9 \Omega} = 1 A$ current.

In reality we measure $0.8 A$.
Therefore the total resistance ${R}_{\text{tot}} = \frac{9 V}{0.8 A} = 11.25 \Omega$

So the internal resistance of the battery in this case:

${R}_{\text{int" =R_"tot"-R_"ext}} = 11.25 \Omega - 9 \Omega = 2.25 \Omega$

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