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#

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