Question

What is the internal resistance of a 12.0-v car battery whose terminal voltage drops to 8.3 v when the starter motor draws 89 a ?

Answer 1

The relationship between the terminal voltage V of a battery and its electromotive force
`\epsilon` is given by

`V=\epsilon - Ir`
where I is the current and r the internal resistance.
In our problem
`\epsilon=12 V`, and when a current of
`I=89 A` flows, the terminal voltage is
`V=8.3 V`. So we can find the value of the internal resistance:

`r= (\epsilon - V)/(I)= (12 V-8.3 V)/(89 A)=0.042 \Omega`