Question

What is the internal resistance of a 12.0-v car battery whose terminal voltage drops to 9.0 v when the starter motor draws 75 a ?

Answer 1

The terminal voltage of a battery is given by

`V=\epsilon - Ir`
where
`\epsilon` is the e.m.f. of the battery, I the current, r the internal resistance.
In our problem, V=9.0 V when
`I=75 A`, while the emf is
`\epsilon =12.0 V`, so we can find the internal resistance r by re-arranging the formula:

`r= (\epsilon-V)/(I)= (12.0V-9.0V)/(75 A)=0.04 \Omega`