The terminal voltage of a battery is given by
[tex]V=\epsilon - Ir[/tex]
where [tex]\epsilon[/tex] is the e.m.f. of the battery, I the current, r the internal resistance.
In our problem, V=9.0 V when [tex]I=75 A[/tex], while the emf is [tex]\epsilon =12.0 V[/tex], so we can find the internal resistance r by re-arranging the formula:
[tex]r= \frac{\epsilon-V}{I}= \frac{12.0V-9.0V}{75 A}=0.04 \Omega [/tex]
