Answer
0.034 $\Omega$
0.093 $\Omega$
Work Step by Step
Figure 19–2 shows the circuit diagram for this problem.
First find the internal resistance using $V_{ab}=\epsilon-Ir$.
$$8.8V=12.0V-(95A)r$$
$$r=0.034\Omega$$
Next solve for the resistance of the starter.
$$V_{ab}=IR$$
$$R=\frac{V_{ab}}{I}=\frac{8.8V}{95A}=0.093\Omega$$
