Question

what is the internal resistance of a 12.0 - v car battery whose terminal voltage drops to 8.4 v when the starter motor draws 83 a? express your answer to two significant figures and include the appropriate units.
part b
what is the resistance of the starter? express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Find voltage drop across internal resistance

The emf of the battery $E = 12.0\ V$ and terminal - voltage $V = 8.4\ V$. The voltage drop across the internal resistance $\Delta V=E - V$.
$\Delta V=12.0 - 8.4=3.6\ V$

Step2: Calculate internal resistance using Ohm's law

According to Ohm's law $V = IR$, for the internal resistance $r$, where $I = 83\ A$ and $\Delta V$ is the voltage drop across the internal resistance. So $r=\frac{\Delta V}{I}$.
$r=\frac{3.6}{83}\approx0.043\ \Omega$

Step3: Calculate resistance of the starter

Using Ohm's law for the starter, $V = 8.4\ V$ and $I = 83\ A$. So $R=\frac{V}{I}$.
$R=\frac{8.4}{83}\approx0.10\ \Omega$

Answer:

Part A:
$r = 0.043\ \Omega$
Part B:
$R = 0.10\ \Omega$