Question

A 2-horsepower automobile electric starter motor operates at 90 percent efficiency from a 12-V
battery. What is the battery internal resistance if the battery terminal voltage drops to 10 V
when energizing the starter motor?

Answer 1

2-horsepower:

$2\cdot 746=1492$ W

With efficiency 90% motor uses:

$P=1492/0.9=1658$ W

The current:

$I=P/V=1658/12=138$ A

However that hypothetically loaded motor is not actually running at its rated voltage and power - its at 10 volts. So the current should be scaled by that:

$I=138\cdot 10/12=115$ A

So, the battery internal resistance:

$r=\frac{12-10}{115}=0.017\ \Omega$