Question

An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï internal resistance. 33% part (a) what is the current to the motor in a?

Answer 1

12 V is the f.e.m. $\epsilon$ of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
$\epsilon - Ir$
this is equal to the voltage drop on the resistance of the motor R:
$RI$
so we can write:
$\epsilon - Ir = RI$
and using $r=0.0305~\Omega$ and $R=0.055~\Omega$ we can find the current I:
$I= \frac{\epsilon}{R+r}=140~A$