12 V is the f.e.m. [tex]\epsilon[/tex] 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:
[tex]\epsilon - Ir[/tex]
this is equal to the voltage drop on the resistance of the motor R:
[tex]RI[/tex]
so we can write:
[tex]\epsilon - Ir = RI[/tex]
and using [tex]r=0.0305~\Omega[/tex] and [tex]R=0.055~\Omega[/tex] we can find the current I:
[tex]I= \frac{\epsilon}{R+r}=140~A [/tex]
