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?
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]
