Archived Calculating Back EMF of an Automobile Starter Motor

[Augustus]

Homework Statement

An automobile starter motor draws a current of 1.9 A from a 5.5 V battery when operating at normal speed. A broken pulley locks the armature in position, and the current increases to 8.6 A.

What was the back emf of the motor when operating normally? Answer in units of V.

Homework Equations

V=IR is the only one I can think of

The Attempt at a Solution

I calculated the resistance. But I don't know why it is at all relevant that the amperage should increase to 8.6 A, if the question asks for the back emf of the motor at normal operating speeds.

 

[cnh1995]

A broken pulley locks the armature in position, and the current increases to 8.6 A.

Since the rotor is locked, there is no back emf in the armature circuit.
Hence, V_battery=I_aR_armature
∴R_armature=5.5/8.6=0.6395Ω
When the motor is running normally,
V_battery=I_aR_armature+E_b
∴E_b=5.5-1.9*0.6395=4.285V
Back emf in normal running condition is 4.285 V.