An electric motor draws a current of 2.50 A from a 15.0-V battery when it runs at normal speed. If the motor's armature is "jammed" so that it cannot rotate, the current suddenly rises to 10.0 A. What is the back emf of the motor when running at normal speed

11.25 V

Explanation:

We know E - IR = V when E = battery emf = 15.0 V and I = current = 10.0 A when the armature is jammed. When the armature is jammed, V = 0. So

E - IR = V

15 - 10R = 0

15 = 10R

R = 15/10 = 1.5 Ω which is the resistance of the windings.

Now, the back emf V when it is running at normal speed of current I = 2.50 A,

V = E - IR = 15 - 2.5 * 1.5 = 15 - 3.75 = 11.25 V