Question

A 120 V electric motor is running with a current of 3 A. When initially started the motor draws 16 A. What is the back emf of the motor at normal operating speed?

Answer 1

Ir=Ea-Eb/R=120-Eb/R=3

Ea/R=16

R=120/16=7.5 A

(120-Eb)/R=3

(120-Eb)=3*7.5=22.5

Eb=97.5 V

Answer 2

When the motor starts up there is no back emf, so the resistance of the armature is 120V/16A = 7.5 Ohm.
At normal speed the pd across the armature is 3 A x 7.5 Ohm = 22.5 V. The back emf at this speed is 120-22.5= 97.5V.

Answer 3

Pm = Vb * I

Pm = V*I1 = 120*3 = 360 W

I = 16A

Vb = 360/16 = 22.5 V