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?
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
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.
A 120 V electric motor is running with a current of 3 A. When initially started...
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