michaelw
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A motor is designed to operate on 117 V and draws a current of 12.2 A when it first starts up. At its normal operating speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
Please let me know if I am on the right track :)
a) V=IR 117V = 12.2A * R
R = 9.59ohms
b) I = V-emf/R
2.3A = 120V-emf/9.59ohms
emf = 97.9V
But i have no idea how to do 3... would i just divide emf / 3 and find current? (97.9/3)
Please let me know if I am on the right track :)
a) V=IR 117V = 12.2A * R
R = 9.59ohms
b) I = V-emf/R
2.3A = 120V-emf/9.59ohms
emf = 97.9V
But i have no idea how to do 3... would i just divide emf / 3 and find current? (97.9/3)