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.(adsbygoogle = window.adsbygoogle || []).push({});

Please let me know if im 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)

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# Homework Help: A motor question, back emf

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