How Does Back EMF Affect Motor Performance?

AI Thread Summary
The discussion focuses on calculating the resistance of a motor's armature coil, the back EMF at normal operating speed, and the current drawn at one-third normal speed. The resistance was calculated to be 9.59 ohms using Ohm's law. The back EMF at normal speed was determined to be 97.9 volts, but there was confusion regarding the method for calculating the current at reduced speed. Participants noted a discrepancy in the application of formulas, particularly regarding the treatment of back EMF in different scenarios. Clarification on the calculations and consistent application of principles is needed for accurate results.
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)
 
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There is an inconsistency between a) and b). In a) you say V = IR, in b) you say V-emf = IR. Revisit your solution.
 
emf = 0 in a)
 
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