A motor question, back emf

  • Thread starter michaelw
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  • #1
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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 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)
 

Answers and Replies

  • #2
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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.
 
  • #3
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emf = 0 in a)
 

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