How Does Back EMF Affect Motor Performance?

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SUMMARY

The discussion centers on the impact of back electromotive force (emf) on motor performance, specifically for a motor rated at 117 V and drawing 12.2 A at startup. The resistance of the armature coil is calculated to be 9.59 ohms. At normal operating speed, the back emf is determined to be 97.9 V, resulting in a current draw of 2.30 A. The inconsistency noted between the calculations for resistance and back emf highlights the importance of correctly applying Ohm's Law in different operational contexts.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Basic knowledge of electric motors and their operational characteristics
  • Familiarity with back electromotive force (emf) concepts
  • Ability to perform electrical calculations involving voltage, current, and resistance
NEXT STEPS
  • Study the principles of back emf in electric motors
  • Learn how to calculate motor performance metrics at varying speeds
  • Explore advanced motor control techniques to optimize efficiency
  • Investigate the effects of load on motor current draw and back emf
USEFUL FOR

Electrical engineers, motor control specialists, and students studying electromechanical systems will benefit from this discussion, particularly those focused on optimizing motor performance and understanding electrical principles.

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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