Motor back emf/current ?s

  • Thread starter candyq27
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  • #1
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Homework Statement


A motor is designed to operate on 130 V and draws a current of 13.8 A when it first starts up. At its normal operating speed, the motor draws a current of 4.40 A.
(a) What is the resistance of the armature coil?

(b) What is the back emf developed at normal speed?

(c) What is the current drawn by the motor at one-third normal speed?

Homework Equations



V=IR
I=(V-E)/R (E=induced emf)

The Attempt at a Solution



I figured out parts (a) and (b) but I'm not sure how to do part (c).
(a) V=IR 130=(13.8)R R= 130/13.8 = 9.42 ohms
(b) I=(V-E)/R so 4.4=(130-x)/9.42 x=88.6V
(c) I'm not sure how to figure this out. I know at normal speed the I is 4.4A, but I can't just divide that by 3 to figure it out. Can someone help? Thanks.
 

Answers and Replies

  • #2
AlephZero
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In (c), the back emf is proportional to the speed.
 
  • #3
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Ok so the back emf would be 88.6/3 = 29.5
so then I can use the equation I= (V-E)/R = (130-29.5)/9.42 = 10.7 A
Thank you!
 
  • #4
AlephZero
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That looks OK to me.
 

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