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

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Homework Help Overview

The problem involves a motor operating at different speeds and requires calculations related to current, resistance, and back emf. The subject area includes electrical circuits and motor theory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the current drawn by the motor at one-third normal speed but expresses uncertainty about the approach. Some participants discuss the relationship between back emf and speed, suggesting that back emf is proportional to the speed of the motor.

Discussion Status

Participants are exploring the implications of back emf on current draw at reduced speed. One participant proposes a calculation based on the back emf at one-third speed, while another participant indicates that this approach appears reasonable. There is no explicit consensus on the final answer, but a productive direction has been provided.

Contextual Notes

The original poster has successfully calculated the resistance of the armature coil and the back emf at normal speed, but lacks information on how to apply these results to find the current at one-third speed.

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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.
 
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In (c), the back emf is proportional to the speed.
 
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!
 
That looks OK to me.
 

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