1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Current drawn by a motor operating at half speed

  1. Apr 8, 2014 #1
    Hello,

    i'm unsure of how to complete the final part of this question;

    1. The problem statement, all variables and given/known data

    A motor is designed to operate at 118V and draws a current of 12A when it first starts up. At it's normal operating speed, the motor draws a current of 2A. Compute:

    a) the resistance of the armature of the coil.

    b) the back emf developed at normal speed.

    c) the current drawn by the motor at half the normal speed.

    2. Relevant equations
    3. The attempt at a solution

    I have completed parts a and b.

    a) [itex]V = IR[/itex]

    [itex]R = \frac{V}{I} = \frac{118}{12} = 9.83 \Omega[/itex]

    b) [itex]2A[/itex] flow through the armature with resistance [itex]9.83 \Omega[/itex].

    [itex]V = IR = 19.66V[/itex]

    Supply voltage = back emf + voltage across armature

    back emf = 118 - 19.66 = 98.34V

    But how do I figure out c?

    Thanks!
     
  2. jcsd
  3. Apr 8, 2014 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Perhaps start by working out the back emf at half the normal speed?
     
  4. Apr 8, 2014 #3
    I can't find any material in my text book that deals with the relationship speed has with other known quantities.

    Do you know of any places I should look?
     
  5. Apr 9, 2014 #4
    Bump, I still haven't found anything.

    If the load stays constant, then decreasing the voltage will decrease the current in the armature and hence the speed of the motor, but I don't know how to quantify these ideas.
     
  6. Apr 9, 2014 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Isn't it important to know why it's at half normal speed? I.e. is it because the applied voltage has been lowered or because the load has been increased?
    If you knew which, I think you could use consideration of power to obtain an answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted