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Homework Help: Help understanding 'back emf'

  1. May 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A series dc motor:
    A dc motor with its rotor and field coils connected in series has an internal resistance of 2.00 Ω. When running at full load on a 120-V line, it draws a current of 4.00 A.

    a) What is the emf in the rotor?
    b) What is the power delivered to the motor?
    c) What is the rate of dissipation of energy in the resistance of the motor?
    d) What is the mechanical power developed?
    e)What is the efficiency of the motor?
    f)What happens if the machine that the motor is driving jams and the rotor suddenly stops turning?
    2. Relevant equations
    Vab = E + Ir,
    P = I^2r
    3. The attempt at a solution
    This is a worked example in my textbook so the answers are all there but I'm finding it challenging to understand what is actually going on with some of the equations.

    what exactly is the back emf and why am I adding it to Ir?

    why does the current become V_ab/r only when the back emf goes to zero? what would be in the equation
    I=v_ab/r if the back emf was still there?

    would appreciate help with any of these questions and maybe an explanation of what is actually going on here. thanks
    Last edited: May 19, 2015
  2. jcsd
  3. May 19, 2015 #2


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    The motor rotates. As a result, there are wires cutting through a magnetic field. This generates a voltage, according to Faraday's Law, and this is opposite to the applied voltage. The current that flow depends on the net voltage.
    There are many analogues to this relationship. You can think of the back EMF as like inertia in kinetics, and the resistance like friction.
    In chemistry, there's Le Chatelier's principle.
  4. May 20, 2015 #3
    thank you. how come the emf and Ir are added together in this case and not subtracted if they are opposite? then when the motor stalls the total voltage of 120v is used in the equation? was it 8 before?
  5. May 20, 2015 #4


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    No, I said it opposes the applied voltage. Ir is then what's left after subtracting the back EMF from the applied EMF.
    Was what 8 before?
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