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Back emf

  1. Dec 25, 2011 #1
    Hi,

    I have a question about back emf

    So, my understanding of back emf ([itex]\epsilon_b=-L\frac{dI}{dt}[/itex])is that its the counter voltage that's induced by change in current, whether it be from a battery or magnetic field ([itex]\epsilon_0[/itex]). We can model this as an LR circuit where [itex]\epsilon_0 +\epsilon_b=IR[/itex]

    But when we add the back emf, don't we change the value of dI/dt? i.e. if we differentiated the equation we'd be left with

    [itex]\frac{d\epsilon_0}{dt}-L\frac{d^2I}{dt^2}=R \frac{dI}{dt}[/itex]

    And then, the time derivative of current would be offset by an additional 2nd derivative of current (multiplied by negative L). I'm assuming we calculated change in current as [itex]\frac{dI}{dt}=\frac{1}{R}\frac{d\epsilon_0}{dt}[/itex], hence the offset.

    Does this model then only hold for cases where current changes linearly/not at all? And if this is correct, how would we model it for such cases where current changes with respect to 2nd or higher derivatives of time?
     
    Last edited: Dec 25, 2011
  2. jcsd
  3. Dec 25, 2011 #2
    If the resistance in the circuit = 0 then the induced emf,e, must be equal to the applied emf, E,
    This means that e = E = -L.dI/dt
    This means that the current will increase at a constant rate of dI/dt = E/L
    If there is resistance in the circuit then the emf appearing across the L = E - Ir
    This means that L.dI/dt = E - Ir
     
  4. Dec 25, 2011 #3
    Thanks for the clarification. So the back emf and IR must add up to the applied emf.

    My confusion/question still remains though.
    That is, if we apply a back emf don't we change the rate at which current changes over time (if it has a nonzero second derivative)? If so, wouldn't this induce another back emf because the "first" back emf has induced a change in current?

    Would we then model this as a summaiton of some sort? i.e.
    [itex]\epsilon_b=-L(\frac{dI}{dt}-\frac{1}{R}\frac{d^2I}{dt^3}+\frac{1}{R^2}\frac{d^2I}{dt^3}-...)[/itex]
     
    Last edited: Dec 25, 2011
  5. Dec 26, 2011 #4
    I think your confusionis inthe phrase '.......if we apply a back emf.....'
    A back emf is not something that is applied.
    When a battery (emf) is connected to a circuit and the switch is closed the current begins to increase. AN Increasing current will result in an induced (back) emf as a result of any inductance in the circuit.
    The induced emf = -LdI/dt (it is not an applied emf,it is caused by the increasing current)
    If there is resistance in the circuit you are correct to state that Ir + induced emf = applied emf, E.
    E = IR + LdI/dt
    the solution to this equation is
    I = Io(1-exp-(Rt/L))
    After a long time the final (steady)current will be Io and the initial rate of increase of current will be = E/L
     
  6. Dec 27, 2011 #5
    Oh, okay. Yes, that makes sense.

    I feel sort of silly now. The backemf is generated differentially so it's just a simple diffeq to solve for it..

    Thanks!
     
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