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Average back emf

  1. Feb 9, 2015 #1
    1. The problem statement, all variables and given/known data
    The coil in a loudspeaker has an inductance of L = 56uH (or 5.6 x 10^-5 H). To produce a sound frequency of 20 kHz, the current must oscillate between peak values of +2.2 A and -2.2 A in one half of a period. What average back emf is induced in the coil during this variation? How does this compare to the applied emf of 18V?

    2. Relevant equations
    T = 1/f

    emf = L(ΔI/Δt)

    L = N(ΔΦB/Δl)

    3. The attempt at a solution

    back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/Δt]

    My problem here is I do not know where I can obtain Δt. I am assuming that it has something to do with the frequency of 20kHz in one half period (1/2T).

    T = 1/f --> 1/2T = 1/f --> 2/f -- > T = 2/20000 = 1.0 x 10^-4 secs

    If this is the case then, by subbing, Δt = 1.0 x 10^-4 secs, into the above equation, I get:

    back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/1.0 x 10^-4 secs]
    = (5.6 x 10^-5)[4.4/1.0 x 10^-4 secs]
    = (5.6 x 10^-5)(44000)
    = 2.464 V

    Is this correct?
     
  2. jcsd
  3. Feb 9, 2015 #2

    DEvens

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    Gold Member

    You are nearly there. But remember what happens in one period. It says it goes from +2.2A to -2.2 in a half period. How many half periods in a whole period? How many whole periods per second to get a frequency of 20 KHz? So how many half periods per second?

    If I give you 100 glasses of water per second, then it takes me 1/100th of a second to give you a glass of water. But how long does it take me to give you a half glass of water?
     
  4. Feb 9, 2015 #3
    Okay, so if I am considering a whole period than my currents are actually +4.4 and - 4.4. Additionally, my period is actually just T = 1/f = 1/20000 = 5.0 x 10^-5.

    back emf = (5.6 x 10^-5)[+4.4 - (-4.4)/5.0 x 10^-5 secs]
    = (5.6 x 10^-5)[8.8/5.0 x 10^-5 secs]
    = (5.6 x 10^-5)(176000)
    = 9.856 V
     
  5. Feb 9, 2015 #4

    lightgrav

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    Homework Helper

    What is the maximum current? (2.2A) ... but how long does it take for the +2.2A to become -2.2A ?
     
  6. Feb 9, 2015 #5

    NascentOxygen

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    Staff: Mentor

    That would be correct if the current slides uniformly from one extreme to the other. But it's a sinewave (well, I presume it is) so di/dt is changing all the time.

    You could try using L.di/dt at the moment the sinewave crosses the time axis, because that's where di/dt is a maximum so it's where the voltage will peak, and it's simply the peak voltage that you're after. You can do calculus? See what you get for an answer using that approach.

    Once you've solved it this way, we'll look at another method. ;)

    EDIT: Hmmm, does the question really ask for the "average back emf"??
     
  7. Feb 9, 2015 #6

    lightgrav

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    Homework Helper

    Yes, the question does explicitly ask for the average emf during this half-cycle. (centered on the maximum emf)
    ... we know how V_RMS relates to V_average and V_max .
     
  8. Feb 10, 2015 #7

    NascentOxygen

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    Staff: Mentor

    This question is sufficiently ill-explained that, were it an exam question, I'd accept either of two analyses: either with the current ramping linearly from +2 to -2 (i.e., a triangular wave), or with the current following a sinusoid. It is just not clear to me which the examiner intends.

    Is there any detail you have left out that could decide what is intended? Have you examined a problem like this in class?
     
  9. Feb 10, 2015 #8

    NascentOxygen

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    Staff: Mentor

    Can you attach a sketch of the waveforms you are expecting?
     
  10. Feb 11, 2015 #9
    No I didn't leave anything out, the question is copied verbatim from the question sheet I was given.
     
  11. Feb 2, 2016 #10
    Was this answered correctly? 9.8 V?
     
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