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Homework Help: Maximum Amplitude, 50hx vs 60hz generator

  1. Jan 30, 2008 #1
    1. The problem statement, all variables and given/known data

    We have an AC generator with output current I. I, as a function of time, is a sine wave whose frequency, f, is the same as the rotational frequency of the generator. So,

    I(t) = I[tex]_{0}[/tex]sin(2πft), where I[tex]_{0}[/tex] is the max. amplitude of the current.

    In terms of magnetic flux, [tex]\Phi[/tex][tex]_{B}[/tex]:

    I = - (Constant * Change in [tex]\Phi[/tex][tex]_{B}[/tex]) / time

    When the generator revolves at 50 revs/second, the maximum amplitude I[tex]_{0}[/tex] = 1000 A. If you increase the spead of the generator to 60 revs/sec, what is the maximum amplitude I[tex]_{0}[/tex]?

    2. Relevant equations

    Given above

    3. The attempt at a solution

    This question really stumped me. It's difficult to bring in additional equations since we don't know more about the generator -- it's obviously a concept question. I suspect that the maximum amplitude will increase proportionally.

    We know that f(old) = 50 Hz, since it will be the same as the rotatation frequency. Thus, the period for this sine curve will be 1/50 second, as the amplitude goes from maximum to maximum. When f(new) = 60 Hz, the period will condense to 1/60 second.

    Thinking about flux, there is no change in magnetic field (B) -- though the [tex]\Theta[/tex] between B and the area of the coil will be changing faster as rotations speed up. So change in fluz would be greater over a given unit of time (???). Thus, from the second equation, we know that I must also increase (???).

    At this point, I simply guess that the change will be proportional -- since rotation is increasing in speed by 20%, so will maximum amplitude ??????????????

    Thanks for your help!
    Last edited: Jan 30, 2008
  2. jcsd
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