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Basic Generator problem

  1. Mar 10, 2004 #1
    The question is:
    A simple generator has a 655 loop square coil 19.7cm on a side. How fast must it turn in a .490 T field to produce a 110V peak output? give your answer in radians/second

    I used V=NBAwsin(wt)
    where V = 110, N=655, B=.490, A=.197*.197, w simplifies to 2*pi*f, wt simplifies to 2*pi
    so plugging in, 110=655*.490*.197*.197*2*3.14*f*sin(6.28)

    i get f = 12.85 rev/s * 360/57.3 = 80.76 rad/s

    but this is wrong..

    anyone know where i went wrong?
    places i suspect are: wrong formula, and i am using the sin(6.28) incorrectly..

    any hints?
  2. jcsd
  3. Mar 10, 2004 #2
    Well, sin(2*pi) is equal to zero using trig circles. There is one problem.
  4. Mar 10, 2004 #3
    Also the peak emf is E(o) not simply E so you can use the following equation:

    E(o)* sin(wt) = NAB w sin(wt)= E

    This simplifies to:

    E(o) = NAB*w

    Solve for w.
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