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Confusion over inductors and cemf

Discussion in 'Electronic Basics' started by James W, Dec 5, 2003.

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  1. James W

    James W Guest

    Consider a simple inductive cicuit with a 1v(p-to-p) AC source at
    1Hz,and an inductor with Z=1ohm.

    The inductors value would be 1ohm=2*pi*1Hz*L, so L= 1/2pi

    Looking at the standard Voltage and Current drawings, we see current
    lagging voltage by 90degrees.

    Here's my problem. At 90 degrees, the applied voltage is 1volt. The
    current is zero. di/dt is 1, and we know that the cemf V=(di/dt)L=1/2p
    Volts.

    So... why if applied V is 1v and cemf is less than 1v do we have a
    current of zero?

    I've got the sneaky feeling that I'm trying to add apples and oranges.

    My assumption was that the reason the current lags the voltage is the
    back emf is being added to ( subtracted from ) the applied voltage to
    give us the instantaneous voltage that then drives the instantenous
    current, but perhaps this whole theory is wrong.

    HELP!!
     
  2. ^^^^^^^^^^
    Why do you think di/dt is 1? Its 2pi.
     
  3. James W

    James W Guest

    Yep.. that's where I was going wrong. Thanks
    - jim
     
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