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Alternating current about resonance width

  1. Nov 1, 2006 #1
    Since the question quite long and contains many formulas,
    so i take a photo instead.

    my question is that for part b
    since [itex] Z= \sqrt{R^2+(X_L-X_C)^2} [/itex]

    can i subst [itex] X_L = (W_0+\triangle{\omega})L [/itex]

    and [itex] X_C = \frac{1}{(w_0+\triangle{\omega})L} [/itex]

    into the [itex] Z= \sqrt{R^2+(X_L-X_C)^2} [/itex]

    if it does, can u show me some of steps ?
    I can derive the result from the problem in part b

    By the way, what do the amplitude of current is half the resonance value?
    is it [itex] I=\frac{\omega}{2} [/itex] ?

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    Last edited: Nov 1, 2006
  2. jcsd
  3. Nov 2, 2006 #2

    Andrew Mason

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    I think you meant [itex] X_C = \frac{1}{(w_0+\triangle{\omega})C} [/itex]

    Yes. Work out the expression for Z by expanding:

    [tex]Z = \sqrt{R^2+\left((W_0+\Delta{\omega})L -\frac{1}{(w_0+\Delta{\omega})C} \right)^2}[/tex]

    and ignoring the higher order terms of [itex]\Delta\omega[/itex]
    I am not sure I understand your question here. The amplitude of the current is V/Z. Z is not a linear function of [itex]\omega[/itex]

    Last edited: Nov 2, 2006
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