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Homework Help: AC Source/ Variable Resistance

  1. Sep 29, 2007 #1
    1. The problem statement, all variables and given/known data
    In the figure, a generator with an adjustable frequency of oscillation is connected to a variable resistance R, a capacitor of C = 2.50 μF, and an inductor of inductance L. The amplitude of the current produced in the circuit by the generator is at half-maximum level when the generator's frequency is 2.3 or 2.5 k Hz. What is L?
    2. Relevant equations
    1. [tex]I=\frac{\xi_{m}}{Z}cos(\omega_{d}t-\varphi)[/tex]

    2. [tex]Z=\sqrt{R_{v}^2+\chi^2}[/tex]

    3. [tex]\chi=L\omega-\frac{1}{C\omega}[/tex]

    4. [tex]\omega_{r}=\frac{1}{\sqrt{LC}}[/tex]

    5. [tex]\Delta\omega=\frac{R}{L}[/tex]

    6. [tex]\omega=2\pi\nu[/tex]

    Where [tex]\Delta\omega[/tex] is the spread of [tex]\omega[/tex] at .7 times the maximum current (at [tex]\omega_{r}[/tex])

    3. The attempt at a solution


    At the given frequencies we have


    The voltage amplitudes cancel and we have one equation and two unknowns.

    I was thinking about using equation 5 but that is only for the gap between the two frequencies for [tex]I_{m}=.7I_{res}[/tex] which isn't the case here.
    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 29, 2007 #2
  4. Sep 29, 2007 #3


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    I'm not seeing the picture. I think the link you used is wrong.
  5. Sep 29, 2007 #4
    Problem corrected. Sorry
  6. Sep 29, 2007 #5


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    As per your equations, at half amplitude you have:

    2R = net impedance

    so plugging in your two frequencies into this equation... you get two equations. You have two equations with 2 unknowns (L and R). You can solve for L.
  7. Sep 29, 2007 #6
    The equation I come up with at the end is [tex]L=\frac{1}{C^{2}\omega_{1}^{2}\omega_{2}^{2}}[/tex]. Is this right? I seem to be getting the wrong answer with this equation?

    I wish I had some kind of elf to do tedious algebra for me...
  8. Sep 30, 2007 #7


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    EDIT: wait: shouldn't it be [tex] L = \frac{1}{C\omega_{1}\omega_{2}}[/tex]
    Last edited: Sep 30, 2007
  9. Sep 30, 2007 #8
    Indeed it should. Thanks. That kind of mistake is typical of me...
  10. Sep 30, 2007 #9


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    oops ... you know that (at resonance), LC = 1/w^2 , with w~2*pi*2.3 KHz ...
    at the half-max current (hi_f and lo_f) , chi = (+/-) sqrt(3) R
    (... from #2, 4R^2 is mostly inductive at higher frequencies than resonance.)
    the algebra isn't very tedious when you eliminate the sqrt(3)R first
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