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Parallel RLC Circuit

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data
    2zflno5.png


    2. Relevant equations



    3. The attempt at a solution
    I found ω0 = 1/√LC when IR is at its maximum. (Purely resistive)

    I have a feeling that the last part requires an approximation and then an expansion which gives 2 values of ω.

    But the thing is, (1 + x)n ≈ 1 + nx + .....

    only when x << 1

    In this case it's L/RC << 1 which doesn't really fit..

    530jdy.png
     
  2. jcsd
  3. May 4, 2013 #2

    SammyS

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    Well, what did you get for ω1 and ω2 ?
     
  4. May 5, 2013 #3
    I can't do the expansion, because I can't find the small approximations in the square root!
     
  5. May 5, 2013 #4

    SammyS

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    You can find ω+ and ω- without an approximation.

    However, I do see that the instructions do say to find approximate values assuming RC/L >> 1 .

    Also notice that those values of ω are for (IR)2 = (I0)2/2 , so the square root goes away.

    If they had said R2C/L >> 1, then I could see that helping.


    Sorry. I can't be of more help.
     
  6. May 7, 2013 #5
    Hmmm it's alright, let's leave this question up for other takers.
     
  7. May 8, 2013 #6

    NascentOxygen

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    Given RC/L ≫1, then for R>1Ω it follows that 4R2C/L ≫1.

    Taking R>1Ω seems very reasonable for practical tuned circuits, I'd say.
     
  8. May 8, 2013 #7

    rude man

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    First: it's easier to use admittance and conductance rather than impedance and resistance when dealing with components connected in parallel. So for example for your circuit Y = G + jwC - j/wL where G = 1/R and Y = 1/Z.

    Then: What is V as a function of w, where V is the voltage across your network? V is complex, but take |V|^2 which isn't.

    Now, using |V|^2, what is I thru R? And I^2 thru R?
    Then form I0^2/I^2 where I = Y*V. The V's cancel, giving you what for this ratio?

    More when you get to this point.
     
  9. May 9, 2013 #8
    I have discussed this problem with my tutor, apparently the [RωC - R/(ωL)]2 = 1

    which literally stems from the definition I = I0/√(2)
    Then we get two values of ω.
     
  10. May 9, 2013 #9

    rude man

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    your tutor is right.
     
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