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Missing component to get the impendance ohmic

  1. Jul 19, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi, following graphic is given (attachment)

    2. Relevant equations
    -



    3. The attempt at a solution

    I tried it with the following equation:

    [tex] \omega iL-\frac i {\omega C} + X =R[/tex]

    but I don't get to the solution, is this the right attempt? If yes, I'll try later on again, have to go to the university right now.

    Thanks for the help :smile:
     

    Attached Files:

  2. jcsd
  3. Jul 19, 2011 #2
    what is the question?
     
  4. Jul 19, 2011 #3
    ah sorry, I was in a hurry. I shall find the missing component so that the impendance/resistance of the series connection is just ohmic
     
  5. Jul 19, 2011 #4

    ehild

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    Gold Member

    L and R are parallel. What is their resultant impedance?

    ehild
     
  6. Jul 19, 2011 #5
    Will that work?
    the unknown resistance is also the part of LR circuit
     
  7. Jul 19, 2011 #6
    The impendance is:

    [tex] \frac 1 Z = \frac 1 {i\omega L} - \frac {\omega C} {i} [/tex]

    Do I have to solve it afterwards with the equation

    [tex]R=X+Z[/tex]

    And find the X ?
     
  8. Jul 19, 2011 #7
    what is C?
     
  9. Jul 19, 2011 #8

    ehild

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    That box in the figure is not a capacitor but a resistor R connected parallel to the inductor. And yes, Z+X has to be real.

    ehild
     
  10. Jul 19, 2011 #9
    Sorry everyone, I'm totally asleep at the switch, having exams in a few days and things are getting confused. I'll try it lateron, have so solve some other things first.

    @ Cupid, I don't have any idea why I thought theres a capacity. maybe I should better look at the drawing :grumpy:

    thanks everyone
     
  11. Jul 19, 2011 #10
    I'm stuck, can't solve it.

    I tried to calculate Z, my Z is the parallel connection between R and L

    and I get for Z:

    [tex] \frac 1 Z = \frac 1 R + \frac {1} {i \omega L} [/tex]

    after some conversions I get to:

    [tex] Z= \frac {R\omega^{2}L+iR^{2}\omega L} {R^{2}+\omega L^{2}}[/tex]

    Now I want to get a pure ohmic resistance so

    [tex] Z+X=R[/tex]

    but I just can't solve it. The task says the missing component is a "normal" one, so I guess it should be a capacity, inductor etc. Though I tried to take one of these to test if I get a solution, I don't get to an answer. Can anyone help me:confused:
     
  12. Jul 19, 2011 #11

    gneill

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    Staff: Mentor

    Check your algebra. The ω in the term in the denominator should be squared, and the L in the real portion of the numerator should also be squared.

    Once you've sorted that out, divide the expression into two parts: the real part and the imaginary part. It's the imaginary part that you want to "take care of" with the new impedance X.
     
  13. Jul 19, 2011 #12
    Hope I haven't done any mistakes again:

    [tex] X=R- \frac {R\omega^{2}L^{2}} {R^{2}+\omega^{2}L^{2}} - \frac {iR^{2}\omega L} {R^{2}+\omega^{2}L^{2}}[/tex]

    so in order to get a pure ohmic term my X has to be:

    [tex] X= \frac {iR^{2}\omega L} {R^{2}+\omega^{2}L^{2}}[/tex]

    but what component is that? I don't get it :(
     
  14. Jul 19, 2011 #13

    gneill

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    If you let Z be the impedance of the parallel R and L components, then you've found that

    [tex] Z = \frac {R\omega^{2}L^{2}} {R^{2}+\omega^{2}L^{2}} + i\frac {R^{2}\omega L} {R^{2}+\omega^{2}L^{2}}[/tex]

    In order to cancel the imaginary term you need to put in series an impedance that is its negative, that is,

    [tex] X = - i\frac {R^{2}\omega L} {R^{2}+\omega^{2}L^{2}} [/tex]

    Now, what sort of component has an impedance that yields a negative imaginary term?
     
  15. Jul 19, 2011 #14
    I'd say capcity with:

    [tex] R_{C}=-\frac i {\omega C} [/tex]

    if that answer is right, I still don't understand why to use this one
     
  16. Jul 19, 2011 #15

    gneill

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    Yes, a capacitor gives a negative imaginary impedance. :smile:

    If you set the above expression for a capacitor's impedance equal to the X expression then you can solve for C, the value that the capacitor must have in order to cancel out the imaginary term of the impedance in Z at a given frequency ω.
     
    Last edited: Jul 19, 2011
  17. Jul 19, 2011 #16
    Ok, thank your very much
     
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