1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

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

  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


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


    User Avatar
    Homework Helper

    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.

  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


    User Avatar

    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


    User Avatar

    Staff: Mentor

    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


    User Avatar

    Staff: Mentor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook