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!

Width of Resonance in RLC-circuit

  1. Dec 16, 2016 #1
    1. The problem statement, all variables and given/known data
    In an undriven RLC-circuit, the characteristic time of the capacitor, ie the time taken for the amplitude of the capacitor voltage ## V_c ## to drop by a factor e, is ## T = 2L/R ##.
    We now have a RLC-circuit which is driven by a AC voltage of variable frequency ## \omega ## with amplitude ##V_0##. Show that for small R. the width of the resonance is approximately ##2/T = R/L##.The width of the frequency is defined as the angular frequency range for which the amplitude of ##V_c## is greater than ##1/\sqrt{2}## of its resonance value.

    2. Relevant equations
    Ohm's law: ##V = I Z##
    Impedance: ##X_c = 1/i\omega C ##, ## X_L = i\omega L ##, ##X_R = R##
    Resonance in LC-circuit: ## \omega_0^2 = 1/LC ##

    3. The attempt at a solution
    I have attached a photo of my attempt at a solution. The basic method is the following:

    1. By using the fact that the current through the capacitor must be the same as the total current supplied by the voltage source, I find that the amplitude of the voltage on the capacitor can be expressed as:
    ## |V_c| = \frac{V_0}{\omega C} \cdot \frac{1}{\sqrt{(\frac{R\omega}{L})^2 + \omega^2 - \omega_0^2}}##

    2. I find the resonance value to be: ##V_c(\omega_0) = \frac{V_0L}{R\omega_0^2} ##
    (this uses the assumption that R is small)

    3. I want to find the two values of ##\omega## which yield ##V_c(\omega) = \frac{V_0L}{R\omega_0^2} \cdot \frac{1}{\sqrt{2}} ##. The difference between these values should be the width we are looking for.
    This gives an equation with the solution ## \omega = \pm \sqrt{2} \omega_0##. Something must be wrong, because ##\omega## cannot be negative, and this doesn't produce the required result either.

    What have I done wrong? Many thanks for all help! :)
     

    Attached Files:

  2. jcsd
  3. Dec 16, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    How do you prove this 'fact' ? At resonance, the current through the resistor is ##V_0\over R## I seem to remember.
     
  4. Dec 16, 2016 #3
    Well, all components are in series. Thus, by Kirchhoff's current law (esentially conservation of fundamental charge) the same current must run through all components. This is the "total current supplied by the voltage source", which at resonance when the reactive parts of the impedance cancle is indeed ##V_0/R##.
     
  5. Dec 16, 2016 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    If you make the substitution ω←##\frac 1{\sqrt {LC}}## in your expression for ## |V_c|## what do you get for Vc at resonance?
     
  6. Dec 16, 2016 #5

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, my bad. Focused on the ##V_c## but O2 already pointed at that.
     
  7. Dec 16, 2016 #6
    Oh, there's a misstake, isn't it? I thought it would be ##\frac{V_0L}{R\omega_0^2} ## but it should be ##\frac{V_0L\omega_0}{R} ##, shouldn't it?
    But does that change anything at all? Isn't the basic equation still ## \sqrt{(\frac{R\omega_0}{R})^2 + \omega^2 - \omega_0^2} = \sqrt{2} \cdot \frac{R\omega_0}{R} ##
     
    Last edited: Dec 16, 2016
  8. Dec 16, 2016 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    Umm, is that an R in the numerator and an R in the denominator?? :wideeyed:

    What I intended earlier, exactly: :confused:

    If you make the substitutions ω←##\frac 1{\sqrt {LC}}## and ωo←##\frac 1{\sqrt {LC}}## in your expression for ## |V_c|## what do you get for Vc at resonance?
    Right, that's VC at resonance. And this can be written ##\frac {V_O}Q## where Q is the Q of the circuit.
     
  9. Dec 19, 2016 #8
    No sorry! That was a typo. It should be L in the denominator of course!

    What is Q in this context and how does it help us?
    Isn't the basic equation still ## \sqrt{(\frac{R\omega_0}{L})^2 + \omega^2 - \omega_0^2} = \sqrt{2} \cdot \frac{R\omega_0}{L} ## with soultion ##\omega = \sqrt{2} \omega_0 ## which cannot be right?
     
  10. Dec 19, 2016 #9

    NascentOxygen

    User Avatar

    Staff: Mentor

    Earlier in the maths, have you said that ##{(\omega\ – \omega_0)}^2\ =\ \omega^2 – \omega_0^2## ?
     
  11. Dec 19, 2016 #10
    No I haven't, but I seem to have completely missed the square! And a factor 2 in one of the lines...
     
  12. Dec 19, 2016 #11
    Okay, So I have tried reworking it. If I am not wrong, the equation I have to solve to find at which ##\omega## the amplitude is ##1/\sqrt{2}## of the maximum is:
    ## \sqrt{(\frac{R\omega_0}{L})^2 + (\omega^2 - \omega_0^2)^2} = \sqrt{2} \cdot \frac{R\omega_0}{L} ## . This is a rather ugly equation, which yields:
    ## \omega^2 = \omega_0^2 - \frac{R^2}{2L^2} \pm \sqrt{2\omega_0^4 - \omega^2 + \frac{R^4}{4L^4}} ##. This seems rather far away from proving ##\Delta \omega = 2L/R ##, especially because taking the square-root of that expression will give an algebraic jumble... Have I done something wrong?
     
  13. Dec 19, 2016 #12

    NascentOxygen

    User Avatar

    Staff: Mentor

    We'll go with your first equation in post #11. :smile:

    Your equation has (ω - ωo) so keep that together and try again to isolate that term on one side of the equation. Show your working here.

    BTW, I do obtain their final answer.

    https://www.physicsforums.com/attachments/holly-1756-gif.110502/
     
  14. Dec 22, 2016 #13

    Okay! So we have:
    ## \sqrt{(\frac{R\omega}{L})^2 + (\omega^2 - \omega_0^2)^2} = \sqrt{2} \cdot \frac{R\omega_0}{L} ## .
    (Note: My equation in #11 had a misstake, one of the ##\omega_0## is actually supposed to be ##\omega##... (as seen above) Is this a problem or did you calculate with it?)

    Thus:

    ##(\frac{R\omega}{L})^2 + (\omega^2 - \omega_0^2)^2 = 2 \cdot (\frac{R\omega_0}{L})^2 ##
    ##(\omega^2 - \omega_0^2)^2 + \frac{R}{L})^2 (\omega^2 - \omega_0^2) - \frac{R}{L})^2 \cdot \omega_0^2 = 0 ##
    here we can use the quadratic formula:
    ##\omega^2 - \omega_0^2 = \frac{(\frac{R}{L})^2}{-2} \pm \sqrt{(\frac{(\frac{R}{L})^2}{-2})^2 + (\frac{R}{L})^2 \cdot \omega_0} ##

    now I would like to express this as a single term, to then be able to express ##\omega## in some nice way, but I am stuck...



    That's good to hear! That means I only have to figure out how to get there! :smile:
     
  15. Dec 22, 2016 #14
    I think since we are interested in small range ##\Delta\omega=\omega-\omega_0<<\omega_0## of values around ##\omega_0##, the following approximations are valid ##\omega+\omega_0=2\omega_0+\Delta\omega \approx 2\omega_0##. Hence ##\omega^2-\omega_0^2=2\Delta\omega \omega_0##. Use that and start from equation at post 11, it doesn't matter if it is ##\omega## or ##\omega_0## in there because ##R\omega=R\omega_0+R\Delta\omega\approx R\omega_0## since R is small and ##\Delta\omega## is not very big.
     
    Last edited: Dec 23, 2016
  16. Dec 23, 2016 #15
    Thank you very much! I finally got it now
    ##(\frac{R\omega}{L})^2 + (\omega^2 - \omega_0^2)^2 = 2 \cdot (\frac{R\omega_0}{L})^2 ##
    ## \rightarrow (2\Delta\omega\omega_0)^2 = (\frac{R\omega_0}{L})^2##
    giving ##\Delta\omega = R/2L = 1/T ##, thus the total width of the resonance is ##2/T## as was to be shown. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Width of Resonance in RLC-circuit
  1. Resonant RLC Circuit (Replies: 1)

Loading...