1. Not finding help here? Sign up for a free 30min 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!

Purcell 8.9- Q Value

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    This is to look at the effect of damping on the frequency of a series RLC circuit. Let W=1/(sqrt(LC)) be the frequency of an undamped circuit. Suppose enough resistance is added to bring Q down from infinity to 1000. By what percentage is the frequency, ω, thereby shifted from W?


    2. Relevant equations



    3. The attempt at a solution
    So, I know that ω=sqrt(W^2-(R/2L)^2). I can pull out the W, and say ω=W[1-(R/(2LW))^2]^(1/2). I think I can justify that fraction being small, because let's just assume that the R I added was small (Q is 1000, which is quite large, so that's a fair assumption). Then I'll use a binomial expansion to say that ω≈W[1-(R^2/(8*L^2*W^2)].

    But now I'm stuck: I know that Q=ω*L/R=1000. So, W[1-(R^2/(8*L^2*W^2)]*L/R=1000, but who do I solve this as a percentage of ω?

    Or have I made a mistake?
     
  2. jcsd
  3. Nov 12, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Using just the symbols ω and W, how would you express the percentage change of ω from W?

    Then see if you can use your expression ω≈W[1-(R^2/(8*L^2*W^2)] to express this percentage change in terms of R, L, and W. You'll then need to relate this to Q.
     
  4. Nov 12, 2012 #3
    The percentage change would be ω/W, yes?

    Then, the percentage difference would really just be 1-(R^2/(8L^2W^2)... but R^2/(L^2*W^2) is actually just 1/Q^2?

    And then I plug in 1000 for that Q?
     
  5. Nov 12, 2012 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Note quite. The percent change in a quantity is the amount the quantity changes divided by the initial amount (and then expressed as a percent).
     
  6. Nov 12, 2012 #5
    Whoops, my mistake! Its (Omega-W)/(Omega)?

    I end up getting -1.25e-7, so it decreases by 1.25e-5 percent... very small!
     
  7. Nov 12, 2012 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    That's what I get too. Yes, it is small. I guess a Q of 1000 is practically the same as a Q of infinity (unless you're dealing with very accurate measurements)!
     
  8. Nov 12, 2012 #7
    Thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook