# Purcell 8.9- Q Value

1. Nov 12, 2012

### StandardBasis

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. Nov 12, 2012

### TSny

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.

3. Nov 12, 2012

### StandardBasis

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?

4. Nov 12, 2012

### TSny

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).

5. Nov 12, 2012

### StandardBasis

Whoops, my mistake! Its (Omega-W)/(Omega)?

I end up getting -1.25e-7, so it decreases by 1.25e-5 percent... very small!

6. Nov 12, 2012

### TSny

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)!

7. Nov 12, 2012

### StandardBasis

Thank you very much!