LRC Circuit: Frequency of Oscillation determined by R

[jamesbiomed]

Homework Statement

How much resistance must be added to a pure LC circuit (L = 340 mH, C = 2200 pF) to change the oscillator's frequency by 0.10 percent?

Will the frequency be increased or decreased?

Homework Equations

I'm assuming critical damping: R^2=4L/C, w=((1/LC)-R^2/(4L^2))^1/2

With no R: w=(1/(LC))^1/2

The Attempt at a Solution

I've been at this for a few hours :/ I'm letting w stand in for frequency since w=2pif the percent change needed for f should be the same for w (I hope). I've tried several approaches:

First, I tried w=(1/LC)^1/2 to get w. Then multiplied w by .001 to get .1%. Then made an equation: w-((1/LC)-R^2/(4L^2))^1/2=.001w

That didn't work, but the answer was close to the given one.

I also tried simply setting .001w=(1/LC-R^2/2l^2)^1/2 and that answer was further off.

If I'm missing anything or you need information I'd be happy to give it I'm really enthusiastic to get this done. Thank you in advance!

 

[ehild]

The angular frequency of a damped oscillator is

[tex]\omega=\sqrt{\frac{1}{LC}-\left(\frac{R}{2L}\right)^2}[/tex]

Your first approach is correct, but it would be easier to write ω=K√(1/LC). How much is K? less then 1 or greater then 1? Then you can square both sides. Do it symbolically, do not evaluate. ehild

 

[jamesbiomed]

Hey echild. The problem was I wasn't squaring k, just plugging it in after the fact.

K is less than one, so the frequency should be decreased.

For anyone with same problem:
w' is .1 % w.

w'=k(1/LC)^1/2=> (w')^2=k^2/LC=>(w')^2=(.001)^2/(LC) so omega is the square root of that.

Then w-(w^2+(R/2L)^2)^1/2=w'. Solve for R:

R=(((w-w')^2-w^2)4L^2)^1/2. That solved it.

Thank you!

 

[ehild]

I meant that the new angular frequency is 0.999 ω₀, where ω₀=1/√(LC).

So
[tex]\omega=\sqrt{\frac{1}{LC}-\left(\frac{R}{2L}\right)^2}=0.999\sqrt{\frac{1}{LC}}[/tex]

Squaring both sides:[tex]\frac{1}{LC}-\left(\frac{R}{2L}\right)^2=0.998\frac{1}{LC}[/tex]

[tex]0.002\frac{1}{LC}=\left(\frac{R}{2L}\right)^2[/tex]

ehild

 

[Nickie]

I understand your frustration and the effort you have put into finding a solution. However, I would suggest approaching this problem using the formula for angular frequency, w = 1/sqrt(LC), where L is in henries and C is in farads. This will give you the natural frequency of the LC circuit without any resistance.

To find the change in frequency by 0.10%, you can use the formula w' = w + 0.001w, where w' is the new frequency and w is the natural frequency. Then, you can plug in the given values of L and C to solve for w' and get the new frequency.

To find the amount of resistance needed, you can use the formula R = sqrt(4L/w'^2 - 4L/w^2). This will give you the resistance needed to change the frequency by 0.10%.

As for whether the frequency will be increased or decreased, it will depend on the initial frequency of the circuit. If the initial frequency is higher than the natural frequency, then the added resistance will decrease the frequency. If the initial frequency is lower than the natural frequency, then the added resistance will increase the frequency.

I hope this helps and good luck with your homework!