What is the Relationship Between Bandwidth and RLC Circuit Parameters?

[IBY]

Homework Statement

and

Homework Equations

I am trying to get from:

[tex]I_{max}=\sqrt{2}I=\frac{V}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}[/tex]

to:

[tex]\Delta\omega=\omega_1-\omega_2=\frac{R}{L}[/tex]

The Attempt at a Solution

From the equation above:
[tex]\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}=\frac{V}{\sqrt{2}I}[/tex]

Square the above:
[tex]R^2+(\omega L-\frac{1}{\omega C})^2=\frac{V^2}{2I^2}[/tex]

Subtract from both sides R squared and square root:
[tex]\omega L-\frac{1}{\omega C}=\sqrt{\frac{V^2}{2 I^2}-R^2}[/tex]

Now, I want to put it in the form of quadratic equation, so I multiply omega:
[tex]\omega^2 L-\frac{1}{C}=\omega \sqrt{\frac{V^2}{2 I^2}-R^2}[/tex]

Now I am stuck. It is not like I can turn V/I=R because that I there is root square mean current, and even if I did, I would end up with an imaginary term. The answer is not imaginary, nor complex.

 

[vela]

What do ω₁ and ω₂ represent?

 

[IBY]

They represent the two solutions of angular frequency when the current of the circuit is 1/sqrt(2) of the resonance current (Imax).

The current is resonant when:

[tex]\omega=\frac{1}{\sqrt{LC}}[/tex]

That is because it turns this:

[tex]I=\frac{V}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}[/tex]

into this:

[tex]I=\frac{V}{R}[/tex]

 

[IBY]

Okay, I looked further into it, and my mistake was to assume that the first equation was equal to Imax, so instead, I changed it like this:

[tex]\frac{I_{max}}{\sqrt{2}}=I=\frac{V}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}[/tex]

Doing the same process from above gives me:
[tex]\omega L-\frac{1}{\omega C}=\sqrt{\frac{V^2}{I^2}-R^2}[/tex]

But since I=Imax/sqrt(2), that gives me:
[tex]\omega L-\frac{1}{\omega C}=\sqrt{\frac{2 V^2}{I_{max}^2}-R^2}[/tex]

I can finally turn V/Imax into R:
[tex]\omega L-\frac{1}{\omega C}=\sqrt{2 R^2-R^2}=R[/tex]

I set up the quadratic polynomial:
[tex]\omega^2 L-\frac{1}{C}-\omega R=0[/tex]

And then the formula:
[tex]\omega_1=\frac{R+\sqrt{R^2+4 \frac{L}{C}}}{2L}[/tex]
and
[tex]\omega_2=\frac{R-\sqrt{R^2+4 \frac{L}{C}}}{2L}[/tex]

But now I have another problem. When I subtract both solutions, the R/2L term disappears and that leaves the term in the squareroots.

 

[vela]

I didn't actually work this out, but I can see how this works from the original equation for the impedance. You're looking at the wrong two roots. One frequency should satisfy

[tex]\omega L - \frac{1}{\omega C} = +R[/tex]

while the other satisfies

[tex]\omega L - \frac{1}{\omega C} = -R[/tex]

Somewhere along the way, you took a square root, and you need to consider both the positive and negative roots.

 

[IBY]

Oh right, thanks. One more problem, though, that leaves me with 4 solutions. The two of them is the one mentioned in my last post above, and two of them would be with the Rs in front shifted into negative.

 

[IBY]

Just to help you visualize the four solutions:

[tex]\omega_1=\frac{R+\sqrt{R^2+4 \frac{L}{C}}}{2L}[/tex]

[tex]\omega_2=\frac{-R+\sqrt{R^2+4 \frac{L}{C}}}{2L}[/tex]

[tex]\omega_3=\frac{R-\sqrt{R^2+4 \frac{L}{C}}}{2L}[/tex]

[tex]\omega_4=\frac{-R-\sqrt{R^2+4 \frac{L}{C}}}{2L}[/tex]

 

[vela]

The sign on two of those is negative. You can toss those solutions.

 

[IBY]

I see that if I subtract omega 1 with 2, that gives me the correct solution, but why do I toss out omega 3 and 4, though? Why wasn't the negative R tossed out? It seems completely arbitrary.

 

[vela]

In the expression for ω₃, is the radical less than or greater than R? Similarly, what's the sign of ω₄?

 

[IBY]

Oh, because in those cases, angular frequency would turn out negative, and there is no such thing. It would turn out negative because R^2 is being added by 4L/C, and since without the 4L/C, the square root term would be R, then that square root term must be greater than the R outside the square root. Thanks!