AC Source/ Variable Resistance

[americanforest]

Homework Statement

http://www.imagehosting.com/out.php/i1196840_3326.jpg
In the figure, a generator with an adjustable frequency of oscillation is connected to a variable resistance R, a capacitor of C = 2.50 μF, and an inductor of inductance L. The amplitude of the current produced in the circuit by the generator is at half-maximum level when the generator's frequency is 2.3 or 2.5 k Hz. What is L?

Homework Equations

1. I=\frac{\xi_{m}}{Z}cos(\omega_{d}t-\varphi)

2. Z=\sqrt{R_{v}^2+\chi^2}

3. \chi=L\omega-\frac{1}{C\omega}

4. \omega_{r}=\frac{1}{\sqrt{LC}}

5. \Delta\omega=\frac{R}{L}

6. \omega=2\pi\nu

Where \Delta\omega is the spread of \omega at .7 times the maximum current (at \omega_{r})

The Attempt at a Solution

I_{res}=\frac{\xi_{m}}{R}

At the given frequencies we have

I=\frac{I_{res}}{2}=\frac{\xi_{m}}{2R}=\frac{\xi_{m}}{\sqrt{R_{v}^2+\chi_{1,2}^2}}

The voltage amplitudes cancel and we have one equation and two unknowns.

I was thinking about using equation 5 but that is only for the gap between the two frequencies for I_{m}=.7I_{res} which isn't the case here.

 

[americanforest]

anybody?

 

[learningphysics]

anybody?

I'm not seeing the picture. I think the link you used is wrong.

 

[americanforest]

I'm not seeing the picture. I think the link you used is wrong.

Problem corrected. Sorry

 

[learningphysics]

As per your equations, at half amplitude you have:

2R = net impedance

so plugging in your two frequencies into this equation... you get two equations. You have two equations with 2 unknowns (L and R). You can solve for L.

 

[americanforest]

The equation I come up with at the end is L=\frac{1}{C^{2}\omega_{1}^{2}\omega_{2}^{2}}. Is this right? I seem to be getting the wrong answer with this equation?

I wish I had some kind of elf to do tedious algebra for me...

 

[learningphysics]

The equation I come up with at the end is L=\frac{1}{C^{2}\omega_{1}^{2}\omega_{2}^{2}}. Is this right? I seem to be getting the wrong answer with this equation?

I wish I had some kind of elf to do tedious algebra for me...

EDIT: wait: shouldn't it be L = \frac{1}{C\omega_{1}\omega_{2}}

 

[americanforest]

EDIT: wait: shouldn't it be L = \frac{1}{C\omega_{1}\omega_{2}}

Indeed it should. Thanks. That kind of mistake is typical of me...

 

[lightgrav]

oops ... you know that (at resonance), LC = 1/w^2 , with w~2*pi*2.3 KHz ...
at the half-max current (hi_f and lo_f) , chi = (+/-) sqrt(3) R
(... from #2, 4R^2 is mostly inductive at higher frequencies than resonance.)
the algebra isn't very tedious when you eliminate the sqrt(3)R first