Homework Help: AC Source/ Variable Resistance

1. Sep 29, 2007

americanforest

1. The problem statement, all variables and given/known data
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?
2. Relevant 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 $$\omega_{r}$$)

3. 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.

Last edited: Sep 29, 2007
2. Sep 29, 2007

americanforest

anybody?

3. Sep 29, 2007

learningphysics

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

4. Sep 29, 2007

americanforest

Problem corrected. Sorry

5. Sep 29, 2007

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.

6. Sep 29, 2007

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...

7. Sep 30, 2007

learningphysics

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

Last edited: Sep 30, 2007
8. Sep 30, 2007

americanforest

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

9. Sep 30, 2007

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