# AC Source/ Variable Resistance

1. Sep 29, 2007

### americanforest

1. The problem statement, all variables and given/known data

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