(adsbygoogle = window.adsbygoogle || []).push({}); 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. [tex]I=\frac{\xi_{m}}{Z}cos(\omega_{d}t-\varphi)[/tex]

2. [tex]Z=\sqrt{R_{v}^2+\chi^2}[/tex]

3. [tex]\chi=L\omega-\frac{1}{C\omega}[/tex]

4. [tex]\omega_{r}=\frac{1}{\sqrt{LC}}[/tex]

5. [tex]\Delta\omega=\frac{R}{L}[/tex]

6. [tex]\omega=2\pi\nu[/tex]

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

3. The attempt at a solution

[tex]I_{res}=\frac{\xi_{m}}{R}[/tex]

At the given frequencies we have

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

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 [tex]I_{m}=.7I_{res}[/tex] which isn't the case here.

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# Homework Help: AC Source/ Variable Resistance

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