AC Source/ Variable Resistance

In summary, the conversation discusses finding the value of inductance (L) in a circuit with a generator, variable resistance, capacitor, and inductor, where the amplitude of current is at half-maximum at two given frequencies. Using equations related to impedance and resonance, the value of L can be solved for by setting up equations with known values and using algebraic manipulation. The final equation for L is 1/(C*omega1*omega2), where omega1 and omega2 are the two frequencies at half-maximum current.
  • #1
americanforest
223
0

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. [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])

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.
 
Last edited:
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  • #2
anybody?
 
  • #3
americanforest said:
anybody?

I'm not seeing the picture. I think the link you used is wrong.
 
  • #4
learningphysics said:
I'm not seeing the picture. I think the link you used is wrong.

Problem corrected. Sorry
 
  • #5
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
The equation I come up with at the end is [tex]L=\frac{1}{C^{2}\omega_{1}^{2}\omega_{2}^{2}}[/tex]. 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
americanforest said:
The equation I come up with at the end is [tex]L=\frac{1}{C^{2}\omega_{1}^{2}\omega_{2}^{2}}[/tex]. 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 [tex] L = \frac{1}{C\omega_{1}\omega_{2}}[/tex]
 
Last edited:
  • #8
learningphysics said:
EDIT: wait: shouldn't it be [tex] L = \frac{1}{C\omega_{1}\omega_{2}}[/tex]

Indeed it should. Thanks. That kind of mistake is typical of me...
 
  • #9
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
 

What is an AC source and variable resistance?

An AC source is an alternating current source that provides an electrical current that periodically changes direction. Variable resistance refers to a component in a circuit that can change its resistance, allowing for the control of current flow.

How does an AC source and variable resistance work?

An AC source works by converting mechanical energy into electrical energy through a rotating coil within a magnetic field. Variable resistance works by changing the amount of resistance in a circuit, which in turn affects the flow of current.

What are the applications of an AC source and variable resistance?

An AC source is commonly used in household outlets, power grids, and electronic devices. Variable resistance is often used in dimmer switches, temperature control systems, and audio equipment.

What are the advantages of using an AC source and variable resistance?

The use of an AC source allows for the efficient transmission of electricity over long distances, and the use of variable resistance provides control over the amount of current flowing through a circuit, allowing for flexibility and precision in electronic devices.

What are the potential issues with an AC source and variable resistance?

Some potential issues with an AC source include power outages due to disruptions in the power grid, and potential safety hazards if not used properly. Variable resistance can also cause issues if not calibrated correctly, leading to inaccurate control of current flow.

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