RLC circuit and resonance frequency

In summary, the conversation discusses finding an equation for the resonance frequency and determining the values of R, L, and C given a resonance frequency of 10 Hz. The conversation concludes that the problem is not uniquely solvable and suggests picking values for R and C or using a suggested value for L to solve for the other variables.
  • #1
tennisguy383
4
0

Homework Statement


b735629eaadf7aa02eacc7f0ddf38c3b.png

Find an equation for the resonance frequency. Then find R L and C given that the resonance freq is 10 Hz.

Homework Equations


Wrote below


The Attempt at a Solution


y(jw) = (1 / R + jwl) + jwC is the basic equation.. after simplifying and setting imaginary part to 0 and changing w to f I get
RF = 1/2pi * ((L-R^2C)/(L^2*C))^1/2

I'm pretty sure that part is correct.. how do I find the values for R L and C using this now?
 
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  • #2
I don't believe R, L and C are uniquely derivable from just the resonant freq.

For example, 1/2π√(LC) = 10Hz with R = 0 is one solution. Pick any L or C, then the other is determined.

To include a finite R, adjust L or C slightly.

Dumb problem.
 
  • #3
It's a rather vague problem in its given form, but in a real life situation you might be required to design a circuit where one or more of the parts happen to be "given" because they happen to be in stock or otherwise the only ones available. For example, you might have a 10mH coil in hand, and being a real-life coil would have an inherent winding resistance. That would start you off with L and R as givens.
 
  • #4
rude man said:
I don't believe R, L and C are uniquely derivable from just the resonant freq.

For example, 1/2π√(LC) = 10Hz with R = 0 is one solution. Pick any L or C, then the other is determined.

To include a finite R, adjust L or C slightly.

Dumb problem.

I get what your saying, if we assume that the resonance frequency is given with the following formula

f_o= sqrt(1/LC).

We are asked to find the values of L and C for a resonance frequency of 10 Hz. The solution is not unique as we are given two unknowns and there is one known. For instance 0.1H and 0.1F for the inductor and capacitor respectively satisfy the condition for 10Hz. The problem does not say whether R needs to be finite or not so I'm not quite sure..

b65e842d3883096cd8ed1dd42034e77f.png

That's my expression for the resonance frequency.. So do I just pick a value for C, say 0.1F and then solve for L, and say R is 0?

And to include a finite R, how do I adjust L and C. For example in the basic expression above with 0.1H and 0.1F satisfying the condition for 10Hz, how do you determine what R is?
 
  • #5
tennisguy383 said:
I get what your saying, if we assume that the resonance frequency is given with the following formula

f_o= sqrt(1/LC).

We are asked to find the values of L and C for a resonance frequency of 10 Hz. The solution is not unique as we are given two unknowns and there is one known. For instance 0.1H and 0.1F for the inductor and capacitor respectively satisfy the condition for 10Hz. The problem does not say whether R needs to be finite or not so I'm not quite sure..

b65e842d3883096cd8ed1dd42034e77f.png

That's my expression for the resonance frequency.. So do I just pick a value for C, say 0.1F and then solve for L, and say R is 0?
You can certainly do that.

And to include a finite R, how do I adjust L and C. For example in the basic expression above with 0.1H and 0.1F satisfying the condition for 10Hz, how do you determine what R is?

Pick any R and C and solve for L. Or, more realistically, as gneill suggests, pick his L, add a few R and compute C using your formula again.
 
  • #6
rude man said:
Pick any R and C and solve for L. Or, more realistically, as gneill suggests, pick his L, add a few R and compute C using your formula again.

Alright thanks for the help. Choosing R = 1 ohm and L = 0.1H I get 0.002F for C. The professor just sent out an email a bit ago saying that the solution is not unique as well so that clarifies the problem.
 

1. What is an RLC circuit?

An RLC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C). These three components are connected in series or parallel and can be used to filter and manipulate electrical signals.

2. How does an RLC circuit work?

An RLC circuit works by utilizing the properties of the resistor, inductor, and capacitor to control the flow of electrical current. The resistor limits the current, the inductor stores energy in the form of a magnetic field, and the capacitor stores energy in the form of an electric field.

3. What is resonance frequency in an RLC circuit?

Resonance frequency in an RLC circuit is the frequency at which the circuit naturally oscillates with the least amount of resistance. This is when the reactance of the inductor and capacitor cancel each other out, resulting in a peak in the current and a minimum in the impedance.

4. How is resonance frequency calculated in an RLC circuit?

The resonance frequency of an RLC circuit can be calculated using the formula fr = 1/(2π√(LC)), where fr is the resonance frequency, L is the inductance, and C is the capacitance.

5. What are some practical applications of RLC circuits and resonance frequency?

RLC circuits and resonance frequency are used in many practical applications, including radio and television receivers, high pass and low pass filters, and in electronic tuning circuits. They are also used in medical devices such as MRI machines and pacemakers.

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