How Does Resonance Frequency Calculation Work in LC Circuits?

In summary: R_e}{R_p}\frac{1}{1+jx}In summary, the conversation discusses trying to figure out resonance in a circuit that has not been taught yet. The circuit in question is Rp || L || C with a sine source and load connected. The goal is to measure the gain, with Vi being across the sine source and Vo across the load resistor. The equation for gain is given as H = Vo/Vi. The conversation then discusses taking the derivative with respect to j*w and solving for j*w, but realizing that the desired solution is actually for w. The correct formula for the circuit's resonant frequency is w_o = 1/sqrt(L*C). The conversation ends with a suggestion to try taking
  • #1
Little Dump
19
0
This isn't really a homework question. I'm just trying to figure things out for myself about resonance in a circuit that they haven't taught us yet.

Say you have a circuit

Rp || L || C and you have a sine source and some load connected to it call it Re and you want to mesaure the gain. Vi being across the sine source, Vo being across the load resistor.

So you get somehing like this:

[tex]
H = \frac{V_o}{V_i} = \frac{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p}{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p + j*w*L*Rp}
[/tex]

So you take the derivative with respect to j*w and get something even uglier then that.

Set it equal 0 and solve for j*w and you get

[tex]
j*w = \frac{1}{\sqrt{L*C}}
[/tex]

But don't you actually want to solve for w? Not j*w? In which case if you solve for w you get a complex resonant frequency. Which to me makes no sense. And is wrong because I know the formula for the circuit in question is:

[tex]
w_o = \frac{1}{\sqrt{L*C}}
[/tex]

What did I over look?
 
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  • #2
Little Dump said:
This isn't really a homework question. I'm just trying to figure things out for myself about resonance in a circuit that they haven't taught us yet.

Say you have a circuit

Rp || L || C and you have a sine source and some load connected to it call it Re and you want to mesaure the gain. Vi being across the sine source, Vo being across the load resistor.

So you get somehing like this:

[tex]
H = \frac{V_o}{V_i} = \frac{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p}{R_e*R_p + R_e*j*w*L - R_e*w^2*L*C*R_p + j*w*L*Rp}
[/tex]

So you take the derivative with respect to j*w and get something even uglier then that.

Set it equal 0 and solve for j*w and you get

[tex]
j*w = \frac{1}{\sqrt{L*C}}
[/tex]

But don't you actually want to solve for w? Not j*w? In which case if you solve for w you get a complex resonant frequency. Which to me makes no sense. And is wrong because I know the formula for the circuit in question is:

[tex]
w_o = \frac{1}{\sqrt{L*C}}
[/tex]

What did I over look?

Did you take the derivative with respect to jw properly... wherever
you have w^2, rewrite it as -(jw)^2. Also, I'd then replace all the jw, with a variable x, and take the derivative with respect to x. This just makes things a little easier to look at...
 
  • #3
That's exactly how I did it. I got maple to do it for me, and then I did it by hand. Same thing.
 
  • #4
Try again with
[tex]H=\frac{|V_0|}{|V_i|}[/tex]
 

Related to How Does Resonance Frequency Calculation Work in LC Circuits?

1. What is resonance in an electric circuit?

Resonance in an electric circuit occurs when the inductive and capacitive reactances are equal, resulting in the maximum flow of current through the circuit.

2. How does resonance affect the behavior of an electric circuit?

Resonance causes the circuit to become highly efficient, allowing for a greater transfer of energy. It can also cause the voltage and current in the circuit to become out of phase, which can lead to power loss.

3. What factors can affect the resonance frequency of an electric circuit?

The resonance frequency of an electric circuit can be affected by the values of inductance and capacitance, the type of components used, and any external factors like temperature or interference.

4. How is resonance used in practical applications?

Resonance is used in a variety of practical applications, such as in radio and television broadcasting, where it allows for efficient transfer of signals. It is also used in electrical filters and oscillators.

5. What are the potential dangers of resonance in an electric circuit?

If not properly controlled, resonance can lead to overheating of components and damage to the circuit. It can also cause interference with other nearby circuits and devices.

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