Find the voltage across a capacitor in a LRC circuit

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Homework Help Overview

The discussion revolves around finding an equation that expresses the ratio of voltage across a capacitor to the input voltage in a series LRC circuit, as a function of inductance (L), resistance (R), capacitance (C), and angular frequency (omega).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between voltage and impedance in the circuit, with attempts to express the voltage ratio using complex impedance. Questions arise regarding the correct formulation and understanding of the circuit's behavior.

Discussion Status

Some participants have made progress in understanding complex impedance and have proposed new equations. There is ongoing clarification regarding the correct interpretation of the voltage across the capacitor and how it relates to the overall impedance of the circuit. Multiple interpretations of the problem are being explored.

Contextual Notes

Participants are working under the constraints of not having access to solution keys and are seeking guidance on their reasoning and calculations. There is a focus on ensuring the correct application of concepts related to impedance and voltage ratios.

splatcat
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thank you for any help in advance.

The question is; In a series LRC circuit, find an equation that shows the ratio of Voltage out [the voltage measured over the capacitor] over the voltage in from an A/C source as a function of L R C and omega

Starting with this equation:

Vs = sqrt(VR2 + (VL - VC)2)



The attempt at a solution, I have no idea how close/right this is as I do not have the solutions,

Vout / Vs = 1/(C*omega*sqrt(R2 + (omega*L - 1/C*omega)2))

[/b]

Any help/ advice would be greatly appreciated :) thanks

Stu
 
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splatcat said:
thank you for any help in advance.

The question is; In a series LRC circuit, find an equation that shows the ratio of Voltage out [the voltage measured over the capacitor] over the voltage in from an A/C source as a function of L R C and omega

Starting with this equation:

Vs = sqrt(VR2 + (VL - VC)2)



The attempt at a solution, I have no idea how close/right this is as I do not have the solutions,

Vout / Vs = 1/(C*omega*sqrt(R2 + (omega*L - 1/C*omega)2))

[/b]

Any help/ advice would be greatly appreciated :) thanks

Stu

Welcome to the PF.

Do you know how to express the complex impedance of each element in this circuit?
 
thank you for your response, your question helped direct my reading. I think I have a better idea of what I am doing now. I have come to a new solution. Is it any closer?
I have some understanding of complex impedance now.


w = omega
Vout/Vin = (R + iLw) / (R + i(wL + 1/wC) )
 
splatcat said:
thank you for any help in advance.

The question is; In a series LRC circuit, find an equation that shows the ratio of Voltage out [the voltage measured over the capacitor] over the voltage in from an A/C source as a function of L R C and omega

Starting with this equation:

Vs = sqrt(VR2 + (VL - VC)2)
The attempt at a solution, I have no idea how close/right this is as I do not have the solutions,

Vout / Vs = 1/(C*omega*sqrt(R2 + (omega*L - 1/C*omega)2))Stu

That is the correct solution if the ratio of the magnitudes was the question.

ehild
 
splatcat said:
thank you for your response, your question helped direct my reading. I think I have a better idea of what I am doing now. I have come to a new solution. Is it any closer?
I have some understanding of complex impedance now.


w = omega
Vout/Vin = (R + iLw) / (R + i(wL + 1/wC) )

Wrong. The voltage across the capacitor was the question. And the complex impedance of the whole circuit is R + i(wL - 1/wC).

ehild
 
(Voltage amplitudes was the quesion :) ) Sorry I did mean to put a minus sign in there instead of a + sign. However does the system not work like a potential divider? Only resistances are impedances?

Stu
 
splatcat said:
(Voltage amplitudes was the quesion :) ) Sorry I did mean to put a minus sign in there instead of a + sign. However does the system not work like a potential divider? Only resistances are impedances?

Stu

Yes, but you need add up the square of the real and imaginary parts, and then take the square root to get the resultant impedance, as you did in your first post.

ehild
 
Thank you very much for your help. I am pretty sure I understand where I have gone wrong :)
 

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