Find the voltage across a capacitor in a LRC circuit

AI Thread Summary
In a series LRC circuit, the goal is to find the voltage ratio across the capacitor relative to the input voltage from an AC source, expressed in terms of inductance (L), resistance (R), capacitance (C), and angular frequency (omega). The initial equation provided for voltage output was incorrect, but after discussions, a new expression for the voltage ratio was derived using complex impedance. The correct formula for the voltage across the capacitor involves understanding the total impedance as R + i(wL - 1/wC). The conversation emphasized the importance of calculating the resultant impedance by combining the real and imaginary components. Overall, the participants clarified the approach to solving the problem and improved their understanding of complex impedance in LRC circuits.
splatcat
Messages
8
Reaction score
0
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
 
Last edited by a moderator:
Physics news on Phys.org
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 :)
 
Back
Top