Homework Help: Find the voltage across a capacitor in a LRC circuit

1. Aug 2, 2010

splatcat

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))

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Any help/ advice would be greatly appreciated :) thanks

Stu

Last edited by a moderator: Aug 2, 2010
2. Aug 2, 2010

Staff: Mentor

Welcome to the PF.

Do you know how to express the complex impedance of each element in this circuit?

3. Aug 3, 2010

splatcat

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) )

4. Aug 3, 2010

ehild

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

ehild

5. Aug 3, 2010

ehild

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

ehild

6. Aug 3, 2010

splatcat

(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

7. Aug 3, 2010

ehild

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

8. Aug 3, 2010

splatcat

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