1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Aug 2, 2010 #1
    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: Aug 2, 2010
  2. jcsd
  3. Aug 2, 2010 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    Do you know how to express the complex impedance of each element in this circuit?
     
  4. Aug 3, 2010 #3
    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) )
     
  5. Aug 3, 2010 #4

    ehild

    User Avatar
    Homework Helper

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

    ehild
     
  6. Aug 3, 2010 #5

    ehild

    User Avatar
    Homework Helper

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

    ehild
     
  7. Aug 3, 2010 #6
    (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
     
  8. Aug 3, 2010 #7

    ehild

    User Avatar
    Homework Helper

    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
     
  9. Aug 3, 2010 #8
    Thank you very much for your help. I am pretty sure I understand where I have gone wrong :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook