Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Impedance of an RLC series circuit when in resonance

  1. Dec 10, 2009 #1


    User Avatar
    Gold Member

    I don't find my notes right now, I'll try to use my memory on this.
    According to my notes, the impedance of an RLC series circuit is given by [tex]\frac{1}{\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C}\right )^2 }}[/tex].
    So when in resonance, [tex]Z=\frac{1}{R}[/tex] instead of [tex]Z=R[/tex].

    Also if I recall well, for an RLC parallel circuit, [tex]Z=\frac{1}{\sqrt { \frac{1}{R^2} } +\left ( \omega C-\frac{1}{\omega L}\right )^2 }[/tex] or something close to this, meaning that in resonance Z=R instead of Z=1/R.
    Are my notes wrong? Or am I doing something wrong?
  2. jcsd
  3. Dec 10, 2009 #2


    User Avatar
    Science Advisor
    Gold Member

    Your notes have to be wrong because all your impedances are real. The reactance caused by the inductive and capacitive elements give rise to imaginary impedances. The wikipedia article gives the results but they are trivial to work out yourself too.

    Series: [tex]Z = R-i\omega L+\frac{i}{\omega C}[/tex]
    Parallel: [tex]\frac{1}{Z} = \frac{1}{R}+\frac{i}{\omega L}-i\omega C[/tex]

    where the time dependence is [tex]e^{-i\omega t}[/tex].
  4. Dec 10, 2009 #3


    User Avatar
    Homework Helper

    Actually, it is possible to define the impedance as a real quantity, the magnitude of the complex impedance,
    [tex]Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}[/tex]
    (for series circuits) This is sometimes done in introductory EM courses. It's not as powerful a concept as the complex impedance but it still allows you to do calculations.

    fluidistic, I would imagine you should know that impedance is a quantity analogous to resistance, and specifically, it has the same units as resistance, which should have told you that
    [tex]Z = \frac{1}{\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}}[/tex]
    couldn't be right.
  5. Dec 10, 2009 #4


    User Avatar
    Gold Member

    Ok thank you both.
    I'll have to check out my notes.
    Indeed I know that in resonance the circuit is purely resistive and the impedance have ohm's units. That's why I doubted about my notes.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook