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Parallel resonance LC circuit question?

  1. Sep 1, 2009 #1
    "At resonance frequency, the circuit impedance is maximum. It is resistive and its value is given L/Cr..."
    The book doesn't explain how impedance equals L/Cr, so I'm confused here. How is it derived?
  2. jcsd
  3. Sep 1, 2009 #2
  4. Sep 1, 2009 #3
    Here are the basic equations for a parallel L and C, and no resistance. (w = 2 pi f)

    1/Z = 1/jwL + jwC = jwC - j/wL = j(w2LC-1)/wL

    So Z = jwL/(1-w2LC)

    When LC=1/w2, Z= infinity

    [Edit] If you also have a parallel resistor such that

    1/Z = 1/jwL + jwC +1/R. then

    Z =~ R at resonance.
    Last edited: Sep 1, 2009
  5. Sep 1, 2009 #4
    But what about Z=L/C*R where R is the resistance of the coil if we don't neglect it?
  6. Sep 1, 2009 #5
    How do we get that?
  7. Sep 1, 2009 #6
    Exactly right.
    Using the same equations I posted earlier, , and the inductance has a small series resistance R, then

    1/Z = 1/(jwL + R) + jwC

    et cetera (like in prev. post), leading to (at LC resonance)

    Z= (jwL + R)/jwCR.

    If R<< jwL, then Z = jwL/jwCR = L/CR

    (note: L/C has units ohms2. so L/CR has units ohms)
  8. Sep 1, 2009 #7
    :-) Thanks!
  9. Sep 3, 2009 #8
    uh, can you tell me why/how j2=-1?

    Actually I'm studying high school physics so I'm studying the equations without the "j" operator or whatever it is..
  10. Sep 3, 2009 #9


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    Science Advisor

    By definiton, j = sqrt(-1). So, obviously, j2 = -1 since that's the inverse operation. It's just the imaginary number; in mathematics sqrt(-1) = i. EEs like to use j because i already means current.
  11. Sep 3, 2009 #10
    Hi uzair
    There is a real physical basis for j2 = -1. As I showed in an earlier post, the time derivitaves of a circular function V(ωt) = V0 sin(ωt) are

    d V(ωt)/dt = ω V0 cos(ωt) = jω V(ωt)

    d2 V(ωt)/dt2 = -ω2 V0 sin(ωt) = (jω)2 V(ωt) = j2 ω2 V(ωt) = -ω2 V(ωt)

    So you can see in red the shorthand notation for a 90 degree phase shift caused by taking the time derivative, and why j2 = -1.
    Bob S
  12. Dec 4, 2010 #11
    Sorry for the bump, but I'm working on the problem including the series R with L, and can't seem to figure out how to derive the resonance frequency w_o in that case. Wikipedia lists it as w_0 = sqrt((1/LC) - (R/L)^2), where the impedance is only real.

    I have the equation Zin = (R + jwL)/(jwRC-w^2*LC+1), which I have rearranged as

    Zin = ((L/C) - jR(w/C))/(R + j(wL - 1/wc))

    When I split up the fractions in the rearranged form, I can easily see where the 1/LC term comes from in w_0, but I'm having difficulty seeing how the (R/L)^2 term is derived. Any advice would be appreciated.
  13. Dec 4, 2010 #12
    Use Zin = -(R + jωL)/(1 + ω2LC -jωRC)

    and multiply both numerator and denominator by 1 + ω2LC +jωRC

    This will get the j out of the denominator. The denominator will be (1 + ω2LC)2 + (ωRC)2.

    Bob S
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