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LRC AC circuit explanation

  1. Feb 24, 2010 #1
    How do you express impedance in terms of w(omega), R, L and C knowing the circuit diagram.

    I understand that for
    R Z=R
    L Z=iLw
    C Z=-i/wC

    What is the effect of having components in parallel/series, could someone please give an example of a circuit and it's impedance expression.
    Thank You
     
  2. jcsd
  3. Feb 24, 2010 #2

    vela

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    Just treat the impedances like resistors. If they're in series, they simply add. If they're in parallel, you use 1/Z = 1/Z1+1/Z2+...+1/Zn.
     
  4. Feb 24, 2010 #3
    So how would you draw the diagram for a circuit with impedance 1/Z,
    i.e. 1/Z=(1/iwL+R1)+(1/R2+1/iwC)
     
  5. Feb 24, 2010 #4

    vela

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    1/Z isn't the impedance; Z is. That formula for 1/Z just tells you how Z is related to the individual impedances.

    If you had, say, a capacitor C and inductor L in series, their combined impedance would be [itex]1/(i\omega C)+i\omega L[/itex]. If they were in parallel, you'd have

    [tex]\frac{1}{Z}=i\omega C+\frac{1}{i\omega L}[/tex]

    so the combination's impedance would be

    [tex]Z=\frac{1}{i\omega C+1/(i\omega L)}=\frac{i \omega L}{1-\omega^2 LC}[/tex]
     
  6. Feb 24, 2010 #5

    vela

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    You can't. The quantity (1/iwL+R1) doesn't make sense. It's like trying to add 1 ohm-1 to 2 ohms. You can't do it because the units don't match.
     
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