How to Express Impedance in Terms of w, R, L, and C in an LRC AC Circuit?

[scarlets99]

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

 

[vela]

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.

 

[scarlets99]

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)

 

[vela]

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 1/(i\omega C)+i\omega L. If they were in parallel, you'd have

\frac{1}{Z}=i\omega C+\frac{1}{i\omega L}

so the combination's impedance would be

Z=\frac{1}{i\omega C+1/(i\omega L)}=\frac{i \omega L}{1-\omega^2 LC}

 

[vela]

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)

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.