Impedance of Circuit (Complex Numbers Question)

[Irishdoug]

Homework Statement

Find the impedance of the circuit shown (R and L in series and then C in parallel with them).

Relevant Equations

$Z = R +i(\omega L - \frac{1}{\omega C})$
$V_{R} = RI_{0}sin(\omega t)$
$V_{L} = \omega LI_{0}cos(\omega t)$
$V_{C} = -\frac{1}{\omega C}I_{0}cos(\omega t)$

Question is from Boas Ch. 2 Q.41 (I've the first edition) or Q 16.8 in the 3rd edition.

Find the impedance of the circuit shown (R and L in series, then C in parallel with them). Circuit is essentially this (it is a closed circuit which I can't easily draw).
-----R------L---
-------C---------

I know $Z = R +i(\omega L - \frac{1}{\omega C})$ = $\frac{V_{R} + V_{L} + V_{C}}{I}$ but I'm not sure exactly what I need to do, or where to start, other than sub in the equations for $V_{R}, V_{L}$ and $V_{C}$. Any help appreciated.

 

[BvU]

I'm not sure exactly what I need to do

1. Post a complete homework problem statement
2. Read the text of the chapter
3. Reconsider your relevant equations -- two out of three are unusable in this context
4. $Z = R +i(\omega L - \frac{1}{\omega C})$ is not correct (that equation is for RLC in series)

Are you familiar with complex numbers ?

 

[Irishdoug]

1. Post a complete homework problem statement
2. Read the text of the chapter
3. Reconsider your relevant equations -- two out of three are unusable in this context
4. $Z = R +i(\omega L - \frac{1}{\omega C})$ is not correct (that equation is for RLC in series)

Are you familiar with complex numbers ?

Yes I am familiar with complex numbers. It is the electronics part I'm struggling with. I'm aware then that the formulas for resistors in parallel and series apply to impedance, Z. I realize I don't need the Voltage equations. As such, I'm trying to figure out the correct equation for Z. I thought it applied to circuits in general. It is not stated that it is only for RLC in series in the text.

I'm thinking so far:

$Z_{R} = R$
$Z_{L} = i\omega L$
$Z_{C} = \frac{1}{i\omega C}$$Z_{R} + Z_{C} = Z_{series} = R+i\omega L$.

Then, $Z_{parallel} = \frac{1}{Z_{series}} + (\frac{1}{i\omega C})^{-1}$

Does this look any better?

 

[BvU]

No ! $$ {1\over Z_{\sf parallel} }= \frac{1}{Z_{\sf series}} + \left (\frac{1}{i\omega C}\right )^{-1} $$

looks better, but we still don't know what we are looking for in this exercise.

 

[Irishdoug]

No ! $$ {1\over Z_{\sf parallel} }= \frac{1}{Z_{\sf series}} + \left (\frac{1}{i\omega C}\right )^{-1} $$

looks better, but we still don't know what we are looking for in this exercise.

Ah you are right. I am just looking for the equation for the total impedance for the circuit. So $Z = R+i(\omega L - \frac{1}{\omega C})$ is the impedance for an RLC in series as BvU pointed out. I need to find it for the circuit I have.

 

[vela]

I am just looking for the equation for the total impedance for the circuit.

Without a circuit diagram, it's not clear what "total impedance for the circuit" means. Here's a reproduction of the diagram of the circuit in the book.

Those two little wires at the end clears up what you're looking for. The implication is that if an AC voltage V is applied across those two wires, the circuit will draw a current I, and the impedance, by definition, is V/I.