# Impedance network

1. Oct 29, 2006

### Valhalla

Hi all, I'm stuck on this part of a problem. I have gotten the net impedance between two terminals A and B. The impedance I got was

$$Z=8.39+2.22i$$

Now the question is...

The network is equivalent a to a resistor and an energy storage element connected in parallel. Find their values.

I thought this was going to be easy then I realized that it was parallel not in series. So then I solved the general equations for the impedance of a resistor and a capacitor and a resitor and inductor in parallel. This is what I got:

Note: w is the angular frequency (the given frequency is 60Hz)
R//C
$$\frac{R-R^2wci}{R^2w^2c^2+1}$$

R//L
$$\frac{RwL(wL+Ri)}{R^2+w^2L^2}$$

Then I figured that i could equate the Real and Imaginary parts of the impedance together and then use the magintude of the impedance to solve for R or the energy storage element. I can't seem to get that to work out. Is this the right track?

Last edited: Oct 29, 2006
2. Oct 29, 2006

### Swapnil

This is certainly the right track. Let me see what I can do...

3. Oct 29, 2006

I think it's the right track, you just need to get the $j$ part separated.

$$Z_C = \frac{1}{j \omega C}$$
$$Z_L = j \omega L$$

$$R || Z_C = \frac{R \left(\frac{1}{j \omega C}\right) }{R+\frac{1}{j \omega C}} = \frac{R}{R j \omega C + 1}$$

Now if you multiply the top and bottom by $Rj \omega C - 1$

$$\frac{R}{R j \omega C + 1 } \frac{(Rj \omega C -1)}{(Rj \omega C -1)} = \frac{R}{C^2R^2 \omega^2 + 1}- \frac{CR^2\omega}{C^2R^2 \omega^2 + 1}j$$

Now obviously this expression will not allow you to get, $Z=8.39+2.22i$.

So if you do the same for the inductor you should be good. I got,
$$R \approx 9 \Omega$$
$$L \approx 90 mH$$

4. Oct 29, 2006

### Swapnil

I think you are suppose to multipy the top and bottom by the complex conjugate which is $-Rj \omega C + 1$ rather than $Rj \omega C - 1$

Last edited: Oct 29, 2006
5. Oct 29, 2006

I'm pretty sure it will come out to the same thing. Using the complex conjugate will save some time factoring out the negatives though.

EDIT:
Since,
$$R j \omega C - 1$$
$$-R j \omega C + 1$$

$$R j \omega C -1 = -1(-R j \omega C + 1)$$
So you are still multiplying by the complex conjugate.

Last edited: Oct 29, 2006
6. Oct 29, 2006

### Valhalla

thanks guys Ill keep working on, I figured the cap one looked like it wouldn't work.

7. Oct 29, 2006

### Valhalla

I got the 9 ohms and the 90mH as well. I ended up with a couple of ugly quadratic equations and then used my TI-89 to solve them. Thanks for all the help guys!

8. Oct 29, 2006

used my 89 to solve them also ;)

9. Nov 17, 2006

### neka

You can do this finding admittance first as Y=1/Z=1/(8.39+j2.22) = 0.1114-j0.0295.
Next observe that, if you have R and L or C in parallel, then the admiitance has real part as 1/R and imaginary part as -j/Lw, that is
Y=(1/R) -j(1/Lw) with w=120pi.
Therefore, you get 1/R= 0.1114 giving R=8.97 and 1/Lw=.0295 thus L=90mH. There is no need to solve quadratic equations.

xxxx