Solve Impedance Network: Resistor & Energy Storage Element in Parallel

[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?

 

[Swapnil]

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

 

[FrogPad]

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,
$$R \approx 9 \Omega$$
$$L \approx 90 mH$$

 

[Swapnil]

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

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

 

[FrogPad]

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

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.

 

[Valhalla]

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

 

[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!

 

[FrogPad]

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!

used my 89 to solve them also ;)

 

[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