# Solve Impedance Network: Resistor & Energy Storage Element in Parallel

• Valhalla
In summary, the network is equivalent a to a resistor and an energy storage element connected in parallel. Find their values. The impedance I got was Z=8.39+2.22i. However, the impedance for the inductor needs to be solved using the complex conjugate.
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:
This is certainly the right track. Let me see what I can do...

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

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 $Rj \omega C - 1$

Last edited:
Swapnil said:
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.

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

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!

Valhalla said:
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 ;)

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

## 1. What is an impedance network?

An impedance network is a combination of resistors and energy storage elements, such as capacitors or inductors, connected in a parallel or series configuration. It is used to analyze and calculate the electrical properties of a circuit.

## 2. How do I solve an impedance network with resistors and energy storage elements in parallel?

To solve an impedance network with resistors and energy storage elements in parallel, you can use the parallel impedance formula: Zeq = (Z1 * Z2) / (Z1 + Z2), where Zeq is the equivalent impedance, Z1 is the impedance of the resistor, and Z2 is the impedance of the energy storage element.

## 3. What is the purpose of solving an impedance network?

The purpose of solving an impedance network is to determine the total impedance of a circuit, which is important for analyzing the flow of current and voltage in the circuit. It also helps in designing and optimizing circuits for specific applications.

## 4. What are some common energy storage elements used in impedance networks?

The most commonly used energy storage elements in impedance networks are capacitors and inductors. Capacitors store electrical energy in an electric field, while inductors store energy in a magnetic field.

## 5. Can an impedance network with resistors and energy storage elements in parallel be simplified?

Yes, an impedance network with resistors and energy storage elements in parallel can be simplified by using the equivalent impedance formula mentioned in question 2. This can make circuit analysis and calculations easier and more efficient.

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