# Find Equivalent Impedance - AC Steady State Analysis

Determine the driving-point impedance of the network at a frequency of $2$kHz: # Solution

Lets first find impedance of elements one by one:

## Resistor $R$

The resistor impedance is purely real and independent of frequency.

$Z_R=R=20 \Omega$

## Inductors $L_1$ and $L_2$

The inductor impedance is purely imaginary and directly proportional to frequency:

$Z_{L_1}=j\omega L_1$

We need to find the impedance in $2$kHz. Therefore:

$\omega=2 \pi f=2 \times \pi \times 2000 = 12566 \frac{rad}{s}$

$Z_{L_1}=j\omega L_1= j \times 12566 \times 0.002=j25.132 \Omega$

$Z_{L_2}=j\omega L_2= j \times 12566 \times 0.001=j12.566 \Omega$

## Capacitor $C$

The capacitor impedance is purely imaginary and inversely proportional to frequency:

$Z_C=\frac{1}{j\omega C}= \frac{1}{j \times 12566.4 \times 0.0001}=\frac{1}{j \times 1.2566}$

To get the standard representation of complex numbers, we need to bring $j$ to numerator and this can be done by multiplying by $j$:

$Z_C=\frac{j}{j \times j \times 1.2566}=\frac{j}{-1 \times 1.2566}=\frac{-j}{1.2566}=-j0.796 \Omega$

Note how capacitor acts in this frequency. The value of impedance is less than $1 \Omega$. Compare this value to values of other components. It is almost equivalent to a short circuit!

## Equivalent Impedance

Let's replace the values in the circuit: $j25.132$ and $-j0.796$ are parallel.

$j25.132 || -j0.796=\frac{j25.132 \times (-j0.796)}{j25.132 + (-j0.796)}=\frac{20}{j24.336}=-j0.823$

One interesting point here is that unlike pure resistive circuits where the equivalent resistance of parallel elements is always less than the resistance of each element, the value of the equivalent impedance of parallel elements can be greater than the value of the impedance of elements. Here the capacitor impedance value is $0.796$ but the equivalent impedance, $0.823$, is higher. Three components are in series. Therefore:
$Z=20+(-j0.823)+j12.566=20+j(-0.823+12.566)=20+j11.743=23.193 \angle 30^{\circ} \Omega$

Now, determine the impedance at $20$Hz and $200$kHz and share it with others below in comments section.

Hi! Yaz is here. I am passionate about learning and teaching. I try to explain every detail simultaneously with examples to ensure that students will remember them later too.

## Join the Conversation

1. 2. 3. 4. 5. 6. 1. Determine the driving-point impedance of the network at a frequency of 2 kHz:
Zc was calculated as -j0.796 Ohms but the correct calculation is -j0.0796 ohms
The final answer will be 20+j12.52 = 23.6 with angle of 32.05.
Thanks

2. Nick says:

In solving for the impedance of the Capacitor, you use .0001 as the value for 1000uF.

3. Alec Poulin says:

The question asks for a 1000 µF (1 mF) capacitor, but the solution is made with a 100 µF (0.1 mF) capacitor.

1. Thank you for reminding. It is fixed now.

4. Aditya says:

Thanks sir for the content you have provided which really help me to improve my concept...

5. Marco Polo says:

Really nice to see the different steps. Thanks!

6. Abdullahi Suleiman Ingawa says:

Why you didn't drop the solution