Start of the new chapter of Electrical Circuits: Impedances (AC Analysis)

IMPEDANCES:

Let’s start the lessons from here — one of the major (or maybe minor) changes when it comes to analyzing an AC circuit from a DC circuit: the impedance.

Now, the impedance of an electronic circuit is the ratio of the phasor voltage V to the phasor current I. It is declared with the variable Z and is measured in Ohms (Ω). In some ways, you could say that the impedance acts the same as the resistance in a circuit, but not completely.

Resistors, capacitors, and inductors are the three elements that cause impedance. Though you have to convert the capacitance C and the inductance L to get their equivalent impedances Z, by applying these formulas:

RESISTORS: 

 Z=R

INDUCTORS: 

Z=j•ω•L

Z=s•L , s=j•ω

CAPACITORS: 

Z=1/(j•ω•C)

Z=1/s•C , s=j•ω

From the stated formulas above, you can observe that the resistance of a resistor can already be considered as its impedance. While for inductors, you have to multiply its inductance by j (imaginary number*) and ω (angular frequency). And lastly for capacitors, you have to divide 1 by the capacitance, j, and ω.

*equal to the square root of -1.

Here are some important notes to remember for inductors and capacitors when it comes to analyzing circuits at DC and high frequencies:

Inductors are considered as short circuit at DC equivalent circuits, while as open circuit at high frequency. The opposite goes for the capacitors, since they are considered as open circuit at DC, while as short circuit at high frequency.

Don’t forget that impedances Z are also measured in Ohms (Ω) like resistances R. So, when you solve for impedances in series and in parallel, all you have to do is to follow the same formulas used when solving for resistors in series and in parallel.

SERIES IMPEDANCES: 

Zeq=Z1+Z2 + ... +Zn

PARALLEL IMPEDANCES: 

1/Zeq = 1/Z1 + 1/Z2 + … + 1/Zn

That’s how impedances look like in a schematic circuit diagram. Just rectangles. Pretty creative, huh?

In solving for the voltage V and current I, you will still apply the same formulas but replacing the resistance R with impedance Z. So, they would now go like:

V = I · Z

I = V / Z

Though for solving power P in AC circuit analysis, it’s a different story compared to solving it in DC. We’ll deal with that sooner or later. Or not.

And that’s about everything there is in the basics of analysing the impedance in an electronic circuit! Though its only the start for another chapter of our life learning the field we chose to walk.