Understanding the Complex Impedance of Resistors and Capacitors

[dionysian]

I am trying to figure out why impedence of a resistor and a capacitor is complex. I am sure it has something to do with lenzs law and and inductors resistance to changing circuits, but i am unable to follow the steps in mathmatical reasoning between lenzs law and $$\frac {1} {j \omega C }$$ for capacitors and $$j \omega L$$ for inductors.

 

[Astronuc]

Impedance of a resistor is real, while impedance of an inductor or capacitor is considered complex or imaginary.

It's really the relationship between voltage and current that is important to understand.

In a purely resistive ciruit, the voltage and current are in phase, while in a purely reactive circuit, the voltage and current are 90° out of phase.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/phase.html

http://www.physclips.unsw.edu.au/jw/AC.html

http://www.walter-fendt.de/ph11e/accircuit.htm - nice little java applet to show the effect of resistance, capacitance and inductance.

I'm working on a tutorial in the PF tutorial section to explain this material, so please bear with me.

 

[stewartcs]

Impedance is complex because it has an imaginary component (reactive or inductive). This results in the need to use the mathematical concept of complex numbers in order to model it. Hence, impedance is deemed complex.

 

[Astronuc]

For inductors, the voltage current relationship is

$$v(t)\,=\,L\frac{di(t)}{dt}$$

and if

$$i(t) = A \exp{j\omega{t}}$$

then di(t)/dt = j$\omega$ A exp(j$\omega$t).

Similarly for capacitors,


$$v(t)\,=\,\frac{1}{C}\int{i(t)\,dt$$

and

$$\int{e^{j\omega{t}}}\,dt\,=\,\frac{1}{j\omega}{e^{j\omega{t}}}$$

 

[dionysian]

Thank for your reply Astronuc.

I look forward to your tutorial on this subject.

However, i didn't follow why you used $$i(t) = A e^{j \omega t}$$ for the current in your explanation. I am would suspect that it has something to do with the phaser representation of current. But i am unable to follow the reasoning why this can be done.

 

[Astronuc]

$$i(t) = A e^{j \omega t} = A\,cos\,\omega{t}\,+\,jA\,sin\,\omega{t}$$ is just a general expression indication a real and reactive component to the current. The same applies to the voltage.

Impedance can be real (resistance) or reactive (capacitor or inductor) and that simply refers to the relationship between voltage and current.

It's just very handy to use complex numbers to represent the behavior of AC voltage and current.

 

[trambolin]

The reason is also intuitive. Since we are trying to find an equivalent "generalized resistance", we model(!) the capacitor and the inductor as a resistance which their behavior is given as a transfer function or a magnitude and phase response at each frequency. Then, using these polynomials, we can simply manipulate these dynamics to simplify, simulate,...

So, don't think like that they are naturally complex and we manipulate them accordingly (everything is real in the circuit!) , but the way we handle them, technically, requires them to be complex numbers. And the real and imaginary parts carry the information as explained above...

 

[bchung]

I've had the same original question myself that dionysian posed originally, and the way I reached an understanding is to look at the impedance of the capacitor and the inductor, both with the "j" in the numerator. So, instead of 1/(jwC), for the capacitor, I put the j on top and negate.

Impedance of inductor: jwL
Impedance of capacitor: -j[1/(wC)]


So, in this comparison an inductor provides a postive imaginary part, and a capacitor provides a negative imaginary part. Now, a positive imaginary impedance simply means that in a current sourced AC circuit, the voltage across the device will lead the current by 90 degrees in phase (inductor). A negative imaginary impedance means that the voltage will lag by 90 degrees.

Hope this helps. There's a little applet out there that shows this interactively if the logic still seems cloudy.

http://picomonster.com/Lesson%202/Lesson%202.html

Hope this helps