# Meaning of zero/infinite impedance

1. Dec 14, 2015

### TheCanadian

I was calculating the impedance of a capacitor and inductor of $1 \mu {\text F}$ and $1 {\text H}$ respectively, at $1000 \frac {\text rad}{\text s}$ in both series and parallel configurations. I found the impedance to be 0 and infinity in the two cases, which is correct. But now I'm just wondering, what exactly does this mean for a real circuit? Since there are individual components in this circuit with a finite impedance, but zero (or infinite) total impedance in the equivalent circuit, what exactly are the properties of this circuit? I guess I'm just trying to understand: what exactly is the meaning of an impedance in this case?

2. Dec 14, 2015

### Luis Obis

Let me try to answer.

The meaning of impedance in this case is the same as always! you can see it was the relation between the voltage between the terminals and the current going into your (equivalent) circuit (or two-terminal element). It doesn't really matter of it's complex or not, it just makes things a bit more complicated at first.

The important thing to notice is that, in this case, impedance is dependant on frecuency, which doesn't happen when you have a purely resistive circuit.

As you have correctly guessed the impedance corresponds to 0 or infinity at that particular frecuency and as you know its a continous function (expect for the point where it goes to infnity) so for frecuencies close to 1000 rad/s, we call this the resonant frecuency of the system, for obvious reasons, it'll be very close to 0 or very large, respectively.

If you add a resistance in series to the circuit, to account for the fact that your components are not perfect capacitors/inductors or for the resistance of the wire itself. You can see that the impedance of the new circuit won't go to zero or infinity but will go to a very low value or to a very high value for that particular frecuency. You have just made a band-pass filter or a band-stop filter!

One typical aplication of this circuit, which you can (and should!) try for yourself is an AM radio. AM stands for amplitude modulation, which means that the message is coded in the amplitude of the wave, which has a constant frecuency, which is perfect for us. If you use the circuit as a band-pass filter (impedance 0) and you feed it with multiple frecuencies waves hooking it to an antenna (radio station waves) the circuit will "select" only the waves which have a frecuency sufficiently close to the resonant frecuency and with very little extra components, in fact you can do this with only a diode on top of the inductor and capacitor (of course you will need some way to listen to it, like a headphone speaker).

Here is the simplified schematic for the AM radio. You can tune the station if you have a varying capacitor or inductor.

I hope this helps you at least a bit, I may have divagated a bit, can't help it!

3. Dec 14, 2015

### TheCanadian

Thank you for the response! I realize this is a theoretical case (where R = 0), but if a current source was hooked up to these two configurations, where Z = 0 and Z = infinity, what exactly happens to this energy/current supplied? For Z = infinity, for example, does the energy just constantly transfer back and forth purely between the inductor and capacitor, thus not making a full loop?

Looking at it as two elements that can store and release energy, the case of Z = infinity despite finite individual impedances is a strange concept to me that I'm just trying to wrap my head around.

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