circuits: What do I do when impedance = undefined?

Question

I have j5ohms in parallel with -j5ohms. this gives me a fraction with zero as the denominator. How do I treat this parallel combination?

Answer 1

you get cases like this in ideal circuits under resonance. A plus j5 and minus j5 in parallel is an open circuit, while the same two impedences in series is a short circuit. If capacitors and inductances are used to derive these impedences, since the value is dependent on the frequency, the condition only occurs at one frequency.
This is used to build filters, oscillators, and such.

In reality, there is always a real component (resistance) which makes the open circuit a large but finite impedence, and the series circuit something with a low impedance.

Answer 2

Mathematics can only go so far in circuit analysis - usually resulting in a prototype or initial circuit design.
The real tests come when you subject the circuit to real world operation under power.

As an example, if you short a battery (don't ever do it - burn and explosion hazard!) with a short, very heavy cable, and apply I=E/R, again, you are dividing by zero - (the amps appear to be an undefined value - perhaps infinite) but in the real world, the battery or source can only produce so much current and Ohm's law doesn't know this.

In your case, logging or graphing the actual current vs. frequency on your circuit in operation is the only valid and real way to determine your "undefined" values. The math has given you a good idea of what to expect - and it won't be undefined once you know the real characteristics of the circuit under actual operation.

Answer 3

Well, I believe what it usually means when you have to divide by zero in this kind of a circumstance is your value is infinite. Your impedence is effectively infinity.