I'm writing up a summary about classic impedances (resistors \$Z_R=R\$, capacitors \$Z_C=1/j\omega C\$ and inductors \$Z_L = j\omega L\$) for students starting out with AC networks.

I was about to tabulate the behaviour of \$|Z|\$ (magnitude), \$\angle Z\$ (phase), \$\textrm{Re}\,Z\$ (resistance) and \$\textrm{Im}\,Z\$ (reactance) for when \$R\$ and \$C\$ and \$L\$ and \$\omega\$ approach either \$0\$ or \$\infty\$, where I'd mention that a capacitor acts like an open circuit for DC signals and an inductor acts like an open circuit when approaching infinitely fast oscillations, since this is standard knowledge. However, it occurred to me: why do we claim this?

An open circuit, as far as I know, is an infinitely resistive path. If that is true, however, then a capacitor and an inductor could never be open circuits, since they have no resistance. It stands to reason that an open circuit is a path for which \$|Z|\$ is infinitely large, which is true when either or both of \$\textrm{Re}\,Z\$ and \$\textrm{Im}\,Z\$ is infinite. But why?

Intuitively, resistance measures how much electrons in a current are scattered and hence reduced in potential, so a big resistance means only a small trickle of current enters and exits a device. Reactance, on the other hand, I have no mental model for. Generally, I associate it with phase shifting (although of course, that's not the full story, since \$1/j\omega C\$ and \$j\omega L\$ can reach any reactance whilst having an ever-fixed phase). What's the intuition about a reactance blocking current when it gets big?

  • \$\begingroup\$ Impedance of infinite resistance is infinite. Impedance of infinite inductance is also infinite. So they both block current. \$\endgroup\$
    – Justme
    Jul 5, 2021 at 17:41
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    \$\begingroup\$ Your idea about resistance, particularly the "scattered and hence reduced in potential" is probably one of those "some truth mixed with fiction" things. I recommend you start with a good reading of the Drude model. It's imperfect, but a lot better than what I'm reading from you. Also, currents are set up by charges on the surfaces of conductors. So more to study there. As far as self-inductance goes, I almost can't think of a better place to go but to volume 2 of Feynman's Lecture series or else to "Matter & Interactions" by Chabay & Sherwood. Intuition like this is found in physics. Not EE. \$\endgroup\$
    – jonk
    Jul 5, 2021 at 17:58
  • \$\begingroup\$ @jonk Oh yes, I did realise that my wording wasn't fully accurate, but I thought the hand-waving would do. I intended a Drude-esque picture of colliding charges, though perhaps I should've been more careful regarding the potential drop: it's probably slightly better to state that charges move from one potential to the other, with collisions tempering the flow rate of the charges. Nevertheless: are you suggesting the Drude model can give me intuition about reactance? \$\endgroup\$
    – Mew
    Jul 5, 2021 at 19:11
  • \$\begingroup\$ @Mew Not directly. Charges still must be set up on the surface of an inductor's conducting material in order for current to flow. Similar that way, I suppose. Think about the conduction band charges as they rotate around a loop. If mean speed is the same (current is the same) then they are under acceleration because their vector direction is always changing, even if the magnitude of that vector isn't. This sets up a fixed magnetic field. But they also accelerate (the current rises) so the speed is changing also, so the magnetic field is changing, which induces a non-Coulomb electric field. \$\endgroup\$
    – jonk
    Jul 5, 2021 at 19:31
  • \$\begingroup\$ @Mew To complicate your life, think then two more ways: (1) superconductor; all charges are Cooper pairs and bosonic, and therefore no possibility for collisions and the concept of mobility is gone as charges accelerate without apparent limit; and, (2) conductor; mobility applies, as usual. What would this difference in behavior suggest to you? Here, what I'm trying to do is point out that if you really want an intuitive understanding, then you will need to be able to answer these kinds of questions. Your mental model has to include these possibilities, and more. \$\endgroup\$
    – jonk
    Jul 5, 2021 at 19:34

4 Answers 4


Look at a capacitor or inductor's response to a voltage step function. An ideal capacitor is "instantaneously like a short circuit but steady-state like an open circuit"; it passes current without resistance, but as it accepts charge, a voltage develops across it which opposes the current, which eventually falls to zero (or if you want to calculus-nerd about it, the current falls below any finite value within finite time). The amount of charge a capacitor can accept before the voltage rises by a certain amount is proportional to its value. In the infinite-reactance limit, the capacitor has a value of zero, so it takes zero coulombs to charge it up and turn it into an open circuit.

An ideal inductor is the opposite; it's "instantaneously like an open circuit but steady-state like a short circuit". The current is initially zero, and ramps up over time as the voltage across the inductor falls to zero. The larger the value of the inductor, the more slowly the current grows for a given voltage. In the infinite-reactance limit, the inductance has an infinite value, so the current stays at a value of zero forever — an open circuit.

Both devices look like an open circuit at one timescale limit, and a short circuit at another timescale limit. Both of them have a parameter that makes their timescale shorter or longer. Low reactance is the direction that brings them closer to the limit where they're short-circuit (high capacitance or low inductance); high reactance is the direction that brings them closer to the limit where they're open-circuit (low capacitance or high inductance). Infinite reactance is the limiting case where no current flows over a finite length of time, just as no current flows through an open circuit.

  • \$\begingroup\$ These are only examples of reactances, but I guess "size of rate of change" is a better way to think about it than "impact on phase" as I suggested in the question. You're basically describing in words what the differential equations for these components say, instead of their impedance formulas, so transient behaviour does seem to be the key. \$\endgroup\$
    – Mew
    Aug 2, 2022 at 16:29

Imagine this.

Now imagine infinite Inductance. It doesn't ramp up but if it is finite, it will but very slowly... If you let it separate really fast before the contacts ionize (~1 us), voltage will eventually arc and big bang then it will oscillate with the dielectric medium until the losses absorb all the energy. E= ½L I^2

Small experiment

I am working as an investigator in a large transformer factory and I have a 5 MVA transformer to debug. I want to confirm the primary inductance, which my RLC meter says is 22 Henries.

So I pull out a Lithium-ion battery and put it across the 25kV rated line input.

dI/dt = V/L = 3.7V/22H so I expect the current to ramp up 1 Amp every 6 seconds.

It does.

After 30 seconds at 5 A, I stop and did not want to wait for saturation for the magnetic core to short out the battery at a faster rate. ;)

But to demagnetize the core, I reverse the battery for the same length of time and quit.

Experiment confirmed intuition.

You can try similar on a solenoid or big relay. I got a big arc when I disconnected the battery. You will get a smaller zap but current limited by R.

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    \$\begingroup\$ I'm stuck on the Imagine this! \$\endgroup\$ Jul 5, 2021 at 18:15
  • \$\begingroup\$ big invisible infinity @StainlessSteelRat \$\endgroup\$ Jul 5, 2021 at 18:37
  • \$\begingroup\$ I know that \$L\to\infty\$ implies \$di/dt \to 0\$, if that's what you're trying to say. However, you're describing a transient circuit, not steady-state AC, even though the latter is a precondition for speaking of an impedance \$Z_L = j\omega L\$. Edit: by the way, I think you mixed up your paragraphs on accident. \$\endgroup\$
    – Mew
    Jul 5, 2021 at 19:23
  • \$\begingroup\$ Yes 80% of dI / dt is approximately related to 2.2/ω equivalent half power frequency. So if driving much faster with AC, no current. I did mix up on a moment an hour ago … Refresh screen…. \$\endgroup\$ Jul 5, 2021 at 19:33

In the context of AC steady state analysis, impedance is complex-valued. For certain operations on complex numbers (such as the rational polynomials that we use for circuit analysis*), any infinity is identical. That is, it doesn't matter from which direction you approach infinity, so long as the magnitude does. This is exactly symmetrical to the case near zero, because of the reciprocal transformation \$Y = \frac{1}{Z}\$: it doesn't matter from which direction you approach zero impedance, zero impedance is zero impedance. Indeed this transformation maps infinity to zero and vice versa. And we regularly use division -- and nothing more complicated than that -- so this basically covers it.

Finally, we define zero impedance as a short circuit, and infinite impedance as open.

*Resistance is the constant polynomial Z = R, inductance is \$j \omega L\$, capacitance is \$\frac{1}{j \omega C}\$, impedances in series go as \$Z_1 + Z_2\$, impedances in parallel go as \$\frac{Z_1 Z_2}{Z_1 + Z_2}\$, and a few less common, more complex operations, like wye-delta transformation. More generally, any solutions from linear algebra (which are in turn also polynomial). Thus, all lumped element circuit analyses lie in the domain of complex numbers in plain old algebra.

We can also generalize the analysis of a lumped-element circuit using linear algebra (nodal or mesh analysis): writing down all the nodes/loops, and the currents/voltages between them, in a matrix; the elements of which will be zero when there is no direct connection between any given pair of nodes/loops. The solution is a polynomial given by the matrix elements (from element impedances/admittances), and the matrix inversion operation (which is ultimately polynomial as well).

If you aren't big on complex algebra, or calculus, or mathematical analysis in general, this probably doesn't mean all that much to you; suffice it to say, there exist operations where you can get results that depend on what "kind" of infinity or zero you are looking at, or what the operation being performed is. Some of these are resolved by taking limits; other operations aren't symmetrical with respect to infinity/zero. But neither of these operations are used here, so the zero/infinity symmetry remains unbroken for this purpose.

  • \$\begingroup\$ I'm very familiar with all the advanced math required. Your conclusion is basically the same as I posited in my self-answer: it's not $$\textrm{Re}\{Z\}\to\infty$$ that causes zero current, but $$|Z|\to\infty$$. Still don't know what the physical effect is that is symbolised by increasing reactance. \$\endgroup\$
    – Mew
    Aug 2, 2022 at 16:21
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    \$\begingroup\$ @Mew Well, the definition of "open circuit" \$\rightarrow R = \infty\$ works at DC. Generalize to AC. In what ways could an AC circuit be "open"? Two: in-phase and out-of-phase. Apply an AC voltage to some impedance, and some current flows, which can be decomposed into orthogonal components. If the resistance is 0, then current flows, but perfectly out of phase, and while instantaneous power sometimes flows, it's always returned precisely opposite on the return swing. In that case, real power doesn't flow, it's "open circuit", even though a current is flowing. \$\endgroup\$ Aug 2, 2022 at 17:31
  • \$\begingroup\$ Likewise for reactance: if zero, current is fully in phase and the situation reduces to the DC case. If infinite, current goes to zero, and power along with (of any phase). So an open circuit at AC doesn't care about the angle, \$|Z| \rightarrow \infty\$ is really just saying at least one of \$\left( \Re\{Z\} \rightarrow \infty , \Im\{Z\} \rightarrow \infty \right)\$. \$\endgroup\$ Aug 2, 2022 at 17:33

In steady-state AC, voltage over a component is described by \$v(t) = |\hat V|e^{j\angle \hat V}\,e^{j\omega t} = \hat V\, e^{j\omega t}\$, and similarly, current through it by \$i(t) = |\hat I|e^{j\angle \hat I}\,e^{j\omega t} = \hat I\, e^{j\omega t}\$.

If an impedance is ohmic, there exists a constant \$Z\in\mathbb{C}\$ such that $$ Z = \frac{v(t)}{i(t)} = \frac{\hat V}{\hat I} = \frac{|\hat V|}{|\hat I|} e^{j(\angle \hat V - \angle\hat I)} = |Z|e^{j\angle Z} $$ There is an open circuit between two points when there is zero current between them regardless of the voltage, or $$ i(t) = 0 \iff \hat I = 0 \iff |\hat I| = 0 \iff \frac{|\hat V|}{|Z|} = 0 $$ ... and since the voltage phasor's amplitude can be any real number, \$|Z|\to\infty\$ is the most general condition for open circuits, not just \$\textrm{Re}\,Z\to\infty\$ (which is one way to get there, but indeed, infinite reactance, or \$\textrm{Im}\,Z\to\infty\$, is as valid to get zero current).


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