Why is an infinite reactance an open circuit?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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).