Open termination, signal propagation, transmission lines
the propagation speed of the waves is independent of the length of the line or the frequency of the source

The phase velocity, which is probably what you mean by propagation speed, is independent of the length of the transmission line, but varies with the frequency of the source.

Lumped circuit theory says that the voltage at the open circuit will follow the oscillating source (instantaneously) and there will be 0 current in the branch at all time.


I think the answer might be that the period of the new source is 10 ns, which is 100 times longer than the one-way travel time of an electrical signal through the load (or there are 50 reflections from open end that make it back to source in one period). Each returning reflection &quot;tells&quot; the source to decrease the amount of current going into the load since the reflection has information about the open circuit.

When the AC signal is first applied, there is a back and forth of reflections, as you imagine. However, over time, &quot;transmission line&quot; consisting of R4 and a ground somewhere approaches an &quot;oscillation mode&quot;. Each point in the resistor/transmission line, will have a sinusoidal instantaneous voltage. The voltage at any other point on the &quot;transmission line&quot; will also have an sinusoidal instantaneous voltage that will generally differ from that of the first point in phase and amplitude.
Yes

In reality the open termination needs at least 100 ps delay between a &quot;new value&quot; being presented at its input (due to changing ac source) and that value having any effect on the open terminated end.

There will be a delay, but again, it is frequency dependent.

Since the load is driven by a constantly changing ac source it seems that there should always be a nonzero current going &quot;into&quot; the load.

I take &quot;the load&quot; to be the R4 in your schematic, and I take &quot;a nonzero current&quot; to mean a nonzero rms current, rather than an instantaneous current which is never zero. With those assumptions, yes, there will be an extremely small AC current into R4 using transmission line theory, rather than &quot;instantaneous&quot; lumped-element theory. The transmission line theory will be correct, there will be some current, but it will be so small that the lumped element model is nearly perfectly correct.

This seems incorrect so I think I'm missing something.

Your reasoning is correct. What you may be missing is that the current will be so small that it hardly differs from the lumped-element prediction.

So if you hook up a low frequency source to this kind of load, although the current is not actually zero it is very small...

Yes

since reflections make it back to the source and &quot;dampen&quot; the incoming current at a rate much faster than the input current increases/decreases.

Yes, when AC is first applied. As time passes, reflections &quot;smooth out&quot;, and an oscillation mode is &quot;excited&quot; (or &quot;entered&quot;) as described above.