Curious How These Nordost Speaker Cables Do What They Do

No, that’s not correct, Steve.

First, keep in mind that a voltage must always be defined relative to some point, i.e., it exists between two points. It is not meaningful to simply think that "the neutral rail is always at zero volts."

In the case of a light bulb the terminal that is connected to AC "hot" is at 120 volts relative to the terminal that is connected to AC "neutral," which in turn is at or very close to zero volts relative to earth ground. In the case of a speaker cable the + wire is at some voltage relative to the - wire, depending on the signal at any given instant, and depending on the design of the amp the voltage on the - wire may or may not be at zero volts relative to the amp’s circuit ground, or relative to AC neutral, AC safety ground, and earth ground. For example, "fully balanced" amps and bridged mono amps both have full amplitude signals on both their + and - output terminals, with the voltages on those terminals differing from each other simply in that they have opposite polarities.

The current that is provided to a speaker by an amp, as well as the current that is provided to a light bulb by the incoming AC, both result from a voltage **difference** that is applied between the two conductors. The fact that one conductor may be at zero volts relative to some reference has no relevance to that.

Also, ieales comment above that relates to this ...

If the current were different in each leg of the cable, energy would either be gained or lost.

Apply a steady 1kHz 1V signal to a speaker. Regardless of where an ammeter is inserted in the circuit, the RMS current is the same.

... is absolutely correct.

Best regards,
-- Al

Steve (Williewonka), your response to ieales about Ohm’s Law was correct, but he is correct in saying that equal currents exist in the positive and negative conductors connecting an amp and a speaker. (Although later in this post I’ll mention an extremely minor exception to that which might occur in the case of an active speaker).

For current to exist, there must be a "complete circuit," from source to load and back. In the case of an AC waveform, such as an audio signal, the direction of the current in both the positive and the negative conductors alternates every half-cycle, and the two conductors play an equal role in the transfer of energy from source to load. Energy is transferred in the form of an electromagnetic wave that is distinct from but is intimately associated with those currents, and that electromagnetic wave in turn propagates mainly in one direction, from source to load. (In the case of an amp/speaker interface some energy may also propagate from speaker to amp, as a result of "back-emf" produced by the drivers, or as a result of the release of energy stored in the inductive or capacitive components of the speaker’s impedance, or as a result of reflection effects that can occur at RF frequencies due to impedance mismatches).

An analogy that can be drawn would be to an ordinary light bulb. It consumes energy, of course, converting some of it into heat and some of it into light. But the amount of current entering it and leaving it via its two terminals, one of which is connected to AC "hot" and the other to AC "neutral," is identical.

The very minor possible exception I mentioned earlier is that in the case of a powered speaker a tiny fraction of the current in the positive conductor could conceivably return to the amp via a ground loop, part of which would consist of the safety ground conductors in the power cords of the amp and the speaker. In other words, to a very small extent, depending on the internal grounding configurations of the amp and the powered speaker, the AC safety ground wiring might be part of a relatively high resistance path that is in parallel with the negative conductor in the speaker cable, and it might thereby conduct a tiny fraction of the current in the positive conductor that would otherwise be returned to the amp by the negative conductor. (When two paths are present between a source and a load, current will utilize both paths, and at a given frequency will divide up between them in inverse proportion to their impedance at that frequency). But even in that case the amount of current leaving the amp will be identical to the total amount of current returning to it via whatever return path(s) is or are present.

Regarding "how Nordost cables do what they do," I have not studied them sufficiently or seen enough information about them to be able to provide meaningful comments at this time.

Best regards,
-- Al