What would happen if?

I don't see any reason why you can't do that, and get reasonably valid results. Provided of course that, as many have said above, you make absolutely sure to avoid the possibility of a short circuit, and as long as connections are changed only with the power off.

The idea of disconnecting only one leg of one of the cables is not correct. It will invalidate your results because on the other polarity (positive or negative; the opposite polarity from the one that you disconnect the one leg from) you will have that polarity conducted in parallel through the corresponding legs of two cables. The current will take advantage of that and divide itself between both paths on that polarity, in inverse proportion to the resistance of each path.

I don't think that concerns about antenna effects and capacitive loading are likely to be relevant to the power amp/speaker interface, due to the extremely low output impedance that power amps have to have (well under one ohm usually), as well as the fairly low input impedance of the speakers themselves. The low impedances would load down any rf or emi that was picked up from the air (that's why speaker cables are usually not shielded, not to mention that the frequency of the interference would be far too high for the speakers to respond to). The added capacitance is likely to be have essentially zero effect because the resulting capacitive impedance (reactance) would be vastly higher than the output impedance of the power amp. It would be a completely different story with line level interconnects.

Regards,
-- Al

Following up on my previous comments about capacitance, consider the following calculation:

Assume as a very rough ballpark that cable capacitance is 50 picofarads (pf) per foot, a typical number. Assume a 20 foot cable length. That gives a total capacitance of around 1000 pf.

The resulting capacitive reactance at 20,000 Hz (the worst case frequency within the audio band, in terms of the potential effects), based on 1/2piFC (pi = 3.14; F = frequency in Hertz, C = capacitance in farads, to get results in ohms), would be a capacitive reactance of around 8000 ohms.

With the output impedance of the amp likely to be well under 1 ohm, and the speaker input impedance likely to be 8 ohms or less, 8000 ohms of capacitive reactance is, in comparison, essentially infinite, and it would seem to the amp/speaker interface that the capacitance is not there.

Regards,
-- Al

Blindjim -- With all due respect, you are missing the point. Your suggestion of disconnecting just one leg of the cable that is not being used DOES provide a complete circuit, that includes the other leg of that cable.

Let's say that one of the two cables under test (call it cable 2) is disconnected on the positive leg, but left connected on the negative leg.

Current will flow from the positive output terminal of the power amplifier, through cable 1 to the speaker, then through the speaker, and then from the negative terminal of the speaker through BOTH cables back to the negative terminal of the power amplifier. It's as simple as that.

As a person with two degrees and several decades of experience as an electrical design engineer and manager, I am fully cognizant of the principle that a complete circuit is required for current to flow. There IS a complete circuit. But on one polarity it has two wires in parallel, and the current will utilize both of them, which, as I said, would invalidate the results of the test.

If for some reason you don't accept that, consider the case where one leg of BOTH cables is disconnected. So that on the positive leg you have one side of cable 1, and on the negative leg you have one side of cable 2. That would obviously provide a complete circuit as well, and would result in normal sound coming from the speakers, which I think helps to get across my point.

Regards,
-- Al

Regardless how any of the pairs of speakers were attached to the amp, only those who had both a positive and negative lead afixed to their binding posts directly from the corresponding ones on the amp would play/work.

Yes. Of course. That is absolutely correct. But we are not talking about having only half the normal complement of cabling. We are talking, in fact, about having 1.5 times the normal compliment of cabling. The normal complement of cabling on one side of the amp/speaker interface (say the positive side), and twice the normal complement of cabling (two parallel conductors) on the negative side. Is that not clear?

As I understood your original suggestion, you were suggesting that to run a comparison between two sets of speaker cables, it was sufficient to just disconnect one conductor of the cable set that was not in use at any given time. So that one side of the speaker would be wired to the corresponding terminal of the amp through one conductor of one cable, and the other side of the speaker would be wired to the other terminal of the amp through the other conductor of that cable PLUS one conductor of the unused cable.

And my point was that you would then have a complete set of connections between the two terminals of the speaker and the two terminals of the amp, but one of those connections would go through two conductors in parallel, rather than just through one conductor.

Which would obviously function fine, but would make the comparison between the two cable sets meaningless, because on one side of the speaker/amp interface the current would be flowing through two conductors, one from each cable set. The current would divide itself between the two conductors in inverse proportion to their resistance.

That is not speculation and it is not rhetoric. It is elementary circuit theory. If you don't agree, then you are envisioning a different configuration of the connections than what is being discussed, or you are not understanding what has been said.

Please do carefully re-read the original poster's question, and my responses to it. The basic point that I can't seem to get across to you is that electrons flowing through a conductor (or two conductors in parallel) between the negative terminal of the speaker and the negative terminal of the amp don't care whether or not those conductors are physically part of the same overall cable assembly that contains the conductor that is connected between the positive terminals of the speaker and amp. Is that not clear?

Regards,
-- Al

... those two fish that swam into a concrete wall. One turned to the other and said, “Dam.”

LOL!

No problem, bud.

Regards,
-- Al

A complete path is certainly not required for current flow. A capacitor connected across two battery posts is a perfect example. We all can agree the di-electric separates CONTINUOUS flow.

Not true. A complete path is always required for current flow. But part of the path can be the electric field that exists across the dielectric of a capacitor, in which the capacitor stores energy.

A capacitor connected across two battery posts would rapidly charge up to the battery voltage, with a time constant determined by the capacitance value and the (very low) source impedance of the battery (plus any stray resistance in the circuit). When the capacitor has charged up to the battery voltage, a state of equilibrium would be established, and no current would flow (ignoring miniscule leakage paths).

The capacitor would continue to store that charge, and if disconnected from the battery and connected to a load, would then discharge into that load, causing a current to flow until discharge was complete.

Basically, charge is stored in the electric field that exists across the dielectric. If the voltage source is at a non-zero frequency (i.e., not a battery or other d.c. source), then continuous current can flow, in effect, proportional to the capacitance times the rate of change of voltage. If the voltage is a sine wave (or a summation of sine waves at various frequencies), with polarity that alternates every half-cycle, the alternate charging and discharging that occurs from each side of the capacitor looks to the rest of the circuit as if the capacitor is conducting current, even though current is not actually flowing through the dielectric. The current flow will be limited or impeded, though, based on the 1/(2piFC) factor I mentioned earlier.

Regards,
-- Al