What would happen if?

Super.

Good luck with that... and I do suppose this rhetoric and theory of your's is something you actually demonstrated recently to prove it?

or is this simple speculation?

I attached two sets of speakers to an SS amp. Two dissimilar types of speakers, I might add. two different types of cabling as well.

Turned on the gear after making up one speaker completely, but only making up the second pair of speakers with one or half the usual compliment of wires in use there....

Played some music.

Found out the speaker set that had only half of the normal compliment of cabling were broken.

They did not play any music... nor did they make any sounds.

I shut down the gear, and reversed the process.... same thing. Sheessh... now I've two sets of broken speakers.

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.

...but I do suggest everyone try this experiement for themselves. Maybe some amp & speaker combinations will run four speakers using one and a half sets of cabling.

Mine don't.

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

Rut ro.

My bad… how I missed that is on me. Sorry.

Naturally my idea focused on TWO pairs of speakers…. Not just the one. It was inconceivable to me for two sets of cables to supply one speakers singular pair of binding posts simultaneously.

Lifting only one side would in fact allow 3 conductors to carry current…. And interfere with the testing.

Lift both legs from the cables you don’t want to hear… and cover the ends so they don’t short together.

I amaze me sometimes.

Thanks for the paitience…. I couldn’t get two pairs of speakers out of my mind.. .. as two set of wires for one pair of speaks, just did not compute.

I feel like one of those two fish that swam into a concrete wall. One trunedd to the other and said, “Dam.”

... 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