What would happen if?


I just recently purchased a new pair of speaker cables and my goal is to hook them up and set them side by side with the old ones for easily comparasion.
My amplify is hidden behind the wall, it's very diff to get behind it to plug in and unplug the speaker cables back and forth. My Q. is what would happen if I get behind it once and tap the new pair on top of the old pair (old one is bananas, new one is spades) and leave the other end to the speakers OPEN. That way I can swap them back and forth for easy comparasion.
As far as I know, both of the cables will receive amplyfied signals at the same time but only 1 selected pair will get hooked up to the speakers. I'm curious what's the damage or any posible senerio would be if I play with them like this for approximately about a month? Will the dynamic of the speakers be decreased?
Thanks,
128x128nasaman
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