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?
If I had a very highly resolving system, I would be a bit wary, thinking that perhaps it would influence the sound, although basically it probably wouldn't. But I would not do it anyway, because having five thumbs on each hand, I'd be afraid of shorting the amp, if the + and - poles of the wires not used, were to touch, while I fumbled around. A tube amp might take it, some SS amps don't like that at all.
I personally would not do it for 2 reasons. 1 - You just added antenna's to the amps output. 2 - Accidentlly touching the ends that are not being used.
I also would advise against it, for the reasons stated by Detlof and Xti16.
I wouldn't do it but also for comparison reasons. Wouldn't the unconnected pair impart unneccessary capacitance at the amp's outputs? Good luck.
when you do it, just be very sure not to let the disconnected ends touch...

alternatively while not as satisfactory but spark free - you might start by wiring one speaker with the old and one with the new to get an initial sense of significant differences.

you will need to run this for a few days to let the new cable settle in - rapid ABing can be very deceptive and is generally not considered to be a good way to tell much of anything - also it will give your ears time to listen and determine what differences you might in fact be hearing

if you decide there are some big differences then you could proceed to a more ambitious comparison

if there is not much difference you can decide which cables you are keeping and which you will be posting on the Gon
Oh go do it. I don't think that the second set of cables left attached to the amp changes any of the parameters of the amp at all. Think about it......

But, shorting is a REAL issue! What I have done when necessary is to tape the wires with spades so that the + and - wires are well away from each other and can't under any circumstances touch. Ditto for banana plugs if they are individual plugs (not two encased in plastic, seperated the proper space to match to many speaker and amp connectors).

Re the taping, hell I'm so anal about the shorting potential I tape all of my loose ends, spades, bananas, whatever, so they can't short if one or both were to come loose from the speakers or amps. Highly recommended!


I'm with Newbie on this one... just do it. it's temp anyhow.

You must have a completed circuit for electrons to move - conduct.

Connect and disconnect whatever, with the amp dead off. Once amp connections are made put whatever cables onto whatever speakers. Only one leg (one positive or one negative) of the cables not being listened to needs be disconnected to prevent any current flow into those 'off for now' speakers.... not both. Of course, insulating the exposed leg so it won't touch it's proposed binding post is important.

A balloon or tape, or a shoe or phone book... anything preventing conductivity will suffice.

Just 'member, power off the amp prior to making up and unmaking ANY connections to the speakers themselves.

As was said above about A/B ing, I'd shy away from instantaneously attemnpting that event. Given the power up and down safety measures you'll want to take, it's not viable anyhow. Just take notes instead.
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.

-- Al

Wouldn't do it. Too many unkowns to say what would happen. You're essentially adding a capacitor across the amp's output. Depending on the capacitance the amp may not like it. However large or small, you're asking the amp to drive an additional needless capacitive load. At the least you're affecting frequency response somewhere. Likely that your evaluation of each cable would be scewed also because of the additional capacitance that wouldn't be there otherwise.


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.

-- Al

The rules for current flow must have changed when I wasn't looking.

At some point long long ago, a completed path for voltage or current, if you would rather, needed to be in place.... or still more simplified, one leg to bring it in on, and one for the return. Things are a bit more involved than that, but in essence that way worked then and now... though perhaps there is some new one conductor circuitry that actually can send and return voltage and current into a component or device without the need for another path to complete the circuitry.

The speakers which has but one leg of the two speaker cables attached to them and to the amp as well have to be there for them to work. For the amp to see an additional load as it were.

Hook one of the two supplied wires in the speaker cables to the speaker, whichever one doesn't really matter. Next, connect the same wire in the second speaker cable to the other speaker. Turn everything on and tell me you can then hear music or even sound. you won't be able to. I assure you.

That could work if a wire was connected to the unlike post on those speakers which were properly connected. That way only one wire would run from the amp to one pair of speakers, and the return path would come via a connection to the playing or properly wired pair.

The load presented to the amp by those speakers that are correctly connected is all the amp will see. The singular existing connection I mentioned still in force on the unused pair can not present any significant load or sufficient resistance, by virtue of having only one lead connected. That circuit is what is called, an 'open' circuit.

There is no completed path for the supplied energy for which to enter the single wire connected speakers, and then return to the amp.

Even if the still attached single leg/lead runs directly into a cap, the effect will be negligible. Unnoticeable. Harmless. No noise.

The note as to having things in parallel doesn't stand up either, as all paralell circuits that in fact operate share both a common or actually a neutral, most likely a ground somewhere, and one hot or feed line, and in EVERY CASE a return path for the signal needs to be in place physically. this can be either the aforementioned common/neutral, or in some instances a common chassis ground or actual ground.

Although you say there is a lets say feed line.... where is the common or return in the speakers which do not have their returns connected, and only one lead afixed to them?

They don't... so coincidentally, they can not be seen as a load by the amp.

it is just like a broken filament in a light buld... break that filament... conduction ceases as the circuit is then, interuppted. A light switch is the same premise.

A return path must be in place always for current flow. AC or DC... it matters not.

If the benighn characters in a speakers xover contributed to some ancillary effect it would be so small as to require a metering device that measured in the thousandths to determine the actual value of it. it would be so infinitesimal as to not be contributary audibly, or mechanically in my proposed premise.

Essentially removing one or both leads at the speakers binding posts will accomplish the same result.

My method is simply easier and quicker, and will not harm either componenet… amp or speaker. If this connecting and disconnecting takes place while the amp is de-energized entirely.

When a person can figure out how to supply energy and return it using a single wire, it will revolutionize the electronic industry, and our world as we know it.

Talk about being 'single ended'! wow.
BlindJim, you sure took the long way 'round John's barn on that one, son! [smile]. But, he's absolutely right - some extra cable hanging off ain't gonna make any difference in the sound, it's not in the circuit...

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.
Two posts?

Try it with only one... that's what I'm saying. Just stick one lead of the cap onto one end of the battery and see how long it takes to charge that cap.

the Sun itself will burn out first.... but I suppose it might keep one off the streets.
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.

-- Al

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?

-- 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.”


No problem, bud.

-- 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.

-- Al