Can an unused cable break in?


I bought a new $300 Audioquest cable about a month ago, hoping to improve the sound of my Cyrus CD transport.  It didn't improve the SQ even after a little  break-in period.   I compared it to another transport system I had and it was quite inferior. So I stopped using the Cyrus but left it plugged in the wall for the month.
Lo and behold, I compared the two transports today and there was virtually no difference in sound between the two of them.
I’m listening to the Cyrus right now and am thrilled with it.
Either it’s my imagination, or the cable broke in while unused!  The difference isn’t subtle.
Is such a thing possible?


128x128rvpiano
@jea48 -  That you were able to even hint at things, such as wave-particle duality, dielectric permittivity and a, "conductor" actually being a wave-guide, without being assailed (thus far) by the resident pseudoscientists, is fascinating.                                                                                        KUDOS!
One of the things that is different about power cords vs ICs and speaker cables is there is always voltage present as long as power is supplied whether or not the device is turned on. If the dielectric is the primary thing that burns in, it seems to me it is the capacitive property of the cable that is affected and current shouldn't matter. What say ye?

I say ye R OTL. A medieval term invented by the Duke of Sandwich meaning "out to lunch": not all there.    

Voltage is a differential. It is in other words measured between two things. Across a circuit. Without which (a complete circuit) there is no voltage, no amperage, no capacitance, no nothing. (Try buying a meter that can read voltage, resistance, capacitance, inductance- you name it - without being connected to anything.) 

Yet another way to think of it, your imaginary voltage, which of the conductors is it "on"? A 120V circuit (there's that word again!) consists of one black "hot" wire and one white neutral wire. Normally voltage is measured across the black and white. But there is in principle no reason it cannot be measured across black and ground. Same 120V either way. But what about white/neutral to ground? Where is your 120V? Not there! So where is the voltage? Not there.  



MC, the parasitics complete the circuit. There are effectively small distributed capacitors along the length of the cable from line to neutral, line to ground, and neutral to ground. For a 2m power cable I would guess these would be in the neighborhood of hundreds of pfs.

My serious question still is as long as the cable is plugged in to AC, and these parasitic capacitors (caused by the insulating material as well as the cable geometry) are being charged and discharged why would the component need to be turned on in order to break in the cable? This is assuming the dielectric is the primary reason for the break in.

If a large current is required is required to break in a power cable it suggests a more complex reason for break in, perhaps one involving the metallurgy of the wire.

Many components cannot be turned off; the off switch merely puts it in standby mode. As long as the PC is plugged into a live wall socket and the device, there will be current flowing. Hence, it will burn in even with the CD player “turned off”. 

In addition, if the PC has capacitance across the individual conductors, the cable will be able to achieve some break in even without it being plugged into a component. It only needs be plugged into a live outlet. Most PCs do have capacitance across its conductors.
Off is off. No current when off. Zero. None. Nada. Zip. It is the very definition of OFF. Off in other words means off.

Now if you want to redefine off as some lesser but still non-zero amount of current is flowing through a closed- not open, closed- circuit, well then you can call it "parasitical" or "not on" but you cannot call it "off" as otherwise you have so destroyed the meaning of the word it loses all utility.

Use your words. Don't suck the life out of them. Don't be a parasite.