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cable break in

had a friend ask me if after you break in/burn in your cables are they more or less conductive? i would have to say less conductive, but not sure why? does anyone have a good answer?
hemidakota
nsgarch's avatar
nsgarch
2,524 posts
 
Hemi: There are three things I'm aware of that affect or improve the conductivity of a given piece of wire. Two have to do with the crystalline structure of the metal:

1.) Working with the direction of the "wire draw", and honestly I don't know which direction has the better conductivity -- in the direction of the draw or against it.

2.) Cryo treatment, resulting in a more compact crystalline structure which improves electron flow through the metal.

3.) The third has to do with the cross-sectional geometry of the conductor -- ribbon vs. square vs. round, etc. and I don't think there is conclusive evidence regarding this issue.
hemidakota OP
20 posts
 
thanks Nsgarch. my friend thinks he knows it all and he will be in town this weekend. this will give me a little better of idea(being this is a little over my head)of how full of crap he is or isnt. thanks guys
inpepinnovations1e75
395 posts
 
Nsgarch, I don't know how you calculate current to a speaker for a known power consumption, but your example using your Levinson is incorrect.
Your example;
"An example would be my Levinson amp which will provide 400W/ch into my 4 ohm (nominal) electrostats, but at the loudest listening levels I can stand, it's only drawing 400W from the wall (or 3.3A) and it's only putting out around 150W rms of audio power, which at its 67V (26dB) gain, is only around 2.2A to the speakers (vs. 3.3A from the wall.)"

If 150 watts are being fed to a 4 ohm speaker, then I2=150/4=37.5. Therefore I (amperage) = 6.12 amps.
Clearly that is higher than the 3.3A pulled from the wall.

At any rate the amperage to the speaker will always be higher than the amperage from the wall to amp, because the voltage to the speaker for the same wattage as pulled from the wall is lower then the wall voltage, therefore the amperage must be higher to be of the same wattage.
nsgarch's avatar
nsgarch
2,524 posts
 
inpep: You are using the formula A = (square root of) W/R.
I am using the formula A = W/V. There is also another formula (ohm's law), A = V/R.

They should all yield the same result, so perhaps we're just plugging in the wrong numbers? Additionally, there are some cofactors when using AC, although I'm pretty sure the output of an amp has no phase angle.

There's a neat formula wheel at:

http://www.sengpielaudio.com/calculator-ohm.htm

and on the following page.
inpepinnovations1e75
395 posts
 
Nsgarch, I am plugging in the correct number. I don't think that the V (I don't know where you get 40 volts) that you are using is correct. Why use V at all, when we have R and P and a formula to calculate I?

24.6 V is the correct voltage at the amp to give 150 watts into 4 ohms.

As I have stated before, the amperage from the amp to the speaker will always be higher than the amperage to the amp from the wall, since the voltage applied to the speaker for the same wattage as the amp will always be lower than from the wall (120 V). I don't know of any speaker requiring more than 120 Volts for them to work!

With respect, Bob P.
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