cable break in

Bob, let's say you've got a SS amp which draws 225W at idle but draws 1200W at rated output (200W/400W into 8/4 ohms). So at let's say a nominal (audio) output of 100W/200W into 8/4 ohms, it's drawing around 600W from the wall. W/V=A, or 600/120=5amps.

Typical amplifier nominal output voltage is around 50V for a 20dB voltage gain (over the preamplifier output) and which is a pretty loud listening level if the speakers are reasonably efficient. Again using W/V=A, you get 2A (@100W) for an 8ohm speaker and 4A (@200W) for a 4 ohm speaker, unless I'm way off somewhere.

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.)

The example you gave assumes an 8V output voltage which would be only about a 3dB voltage gain for the amp. Not very loud, even with a super-efficient speaker.

so you guys agree there is no change in conductivity? only a change in the insulation?

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.

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

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.