cable break in

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, a speaker's volume output varies with voltage and the power consumed is a function of the current and the voltage at the speaker for that voltage.
The amplifier takes in a signal of a certain voltage and increases the voltage to a level which the speaker can respond to. Of course the voltage at the output of the amplifier varies and isn't constant (otherwise the speaker would not get louder if the voltage did not increase) and the speaker draws what ever amps it needs at that specific impedence to produce sound at the given level. So a speaker that has 2.83 V/8ohm/90db sensitivity figure will need 28.3 V for a 100db output and 10 watts consumption. The amp will be suppling about 0.35 amps to the speaker and itself pulling 0.12 amps from the wall, still lower than the speakers.
Your figure of 40V is if the amp at full power at what ever wattage it is capable of supplying (or current) into the speakers at the speakers' impedence. If that power is 400 watts then the current is equal to 10 amps. The current drawn from the wall at 400 watts is, of course, slightly higher than 3.33 amps, still lower than the current going to the speakers.

With respect, Bob P.

Stanhifi, you can also say that any break in time is entirely an opinion and not based on any emprical data, as you put it.
Dave B's opinion of 2 weeks average time is no more erroneous than your opinion that there is no "average" time, and no more misleading than yours.
Salut, Bob P.

Oh, Stanhifi, its the "cable designers'" opinions, then, and other "anecdotal" evidence or experiences that render others' opinions wrong? Interesting, but not very informative, in my opinion, of course!
Bob P.

NSGARCH, are you sure that the current in speaker cables is lower than PCs?
If one is feeding 16 watts to a 4 ohm speaker, you have 2 amps of current. The power amp feeding the speaker, even at 75% efficiency is using about 21 watts, therefore barely 0.2 amps are running through the PC.
I doubt that you use power amps at full power to break in the PC and even then a 120 watt amp would have only 1 amp (2 amps if both channels running) as current, certainly less than a speaker cable running at only 16 watts.

I think that the statement about PC breaking-in more quickly than speaker cables due to higher current, needs to be re-examined.
Salut, Bob P.

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.