2. amps-preamps
  3. 181058-155632

Power conditioner vs use of a audio grade outlet?

If I have a $7000 Power conditioner and really nice power cables is there a need for audio grade outlet? Maybe it's a dumb question but if I didn't have this other good stuff maybe I should get an audio grade outlet.

Pangea seems to offer a good outlet for about 100 bucks. 

jumia
immatthewj
2,094 posts
 

@erik_squires  , thank you for the explanation. 

immatthewj
2,094 posts
 

@erik_squires  , on the same subject, what is it that happens to the voltage carried by the white neutral wire when it reaches the neutral/ground bar in the electrical panel that makes it no longer dangerous?

erik_squires
16,164 posts
 
@immatthewj


The neutral doesn’t carry voltage, it carries current.  A voltage appears only when there is resistance to that current.

Look at the formula:

V = I (or A for amps) x R

In ideal neutral or ideal ground, R is zero, so V is zero. Lets imagine a circuit with a 5A load. Say an idling tube amplifier. The neutral and hot should be balanced, so the neutral wire has this voltage on it:

V = 5A x 0 = 0

Often we’ll see 2 to 4 or 5V on a decently wired neutral due to R not actually being exactly zero.

The problem is that there are lots of connections in a home and some are bound to come loose and increase the relative resistance.

Now lets see why N and G are different in the case of a bad conductor. Lets say instead of 0 resistance you have 10 Ohms on either the neutral or ground wire.

The ground wire carries no current normally, so if you were to touch the ground wire you experience:

V = 0A x 10R = 0V

The neutral on the other hand carries current, so with the same situation:

V = 5 x 10 = 50V

Now we have 50V, already a potentially lethal situation if one touches this with wet hands.

immatthewj
2,094 posts
 

Thank you for sticking with me on this, @erik_squires  , if you could bear with me just a bit longer:

if this was the ideal circuit in a utopian world of circuitry and there was NO resistance

(V=5A x 0R=0V)

then you are saying that although 5 amps of current are flowing through this white neutral wire, it is not dangerous because the formula dictates that 0 volts are created.  In this hypothetical circuit.  Correct?

But you are also saying that in reality this hypothetical circuit doesn't exist, and the example you used of what does exist is

v=5Ax10R=50V

so the white wire (with 10R) is not only carrying current, but because of the 10R it is also carrying volts?  Is that correct?

(And on an aside, doesn't this mean that the old saying, "it's not the volts but the amps that kills you" is not exactly correct, as I am getting out of this that amps without R would not be dangerous?)

But back to my question that prompted this explanation from you: in the real world circuit that you used for an example

V=5Ax10R=50V

what happens to that formula when it reaches the neutral bar via the white wire?  Is it that because of the massive ground cable that ALL resistance disappears at this point and the ingredients for the 50V are gone?

 

 

 

wolf_garcia
6,711 posts
 

I use high quality outlets for their obvious grabbiness as that's sort of comforting. Note that none of the "premium" AC cables I use display that sort of grabbiness on the component end...weird...

More to discover