johnallg wrote:aceonbass wrote:ken_j wrote: Dane you can't think of the wiring as a one way street, electrons follow the easiest path.
Since the "Direct Out" was wired to the same lugs that the pickups were wired to, shouldn't the electrons have gone right back out instead of through any other part of the circuit?
But the tone and volume circuit was still there. The output jack (and the circuit plugged into it) was in parallel with the tone/volume circuit.
If you took the pickup lead to the center of a switch, put the jack on one side of the switch, and the tone/volume circuit on the other side of the switch, you could isolate the pickup to that output.
Yes, the electrical signals don't really care what order things are connected in when they are connected with wires that have practically no resistance or inductance, which is the case with short wires.
Water is often used as an analogy for electric current, but in this case, it doesn't really work. Water in a pipe will flow out of a hole (the output jack) first if it hits that hole first, but this is NOT the case with electrical signals. Since they travel essentially at the speed of light (they don't really, but it's dang fast), they get almost everywhere all at once and go right around the jack and hit the tone circuit too. They also get to the output jack and go out of it too.
Notice I said "electrical signals." We shouldn't really talk about electrons flowing around anyway...
Here we go again...
WARNING: Techno-Nerd™ Alert!!!
Contrary to popular belief, electrons don't go flying through a wire at the speed of light. In reality, it's a lot more like those toys that were all the rage a few years ago (or was it decades?), where you have a series of steel balls hanging from a frame by V-shaped strings. You pull one ball at one end up and let it smack into the series of balls and the one on the other end flies up with all of the ones in the middle just sitting there.
The actual "drift speed" of the electrons can be found with the following equation:
v = I/(nAq)
Where
I is the current in amps
n is number of free electrons per cubic meter of the conductor
A is the cross sectional area of the conductor in meters
q is the charge in "Coulombs"
So, the electron velocity for a current of 3 amps through a 1mm diameter copper wire is (Trust me on the numbers, I could show where they come from too, but that would bore even normal Techno-Nerds™. Any Mega-Techno-Nerds can PM me for the info, but then, they can probably calculate it themselves

):
v = 3/(8.5e28 * 7.85e-7 * -1.6e-19) = -281e-6 m/s = -0.281 mm/s
About 1/4 millimeter per second?!?!?!? WOW, that's just cookin' ain't it! Snails look like they're jet powered in comparison.
Electrical signals don't even go at the speed of light, they travel a bit slower, dependent on the "velocity factor" of the conductor, which is usually on the order of 60% to 99% of the speed of light. But that's a discussion for another time (relieved, aren't ya

).
END of Techno-Nerd™ Alert!!!
Anyway, what really matters is the signal frequency and what effect the attached components will have on it. You can think of the signal as getting everywhere all at once and when it hits the tone circuit, the cap will bleed off the high frequencies as soon as they get there (which is pretty much instantly) so none of those are left to go (be?) anywhere else. And hence, the tone circuit still affects the circuit, even if it's on wires a couple of feet away from the pickup and output jack...