With the negative side of the battery attached to the chassis, more electrons are available to the various devices, scattered all over the vehicle, to be powered, and the control is achieved on the return, not the source.
With the negative side of the battery attached to the chassis, more electrons are available to the various devices, scattered all over the vehicle, to be powered, and the control is achieved on the return, not the source.
I think you have a big conceptual misunderstanding here. It is irrelevant from an electrical point of view whether the positive side of the battery or the negative side is attached to the chassis. Everything will work just the same either way. There is no "source" or "return" in a circuit.
If you connect a conductor to the metal frame of a vehicle, the size of which is tens to hundreds of times larger than a 4 mm wire, you must agree that the voltage drop from one end of the vehicle frame to the other will be less than it will be on the 4 mm wire, especially as current requirements increase. It is. Not by very much, but it is. If the voltage drop is less on the frame, then the source of the power (Electrons), (Electron theory) available must be higher than it would be on the 4 mm wire.
because the bean counters didn't care, they were only concerned with consensus of the scientific community
then the source of the power (Electrons)
In the late sixties, the Hole Theory started to gain some acceptance as a "Out of the Box" way to explain certain things in a different way that some found easier to understand, without getting too deep into the physics
because the bean counters didn't care, they were only concerned with consensus of the scientific community
In the late sixties, the Hole Theory started to gain some acceptance as a "Out of the Box" way to explain certain things in a different way that some found easier to understand, without getting too deep into the physics
Each conductor has resistance, and the resistance of a car body is indeed likely to be less than that of a length of wire, but the choice to connect it to the terminal marked '-' rather than the terminal marked '+' doesn't make any difference whatsoever to the voltage across the load.
Why? Because the load, the wire and the car body are all electrically in series.
The voltage across the load equals the voltage across the battery, minus the voltage drop across the wire, minus the voltage drop across the car body.
Short-form answer: Physics works at both the macroscopic and microscopic level.
"Current" is a macroscopic phenomenon, and can flow in either direction across a specific boundary. Polarity definitions (on current, voltage, etc.) must be consistent and agreed to make useful calculations.
"Electrons" are a microscopic phenomenon. Charge carriers forming current can be electrons with negative charge, holes with positive charge, protons with positive charge, etc. The product of charge density (scalar) times velocity (vector) times boundary area gives current density (vector).
The accepted definitions of polarity are experimentally consistent, and there is no need to change them on a whim.
In my physics education, I have "sensed" (if not "seen") electrons (and other charged particles).
In my physics education, I have "sensed" (if not "seen") electrons (and other charged particles).
1. Millikan oil-drop experiment: measured discrete changes in charge on small oil drops. (At the University of Chicago, I used antique equipment originally built for Professor Robert Millikan himself.)
2. Shot noise in vacuum diodes: statistics of individual electrons vs. diode current.
3. Beta decay: counting individual electrons emitted by nuclear disintegration.
4. Ion beams. Counting individual protons that have been accelerated through a vacuum to a detector.
Of course, one does not see the electrons involved in current through a copper wire. The current is not single electrons zinging down the wire, but a slight tendency for each of a zillion electrons to move in the appropriate direction.
... and you still didn't get it right.Each conductor has resistance, and the resistance of a car body is indeed likely to be less than that of a length of wire, but the choice to connect it to the terminal marked '-' rather than the terminal marked '+' doesn't make any difference whatsoever to the voltage across the load.
Why? Because the load, the wire and the car body are all electrically in series.
Exactly.
But you missed mentioning one - rather critical (especially lethal to your argument) - part of the circuit:QuoteThe voltage across the load equals the voltage across the battery, minus the voltage drop across the wire, minus the voltage drop across the car body.
A point, I might add, you completely ignored in your response to AndyC_772's post.
... and attempt my measurements across those two wires with current flowing.
Your statement:... and attempt my measurements across those two wires with current flowing.is vague.
If you are going to measure the potential difference from one and of wire 'A' to the other and the potential difference from one and of wire 'B' to the other, then your experiment is pointless. The answers can be given using Kirchhoff and Ohm's laws.
What really matters is the voltage across the load.
But your fundamental premise is simply incorrect, so any experimental processes have to be questioned as to how relevant they are in demonstrating the claims.
If there is no clear relevance, then trying to apply some quantitative results is nothing more than useless - unless the intention is to try and bamboozle. Either way, nothing is proven.
The premise of the experiment is that if you use one large conductor and many smaller conductors for multiple devices that are used at the same time, like in an automobile, that the VD would be different on the smaller conductor than on the larger conductor.