Electronics > Beginners
Negative voltage ?
rstofer:
--- Quote from: ebastler on February 20, 2019, 08:03:57 pm ---
So why not use the water model, and even the colloquial term "negative pressure", in the regime where it works nicely to explain voltages, currents, and resistances? As I said in earlier posts, your explanations are "more correct", i.e. they describe a model that will work in a wider range of conditions. (But not under all conditions either.) But that does not make the simpler models wrong or useless within their domain.
--- End quote ---
Because I can demonstrate the arbitrary nature of the reference point (the purpose of this thread) with a couple of batteries. Something that can be done on a tabletop with 2 batteries and a DMM. There is simply no need to get into a discussion of electron flow, majority carriers, charge and a host of other details just to demonstrate something as trivial as 'negative' voltage. And we're describing electrical kinds of things using electrical kinds of things. I think it is terribly important to understand the arbitrary selection of the reference point. This understanding will be vital when the user gets to circuit analysis.
All models have a domain. I don't prefer the water analogy but it has the feature that you can demonstrate it out on the lawn on a warm sunny day. What is overlooked is the discussion of the domain of the water analogy. If we double the current through a resistor, we expect the voltage drop across the resistor to double. That doesn't happen with water.
I prefer to discuss Ohm's Law using batteries and meters.
IanB:
Why the water analogy works for negative voltages:
Voltages are usually measured as a potential difference between two points. Voltages are not absolute single point measurements. You don't measure voltages with a single lead on a voltmeter, you need two leads and two measurement points.
Similarly, consider water flowing through a pipe. You can measure the differential pressure of the flowing water between two pressure taps on that pipe (label them 1 and 2). You will read a differential pressure in pascals (or psi), for example. If you measure from 1 to 2 you may measure a positive pressure difference of say +5 Pa. If you measure from 2 to 1 you will see a reversed or negative pressure difference of −5 Pa.
Analogies work as long as you keep the rules the same and maintain analogous circumstances. If voltages are potential differences, then by analogy you must treat pressures also as potential differences.
IanB:
--- Quote from: rstofer on February 20, 2019, 08:48:13 pm ---If we double the current through a resistor, we expect the voltage drop across the resistor to double. That doesn't happen with water.
--- End quote ---
It may do. If you make the "resistor" a porous block with many small passages in it you will find the pressure difference is proportional to the flow rate. (The system will be operating in the laminar flow regime.)
ebastler:
--- Quote from: rstofer on February 20, 2019, 08:48:13 pm ---If we double the current through a resistor, we expect the voltage drop across the resistor to double. That doesn't happen with water.
--- End quote ---
Sure does. Laminar flow of a non-compressible fluid through a tube follows Poiseuille's law, with flow rate = C * pressure difference. The constant C depends on the tube geometry and the viscosity of the fluid, and plays the role of resistance.
So if you want to force twice the flow rate throug a given "resistor", you have to apply twice the pressure. It's a direct analogy to Ohm's law.
Jwillis:
I was bored .But the water analogy still holds water .As IanB explains it's all in potential whether positive or negative relative to a common steady state.
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