this post was submitted on 18 Mar 2024
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I never got the pipe analogy. Since liquid water can't be compressed, wouldn't the amperes be directly proportional to the volts and to the size of the pipe, assuming there are no air bubbles? Also, supposedly resistance only reduces current, but when I think of hair in a pipe, the pressure after the obstruction would also be lower (because pressure is directly proportional to the amount of water that flows)
Common misconception - it can, just not very much, so the volume change is tiny, and in practice, there's usually something else in the system that is changing volume by a larger amount- like air bubbles, or if there's anything elastic in the plumbing, it will stretch - but regardless, water absolutely can be under pressure.
You are correct, in electronics, resistance drops voltage (assuming the load is in series with the resistance). In fact, a cheap quick and dirty digital to analog converter uses a bunch of resistors to supply different voltages...
Resistance in wire creates a voltage drop, just like hair in a water pipe creates a drop in available pressure.
https://www.youtube.com/watch?v=X_crwFuPht4
AlphaPhoenix also has other fantastic videos explaining and experimenting with all sorts of interesting things.
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He expressed it wrong. Amperes is diameter of the pipe, how much volume (or charge) can be transferred per unit of length at a given pressure; Watt is the amount of water flowing out at the end, which depends both on pressure and diameter.
Shouldn't it instead be the sum of the kinetic energy of all water molecules that come out the other end per unit of time (ie. total amount of energy you use move your volume of water with a certain pressure in a second)?
Yeah, that's not simple anymore then
It would be. By ohm's law, I=V/R and R=V/I, so if V is fixed as V=1, then I=1/R, R=1/I, so it's is effectively the same thing, just measured in reverse.