this post was submitted on 18 Mar 2024
799 points (96.1% liked)

Technology

59030 readers
3053 users here now

This is a most excellent place for technology news and articles.


Our Rules


  1. Follow the lemmy.world rules.
  2. Only tech related content.
  3. Be excellent to each another!
  4. Mod approved content bots can post up to 10 articles per day.
  5. Threads asking for personal tech support may be deleted.
  6. Politics threads may be removed.
  7. No memes allowed as posts, OK to post as comments.
  8. Only approved bots from the list below, to ask if your bot can be added please contact us.
  9. Check for duplicates before posting, duplicates may be removed

Approved Bots


founded 1 year ago
MODERATORS
 
you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 1 points 7 months ago (5 children)

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)

[–] [email protected] 2 points 7 months ago

Since liquid water can't be compressed

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.

resistance only reduces current, but when I think of hair in a pipe, the pressure after the obstruction would also be lower

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...

[–] [email protected] 2 points 7 months ago

Resistance in wire creates a voltage drop, just like hair in a water pipe creates a drop in available pressure.

[–] [email protected] 1 points 7 months ago

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.

[–] [email protected] 1 points 7 months ago (1 children)

https://www.youtube.com/watch?v=X_crwFuPht4

AlphaPhoenix also has other fantastic videos explaining and experimenting with all sorts of interesting things.

[–] [email protected] 1 points 7 months ago

Here is an alternative Piped link(s):

https://www.piped.video/watch?v=X_crwFuPht4

Piped is a privacy-respecting open-source alternative frontend to YouTube.

I'm open-source; check me out at GitHub.

[–] [email protected] 1 points 7 months ago (1 children)

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.

[–] [email protected] 1 points 7 months ago* (last edited 7 months ago) (1 children)

Watt is the amount of water flowing out at the end

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)?

[–] [email protected] 1 points 7 months ago

Yeah, that's not simple anymore then