From one of my favorite books, Code: The Hidden Language of Computer Hardware and Software by Charles Petzold:

7dDVB9Hvgu9qKxP.png

Here, he explains how to setup switches and lightbulbs in two locations with the purpose of communicating by turning the lights on and off according to some previously established protocol/code.

My question: is he oversimplifying - because that isn’t the point of this chapter - which route/path the current would take when both switches are closed? As in:

poKdbhPdpYAALnJ.png

Would the wire, which I awkwardly marked blue, also become charged and/or carry current? Or would there be no current because the cathodes of the two adjacent batteries cancel out any difference/voltage in the “blue wire” that would otherwise cause a current?

  • Natanael@infosec.pub
    3·
    3 days ago

    If you want to be very very hyperpedantic, when the circuit gets closed there is initially a current everywhere reachable and an electric field absolutely everywhere. It then settles down in passive DC circuits when equilibrium has been reached, and with an equal but opposing voltage (“electric field pressure”) on both sides of this wire the current is zero

    It’s like a water pipe with stale water between two water pumps

    • printf("%s", name);@piefed.blahaj.zoneOPEnglish
      1·
      3 days ago

      Thanks! I was just waiting for somebody to point that out! Namely, that there should be a momentary… “Effect”? Of an electric field in all parts of the circuit before it equalizes. Thanks! 😊

      One of these days, I’ll pull myself together and read up on AC too. 😅