this post was submitted on 03 Aug 2026
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No Stupid Questions

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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?

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[–] Natanael@infosec.pub 2 points 1 day ago (1 children)

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

[–] akunohana@piefed.blahaj.zone 1 points 1 day ago* (last edited 1 day 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. πŸ˜…

[–] fullsquare@awful.systems 14 points 1 day ago (2 children)

It's easier to see that this circuit is rather symmetric when you redraw it

Then with both switches closed blue wire (the one in the middle) should only carry small current. If both halves are exactly the same it should be zero

In general when circuit doesn't include nonlinear components (semiconductors most commonly) you can sum two valid solutions and you'll get another valid solution. Note that when one switch is closed, current in blue wire goes one way and when the other is closed, in opposite way

[–] neidu3@sh.itjust.works 6 points 1 day ago

Thanks for drawing it so I didn't have to; I got nerd sniped from seeing what seemed like two identical circuits with shared ground

[–] akunohana@piefed.blahaj.zone 5 points 1 day ago (1 children)

This was an amazing explanation. I'm saving it. Thank you so much! I wanna learn writing schematics or whatever they're called so badly but...

[–] SuiXi3D@fedia.io 3 points 1 day ago

Check out the game Spintronics. It’s kind of expensive for the whole set, but absolutely worth it if you want to play around with circuits without shocking yourself.

[–] marcos@lemmy.world 9 points 1 day ago (2 children)

On a real circuit where every component is different from each other, yes there will be current there.

On that ideal circuit, try specifying the characteristics of those components and writing down how much current goes through each segment. That will let you understand what is happening.

Thanks! In this book, for better and for worse, we are only trying to understand binary logic and how to express parts of that (AND and OR) with serial and parallel switches, so we're not really touching on the electrical characteristics of the individual components. But I'll keep your reply in mind going forward! 😊

[–] Successful_Try543@feddit.org 2 points 1 day ago (1 children)

Try visualizing the current by using arrows of a proportional length directing from+ to - (or in opposite direction).

[–] marcos@lemmy.world 1 points 1 day ago (1 children)

If the book is trying to explain an Wien bridge, the OP will need to actually put numbers there. Just looking at the sizes won't be enough.

It doesn't matter if you use numbers or scaled! vectors.