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Suppose you have an AC circuit and you connect a light "in parallel" on the live wire. Essentially a loop on the live wire. How does electricity flow in such circuit? The light would turn on, but I struggle to understand the potential differential.

I've drawn this horrible schema to (hopefully) clarify my question.

enter image description here

Sorry for making you look at that...

The potential differential between B1 and B0 is obvious, but between B3 and B4? Between J1 and J2 there is potential differential, but so there is between B4 and J2.

I am aware of my lack of knowledge, but I cannot measure the ridiculousness and/or simplicity of my question, so be ruthless, but please someone point me in the right direction.

Much appreciated.

P.S. this question is based on a DIY project that I though would explode but didn't :)

UPDATE

The diagram above is an over-simplification which hides my real question: how is this not blowing up? Thank you to all of you who answered, I'm going to try to be more specific.

enter image description here

In this diagram you can see two lights. R1 is connected on the standard (at least from my experience in the UK) loop circuit. The live wire to R1 is broken by a switch using the neutral cable to connect back to the light. When S1 is ON, R1 is on and the blue/neutral cable coming from S1 is effectively live.

I wanted to try to get another light from the only cable I had access to (I know, don't touch if you don't know...), so I hooked another switch and a light to the cable before realising that the neutral is not actually a neutral.

As I hope the schema shows, now I have an extra switch S2 that bypasses S1 turning on R1 and R2. The behavior I cannot explain is that when S2 is ON, S1 can be flip ON/OFF without apparent effect.

It might be worth mentioning that both lights are LED so extra electronics might be in place.

Tomorrow I'll try to take some pictures but sadly most parts of the cable are inaccessible which theoretically could be hiding some other connections or components.

Thank you again to everyone!!!

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    \$\begingroup\$ That schematic is far better than a lot of the ones we get on here. \$\endgroup\$
    – Hearth
    Oct 16, 2022 at 18:54
  • \$\begingroup\$ The potential difference due to time lag is pretty small at 60 Hz. So in general, the potential from J1 to B3 can be ignored, the potential from J1 to J2 can be ignored, the potential from B4 to J2 can be ignored, and thus the potential from B3 to B4 is zero. And no current will flow through the lamp at B3/B4. \$\endgroup\$
    – user57037
    Oct 16, 2022 at 21:09
  • \$\begingroup\$ I now see your comment where you indicate that you did this and have a light that IS turning on. It might be helpful to see a photograph of your whole setup if that is possible. Also, the specific type of light. Also, the wire from L to B1, what size wire is it, and how much current is flowing through it. Finally, the length from J1 to J2. \$\endgroup\$
    – user57037
    Oct 16, 2022 at 21:12
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    \$\begingroup\$ The reason for the photograph is that there may be some important detail you didn't realize was important. We may spot that in the photograph. Otherwise it can be like playing a game of 20 questions until we figure out the right thing to ask about (or get discouraged and give up). \$\endgroup\$
    – user57037
    Oct 16, 2022 at 21:14
  • \$\begingroup\$ Looking at the updated schematic, I don't see how R2 could be on when S1 and S2 are both on. Maybe you have current flowing through one of the ground wires or something. Andy is right. The update completely changes the question. \$\endgroup\$
    – user57037
    Oct 16, 2022 at 23:50

1 Answer 1

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The op has now altered the question against my recommendations and has now invalidated the authenticity of my answer.

The light connected to B3 and B4 will not light.

Unless you are considering transmission line theory, you must assume that the potential on an interconnecting node will be equal at all points on that interconnection.

If you are considering transmission line theory then the light will illuminate very, very briefly; imperceptibly so. A whole lot more imperceptible than these very small words

$$$$

In other words, at any moment in time if "L" has 100 volts on it then so will B1, B3 and B4.

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  • \$\begingroup\$ Many thanks for your answer. So then I have something else in my system that I naively simplified out, because my light is most definitely on. I'll update the question with the more detailed diagram. \$\endgroup\$
    – lexotero
    Oct 16, 2022 at 21:01
  • \$\begingroup\$ Don't update, ask a brand new question. This isn't a forum it's a question and answer site. This question is now done as far as I'm concerned. \$\endgroup\$
    – Andy aka
    Oct 16, 2022 at 21:33
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    \$\begingroup\$ +1 this answer correctly answered the original question. The OP has now realized she to feedback from this answer and other input that their original impression was wrong. They have thus altered the question to reflect their current understanding. This means that this correct answer dors not now match the revised question. How to best deal with such changes is probably worth considering. \$\endgroup\$
    – Russell McMahon
    Oct 16, 2022 at 22:11

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