# A basic light circuit.

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If you did see an electron from the tv

One swallow doth not a summer make nor one electron a spray.

But then you asked for a spray.

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Yes, but I can't claim expertise, so I might still be wrong about it.

What if we remove lamp L2, creating a break in the circuit. Will L1 and L3 still light up? For how long?

There should still be a current until the potential has been fully equalized to battery voltage to the furthest end of both wires.

(The wires are like a huge capacitor and they will require lots of movement of electrons to even out the charges.)

What happens if you move the switch to immediately before L1?

Then the potential is already at battery voltage at the switch when it gets closed, so L1 lights up immediately.

(L3 would light up as before, but L2 would now also be closer to the switch and lights up close to L3.)

Find a long skinny straw. This straw is so skinny that you can just squeeze an electron in the end of it. Push more electrons in it, and make a string of electrons.

Now form the straw in a circle. Remember that charge is repulsive. Physical confinement is necessary.

Can only one electron move? No. Physical confinement and restraint by the repulsive fields.

The electron shells are not physically touching each others, since they are repulsed by their charges like you agree to, but that also means that an outside higher voltage can push them closer together in one end for a short moment before the potential has evened out through this hypothetical straw of yours. The electrons must start to move in one end before they all can move forward together.

If you have a light year long iron bar and hit it at one end with a hammer it won't move immidiately at the other end, there will be a pressure wave moving through the iron bar at the speed of sound in iron that eventually push the other end forward.

A system (loop or circuit) has to be set up and maintained for current. I can't spay electrons out across the room, like I can with water.

Air is not a good conductor but with high enough voltage you can force a current through it, lightning does this every day on Earth.
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There should still be a current until the potential has been fully equalized to battery voltage to the furthest end of both wires.

(The wires are like a huge capacitor and they will require lots of movement of electrons to even out the charges.)

This was your response to my question: "What if we remove lamp L2, creating a break in the circuit. Will L1 and L3 still light up? For how long?"

I'm still trying to wrap my head around this. If there is still a current then I take that as L1 and L3 will still light up and stay lit "until the potential has equalized".

So while they're lit, they are fully functional in an open circuit (doesn't make sense, but I'll play along). And since the circuit is open the wires are independent of each other. So if they're independent then it shouldn't matter if you completely remove one wire. But now you're left with a single wire connected to a single battery terminal and it lights up fine (until the potential has equalized).

That makes no sense... what am I missing?

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This was your response to my question: "What if we remove lamp L2, creating a break in the circuit. Will L1 and L3 still light up? For how long?"

I'm still trying to wrap my head around this. If there is still a current then I take that as L1 and L3 will still light up and stay lit "until the potential has equalized".

So while they're lit, they are fully functional in an open circuit (doesn't make sense, but I'll play along). And since the circuit is open the wires are independent of each other. So if they're independent then it shouldn't matter if you completely remove one wire. But now you're left with a single wire connected to a single battery terminal and it lights up fine (until the potential has equalized).

That makes no sense... what am I missing?

Yes, as long as there is a current flowing through the lamps they will shine but if the circuit is open at L2 then they will fade when the potential equalizes.

The wires are not independent of each other, if you completely remove one wire then you also completely change the setup.

The two wires are like the opposite sides of a huge capacitor or many small in parallel along the wire length. Place a symbolic capacitor at the end of the wires, in parallel with L2, to see how this affects the circuit. During the creation of the circuit with L2 in place and the switch open, both sides of the capacitor will be charged with the negative side of the battery and when the switch is closed the battery will start to charge the positive side of the capacitor with its voltage.

Removing L2 will not stop the charging but without L2 there will be no current when the charging has finished.

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Thanks Spyman for the explanation - I'm still processing the info. The following will help me:

I removed the switch and L2:

┌─────────────────────────────────────────────────────────┐

I │

+ │

B]

- │

I │

└──────────────────────────(L3)───────────────────────────┘

When you connect the battery, with L3 light (even temporarily)?

┌───

I

+

- │

I │

└──────────────────────────(L3)───────────────────────────┘

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• 2 weeks later...

When you connect the battery, with L3 light (even temporarily)?

That would depend on the potential of the circuit and the charge of the capacitor at the end of the wires. If the battery is connected long enough for the potential to even out and if the battery later is reconnected with the polarity switched, then L3 will light up.

For the second circuit the capacitor is almost entirely removed and to small to light up L3.

Edited by Spyman

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