B Switch-to-lightbulb wire delay

[underworld]

Summary: Question about expectations related to electron flow and conductivity.

Here's a question I have... imagine a scientist in a lab. In front of him there's a battery (with typical red/black +/- poles). Behind him is a normal light socket. Now, between the battery and the light socket there are two very long lengths of wire. (Let's say, about 11 million miles or so).

So, the scientist starts the experiment:

1) He attaches both wires to the socket and the battery. He screws in the bulb. What happens?
Now, we all know what "should" happen; or at least what the casual expectation is - which is that the bulb lights up.

But, the question is - when? (Fyi, 11 million miles is the distance light travels in 5 seconds).

Does it light immediately? If so, what mechanism causes it? Shouldn't there be at least a 10 second delay?
Does it light after 5 seconds? After 10? Never?

2) Now, let's say he unscrews the bulb and unhooks all the wires. This time he starts by screwing in the bulb. He then attaches one side of the socket to the wire, and attaches the other end of the same wire to the battery + pole. He waits 6 seconds, then attaches the other wire to the socket and the - pole. Again, we fully anticipate the bulb lighting. But again, when does it light up?

I'm trying to understand what physically happens to cause the electricity to flow. If you were to replace the above with a water pipe analogy, you can determine that source water creates a pressure in the pipe, which, over time, will reach an equilibrium in the pipe. The process of equalizing when more water is added is what causes the flow. Alternately, if you engineer the pipes just right, gravity will pull the water through. So, is there a "gravity" that pulls electrons through un-level wires at some speed? Or is there an "electron pressure" that equalizes in the wire at some speed? You don't expect the water poured into one end of the pipe to immediately come out the other end.

And what are the limits of this? For example, let's make the wires the size of the milky way (about 52,000 light years). So, does the bulb still come on? If so, when? (in other words, is it just a matter of the time it takes to propagate through the wire? or what if you run out of electrons to before any reach the other end?)

 

[andrewkirk]

In the first case the bulb will illuminate immediately, because the contacts of the socket already have a potential difference between them.
In the second case the bulb illumination will be delayed because the socket contacts have no potential difference up to the point when he connects the second contact in the socket to the battery's negative pole. The delay will be broadly commensurate with the length of the wire that makes that connection.

Also, electrons from the battery don't need to reach the other end. It's a wave that reaches the other end. The electrons at the battery end push/shove those ahead of them in the wire and that shove propagates to the other end of the wire and when it reaches it, it pushes some electrons out of the wire and into the bulb.

 

[Dale]

Let's say, about 11 million miles or so

So that is about 44 kOhms for each wire. How many volts is the battery, what is the resistance of the bulb, and how much power does it take until it lights up and burns out?

 

[Ibix]

See this discussion.

 

[tech99]

In the first case the bulb will illuminate immediately, because the contacts of the socket already have a potential difference between them.
In the second case the bulb illumination will be delayed because the socket contacts have no potential difference up to the point when he connects the second contact in the socket to the battery's negative pole. The delay will be broadly commensurate with the length of the wire that makes that connection.

Also, electrons from the battery don't need to reach the other end. It's a wave that reaches the other end. The electrons at the battery end push/shove those ahead of them in the wire and that shove propagates to the other end of the wire and when it reaches it, it pushes some electrons out of the wire and into the bulb.

The bulb will initially obtain its energy from that stored in the capacitance of the wires. If the resistance of the bulb is less than Zo of the wires, then a wave travels back to the battery, causing it to see a lower resistance and supply more current. The converse occurs if the bulb resistance is higher than Zo. In both cases we will then see the bulb cycling in brightness as waves travel back and forth on the wires, finally settling to a steady value.

 

[jbriggs444]

If the resistance of the bulb is less than Zo of the wires, then a wave travels back to the battery, causing it to see a lower resistance and supply more current. The converse occurs if the bulb resistance is higher than Zo.

The resistance of the bulb is guaranteed to be less than that of the open circuit that pre-dated it. The initial wave from bulb toward battery is going to cause the battery to see the resistance lowered (from infinity) either way.