# I How exactly does electricity "flow" through a circuit? (light year long circuit thought experiment)

1. Nov 15, 2018

### FreeForAll

This thought experiment has been bugging me for a very long time - I'm hoping someone can help me understand it.

Premise:
- Simple DC electrical circuit consisting of a battery, wire, light bulb, and another wire to complete the circuit.
- Each wire has a length of 1 light-year (yes, very long).
- The wires have zero resistance.
- I'm assuming that electricity travels at the speed of light in the wires.

Question:
From when the battery is connected to the circuit, how long will it take for the bulb to illuminate?

I keep thinking the answer must be either 1 year or 2 years. But both answers cause me problems, so expect follow up questions :)

Thanks!

2. Nov 15, 2018

### Grinkle

IMO better to simplify the experiment to radiating power at a photodiode that is 1 LY away and at rest with respect to the energy source. I don't think there is any need to bring Ohms law into it.

I think -

If you send a photon at a target that is 1 LY away at and rest wrt to you, and you wait for a photon to come back to you from the target, it will take 2 years by your clock to see the return photon. You will calculate that it took 1 year since you fired your energy source at the target for the target to have emitted that photon, after you account for how long it took the emitted photon from the target to reach you.

3. Nov 15, 2018

### sophiecentaur

If these wires are close together and more or less parallel then the first sniff of Energy from the battery will arrive after one year (times a factor which will depend on the wires and their insulation - possibly half as long again.
What is being carried by the wires is a 'disturbance in the Electric and Magnetic fields (a guided EM wave). When the bulb starts to get an increase in E field across it, it will conduct a current and get hot.

4. Nov 15, 2018

### FreeForAll

I appreciate the response but I think that alters the experiment too much. In my mind sending a photon is too different than a battery that has 2 terminals.

Also I should have been clear that I'm not concerned at the time it takes for the light to travel from the bulb to the observer's eyes. Only the time it takes for the bulb to illuminate.

5. Nov 15, 2018

### FreeForAll

Let's call it 1 year (I will ignore the factor depending on the wires and insulation). Does the disturbance travel down a single wire, to the bulb, then returns down the other wire back to the battery?

6. Nov 15, 2018

### Grinkle

It is different - but it removes the need for parts of the thought experiment that have no basis in physics, a zero resistance wire, and assuming that propagation through this medium that has no basis in physics is c.

7. Nov 15, 2018

### CWatters

Im a bit rusty but...

If you model it as a transmission line then when you switch it on a voltage step will be launched down the wire. If the source impedance is the same as the transmission line the voltage step will be half the voltage of the source. That voltage step will take 1 light year to arrive at the "bulb".

What happens next depends on the impedance of the "bulb" and the line. The bulb might light at full brightness or at less than full brightness.

The voltage step might be reflected up and down the line several times before the voltage at the bulb matches the source.

So I reckon it will light up after 1 light year but perhaps take a few to reach full brightness.

If the source has zero impedance I think the initial voltage step will be same as the source. But could exceed the source voltage if reflected at the bulb.

8. Nov 15, 2018

### sophiecentaur

If the source has zero impedance, it is a voltage source and nothing can change its output volts. 100% if the reflected pulse will be reflected back to the other end.

9. Nov 15, 2018

### Tom.G

10. Nov 16, 2018

### FreeForAll

The "Christmas Lights" thread is very similar, but most of the responses discuss the practicalities of such an experiment and not the actual results. And then they get into details that are way over my head. That is why I tried to simplify it with a single bulb.

In my experiment, I'm looking for a simple answer measured in time (how long will it take for the bulb to illuminate)? It's OK if it doesn't illuminate at 100%, just at what point does the bulb say "hey, here's some electricity, let me do something"?

11. Nov 16, 2018

### Staff: Mentor

That will be determined by the speed at which changes in the electric field propagate in the wire; that's as close as we can come to the "speed of the electricity" that you reference in the first post of this thread.

That speed will depend on many details of the setup (and of course the assumption of zero resistance in a conductor one light year long is completely unphysical) so there's no exact answer, but it will be an appreciable fraction of the speed of light.

There aren't many interesting things that a light bulb can do (it's basically a variable resistor that glows at some voltage levels) so you might want to look for a different effect: does an oscilloscope with one lead connected to each side of the bulb register something interesting?

Last edited: Nov 16, 2018
12. Nov 16, 2018

### DaveC426913

I'd hazard to guess the OP is less interested in what the bulb does at what time than whether or not a circuit that has not yet completed can actually do some work.
We were taught that a circuit has to be closed for electricity to flow, but this scenario suggests that electricity can flow even though it is not a complete circuit.
(I guess it is somewhat akin to a ground wire - i.e. an infinite sink.)

13. Nov 16, 2018

### FreeForAll

You hit the nail on the head!!! If a closed circuit is not required then why does a battery have 2 terminals if it can be grounded elsewhere?

14. Nov 16, 2018

### DaveC426913

In any real word environment, the propagation of electricity might as well be infinite. Which means there is effectively no delay. If one end is disconnected, the whole circuit drops to zero potential "immediately".

The point is not that a circuit must be closed, but that there is a non-zero potential from one point in a setup to another. That's the key.

As long as current is able to flow from the terminal into the wire, it will continue to flow until the potential reaches zero.

15. Nov 16, 2018

### Staff: Mentor

However, the potential will reach zero very quickly if there isn't a return path to form a closed circuit. Without that return path the only thing sustaining a potential difference is the capacitance of the wire, which is very small for any realistic configuration. It's not small for a wire a light-year long, so the current can flow for an appreciable fraction of a year before an oscilloscope at the far side will twitch.

(DaveC426913 understands this perfectly well - I'm elaborating on his post not disagreeing with it).

Last edited: Nov 16, 2018
16. Nov 16, 2018

### Grinkle

This isn't a great analogy, but its simple -

Consider a water pipe that is connected at one end to a paddle-wheel with a return pipe on the other side of the paddle wheel. There is a faucet on the entry point of the pipe. The return pipe can either be capped or left open as well. Ignore any air in the empty pipe. In this thought experiment, the return pipe is capped at its end, not open.

If your turn on the water at the start point, water will flow to fill the pipe. As it moves past the paddle wheel after filling the first section of pipe, the wheel will turn. Now, if the return pipe is capped at the end, as soon as the water fills the entire circuit and hits the cap, water will cease flowing and the paddle wheel will cease to turn. This is like an open circuit. There was a temporary condition (called a transient) when the water was first turned on and water filled the empty pipe and in doing so it did some work on the paddle wheel. The cap at the end of the return pipe is like the ground terminal of the battery not being connected to the return wire. Once the entire system comes to the voltage of the positive battery terminal, if there is no connection to the low voltage point (voltage being pressure, in the water analogy) then no more work is being done. Filling the pipe with water is the capacitance of the wire filling with electric charge @Nugatory discussed post 15.

17. Nov 16, 2018

### tech99

This is my understading of what happens. It does not matter that the line is long - it is the same as every circuit.
When the connection is made to the battery, say by a switch, an impulse consisting of a step function travels in both directions towards the bulb. On one wire it travels towards the bulb, and on the other it travels via the battery before going on the other wire towards the bulb. The two impulses are 180 degrees out of phase relative to the rest of the Universe. The impulse does not go to the bulb on one wire and back on the other. The impulse travels as a guided EM wave.
The current entering each wire at the switch is the battery voltage divided by the characteristic impedance of the wire.
When the pulses travel along the wires, each wire can carry the pulse alone if required, by surface wave action. The presence of the two wires allows the usually expected TEM mode to exist as well.
As the pulses travel, they will lose energy in the resistance of the wire and to radiation.
If the wire makes a rapid change in spacing anywhere, or enters a dielectric etc, the characteristic impedance will alter. Some energy will then be reflected back towards the switch. When the impulses reach the lamp, the lamp will be excited by a generator having an impedance equal to the characteristic impedance of the wires. If its own impedance differs, then some reflection will take place. Following the initial impulse, the bulb will be powered by the flat part of the step function, and we revert to traditional electricity and the DC model (like the "water" model). However, after the initial fast edge arrives, the lower frequency parts will arrive with less attenuation in the line, so I would expect the bulb to brighten gradually. The reflected energy will return in weakened form about three years later, having been reflected from the switch. Once the DC model is established, it is only the resistance of the wires which will control the supply of energy, rather than the characteristic impedance.
The velocity on the line will of course be less than c as with all transmission lines.

18. Nov 16, 2018

### FreeForAll

I'm pretty sure I get this - thanks for explaining. It sounds similar to post #7.

So in my experiment, the bulb will illuminate until the wire "fills up" then eventually the backpressure (using the water example) will push back and everything will equalize and the bulb turns off.

1) If my assumption above is correct, does everything equalize to 0V or some other risidual value?

2) If the wire after the bulb is infinitely long (again with zero resistance), does that mean zero need for return leg back to the battery?

3) If the battery is connected for 1 minute then disconnected, will the bulb eventually illuminate for 1 minute?

4) If the battery was connected for 1 minute, then both the battery AND bulb disconnected, what would happen to the electricity in the wire with nowhere to go?

Do any answers change if the electricity were AC instead of DC?

I feel like I'm getting closer to understanding this.. and freeing up some space in my brain where this was residing for so long!

19. Nov 16, 2018

### FreeForAll

I'm still digesting the rest of your response but was wondering what happens if 1 impulse reaches the bulb before the other (1 wire is shorter than the other)?

Or in other words, does the bulb need both impulses before it can illuminate?

20. Nov 16, 2018

### Staff: Mentor

The current flow (don't call it an "impulse" - that term already has a meaning, and this isn't it) is driven by the capacitance of the part of the wire that it hasn't reached yet. So there's a potential difference across the light bulb as soon as the change in voltage from the shorter wire arrives.

There's much complexity in the details of exactly what happens at the light bulb, because we cannot make the usual simplifying assumption that voltage changes propagate instantaneously through the wire. Thus, it is possible that all sorts of interesting electrical things (reflections, transients, spikes, ....) happen when the voltage arrives through the sorter wire at the bulb, without the bulb actually lighting. That's why I suggested above that you watch the ends with a oscilloscope instead of a light bulb - the "does the bulb light" question is much harder than the "does the electricity get there" question.