# Solving a Perplexing Thought Experiment: A DC Electrical Circuit

• FreeForAll
In summary, according to the summary, it will take 2 years for the bulb to light up if you send a photon at a target that is 1 LY away at and rest wrt to you.
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!

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.

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.

Grinkle said:
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.
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.

sophiecentaur said:
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.
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?

FreeForAll said:
In my mind sending a photon is too different than a battery that has 2 terminals.

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.

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.

CWatters said:
But could exceed the source voltage if reflected at the bulb.
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.

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"?

FreeForAll said:
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"?
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:
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.)

DaveC426913 said:
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.)
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?

FreeForAll said:
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?
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.

DaveC426913 said:
As long as current is able to flow from the terminal into the wire, it will continue to flow until the potential reaches zero.
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:
Grinkle
FreeForAll said:
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?

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.

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.

Grinkle said:
This isn't a great analogy, but its simple...
@Nugatory discussed post 15.
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!

tech99 said:
... When the impulses reach the lamp, the lamp will be excited ...
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?

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

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

At steady state, an open circuit with a wire and a resistor (the light bulb is basically a resistor) would measure on a voltmeter connected to the battery negative terminal to be at the voltage of the battery high terminal. Your question implies that voltage is absolute. Its a relative measure between two potentials.

edit: Think of the pipe with the end sealed - its full of high pressure water, not zero pressure water, as long as the faucet at the source of the pipe is open. Once the water stops moving, the pressure is the same everywhere in the system.

FreeForAll said:
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?

Its difficult to say - if you are claiming that your infinitely long wire on the return path with zero resistance is equivalent to a ground connection, I guess that is fine, since its a non-physical object in your thought experiment.

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

I think this also depends on the specific properties you assign to the hypothetical wire. If you make the wire transmission properties equivalent to vacuum, and you send energy into it, that energy will continue to propagate.

FreeForAll said:
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?

It would reach the end of the wire and be reflected back to the source.

Last edited:
FreeForAll said:
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?
This is what I think. If the wires are unequal length then a pulse will arrive on one wire first using the single-wire or surface wave mode. The second wire at the bulb will then behave as a current sink, allowing the bulb to light.
Please notice that I am considering the whole system located in free space, with no ground considerations.
I am sorry to cause confusion by using "impulse" but I had not noticed this had been dropped from electronics since my youth.

I am just an old retired TV tech and dropped in here to see what you guys do. You guys are having too much fun with mind-bending analysis of the impossible, but thought provoking "What-If" questions. It does seem to be good information that can have practical application on a test or even in a lab. But then, I retired 15 years ago, and WOW have things changed. It is more fun than injecting a signal into the IF section, and watching a scope to see what the Flying Elvis happened to the chroma signal?

tech99 said:
The second wire at the bulb will then behave as a current sink, allowing the bulb to light.
Hmm, it's more complicated than that, I think. The impedance of the transmission line will be different as the wires have increasing separation; it will increase. That will cause a reflection (or a set of reflections along the taper). Remembering what I used to see in a Time Domain Reflectometer (which uses a similar step function) you get a positive slope on the trace for a time, corresponding to the length of the taper. This means a lot of reflected energy and for the distances and separations involved in the model, there would be significant radiated EM. If you had the time, you would see the trace of reflected signal, go down to a low value.
What happens when the legs are of unequal length is even harder. There will be some energy reaching the light bulb in the 'shortest time' but the majority of the signal in that case will be in an unbalanced mode. The other (longer) wire will behave like a long line in series. "A current sink' could describe it, perhaps but what Impedance" Pretty high, I suggest.
I think the bottom line would be that, if the electrical lengths were sufficiently different then you could treat the originally balanced signal as a pair of unbalanced signals and that would involve significant radiated power from the line.
It could be easier to visualise the result if, instead of separating the wires, you series loaded on wire with inductors or even made it helical over the whole length. Actually you can do this all in the Lab, without needing to wait years for a result. Cheaper but not such fun.

With the case of unequal legs, I suppose we can say that the wire after the bulb presents its characteristic impedance to the bulb.

tech99 said:
With the case of unequal legs, I suppose we can say that the wire after the bulb presents its characteristic impedance to the bulb.
But what "characteristic Impedance are you assuming? I think there's a balanced mode impedance ,referenced to the load but what unbalanced impedance is there?
Department of too hard guv, I think. My transmission line theory doesn't include odd series elements on one leg of a balanced line and the very simplest model would, as I pointed out above, be modeled as a (or several) extra series inductors on one leg.

FreeForAll said:
So in my experiment, the bulb will illuminate until the wire "fills up" then eventually the backpressure

the wire doesn't fill up with anything

sophiecentaur
davenn said:
the wire doesn't fill up with anything
The OP would need to get into some advanced transmission line ideas in order to get any more than the idea of a simple 'delay' to a remote load. One can get through a lot of one's technical life without needing more than that.

davenn
FreeForAll said:
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?

You are confusing two different scenarios. When the switch is first closed the electromagnetic field propagates through the conducting wires. After the switch has been closed for a while (usually a very short time but in your example a much much longer time) there's a steady state. Without the connections to the two battery terminals this steady state would never be achieved, and it is this steady state that you are studying when you first learn about circuits with batteries and bulbs.

FreeForAll said:
Does the disturbance travel down a single wire, to the bulb, then returns down the other wire back to the battery?

The bulb IS a wire! It's just that it's a much narrower wire (called a filament) than the connecting wires. Current is the rate at which electric charge flows, that's true for both fat wires and skinny wires.

sophiecentaur said:
The OP would need to get into some advanced transmission line ideas in order to get any more than the idea of a simple 'delay' to a remote load. One can get through a lot of one's technical life without needing more than that.
Thanks for wording this so eloquently :)

1 more question for anyone, what (if anything) would be different if the battery were replaced by an A/C power source?

Very little difference unless the wavelength involved were very short. More power would be radiated on the way.

FreeForAll said:
1 more question for anyone, what (if anything) would be different if the battery were replaced by an A/C power source?

Imagine the very long wire is now a very long rope or string or slinky (whichever is easier for you to picture) and it is completely lossless. If you whip the front end of the string you will cause a wave to propagate along the string (or rope or slinky). If there is no loss of energy in the wave, it will go all the way to the end of the string and if it is not dissipated somehow at the far end of the string it will then reflect back towards you. If you continuously whip the front end, you will send a continuous wave towards the far end. This is sort of what A/C voltage is like along the lossless / zero resistance wire. Lossless means, among other things (like heat dissipation due to resistance) I am sure, no radiation.

I struggle to come up with a similar analogy for continuous DC current. Strings / ropes / slinkys don't displace due to force in any way that I think of as similar to DC voltage in a wire.

Grinkle said:
I struggle to come up with a similar analogy for continuous DC current. Strings / ropes / slinkys don't displace due to force in any way that I think of as similar to DC voltage in a wire.
A mechanical analogy to a DC signal, after it has been switched on and connected to a transmission line, could be to suspend a stiff rope between two points and move one of the points suddenly sideways. The description of both those would be a Step Function. A step function will propagate along both.

@sophiecentaur The transverse wave on the string is perhaps a bit subtle/deceptive because the analogy to current in a wire is the transverse velocity of the string, not the displacement of the string. I think in your DC proposal you are creating a displacement step, not a velocity step.

Is the below better?

@FreeForAll The below image has the major flaw of not separating the conductor from the current - in the below image the conductor is the current and that is dangerously incorrect in terms of forming early intuitions about electricity. In an actual electric circuit, the picture of a transverse wave moving along a string is much closer to current moving through a wire.

Imagine a long zero-mass rigid cord - this is a zero-inductance long wire. Now let the mechanism that moves the cord be zero-loss, a zero resistance wire.

Instead of the cord moving in a transverse way, it moves along the direction of its own axis. For an A/C analogy, the cord reciprocates. For D/C, the cord moves in one direction at a constant velocity. The D/C analogy requires a reservoir of cord to draw on for a single path conductor analogy. To provide an intuition of the difference between and open circuit and a closed circuit, one needs to bend the rigid cord with a rigid-cord bending device at the far end and continue the path of the still-feeding cord back to the source. A closed circuit will take in the returning cord at the source and let that cord return to the cord reservoir. An open circuit stops moving when the the return cord hits the destination point and is blocked there. The entire loop is present at zero velocity before the "experiment" starts. The rigid cord will have very little compliance in it (capacitance in our electrical wire) so one can intuit that the open circuit will mostly result in a force along the rigid cord with very little actual motion of the cord, and one can see that if the ideal wire has zero capacitance in addition to zero R and zero L, there will be no movement at all.

FreeForAll and sophiecentaur
@Grinkle . There are many possible analogies and many of them fit particular situations better than others. I would say that a step function of position is a good analogy to the step function of Voltage but it's a matter of taste. I just tried to make an analogy with an experiment that we could all do. In fact, a longitudinal wave on a rod would be better because you can get a displacement of the whole length of rod. A rope fails in many ways to fit the requirement; it's just familiar.

<h2>What is a perplexing thought experiment?</h2><p>A perplexing thought experiment is a hypothetical scenario that challenges our understanding of a concept or theory. It often involves a complex problem or paradox that requires critical thinking and creativity to solve.</p><h2>What is a DC electrical circuit?</h2><p>A DC (direct current) electrical circuit is a closed loop of electrical components that allows current to flow in only one direction. It typically consists of a power source, such as a battery, and various components such as resistors, capacitors, and inductors.</p><h2>What are some common challenges when solving a perplexing thought experiment involving a DC electrical circuit?</h2><p>Some common challenges include understanding the principles of electricity and circuit analysis, identifying the relevant variables and parameters, and determining the appropriate mathematical equations to use in the solution.</p><h2>What is the importance of solving a perplexing thought experiment involving a DC electrical circuit?</h2><p>Solving a perplexing thought experiment involving a DC electrical circuit can help us better understand the principles of electricity and circuit analysis. It also allows us to develop critical thinking and problem-solving skills that are applicable in many areas of science and engineering.</p><h2>What are some strategies for solving a perplexing thought experiment involving a DC electrical circuit?</h2><p>Some strategies include breaking down the problem into smaller, more manageable parts, drawing diagrams or schematics to visualize the circuit, and using mathematical equations and principles to analyze the circuit. It can also be helpful to collaborate with others and think outside the box to find creative solutions.</p>

## What is a perplexing thought experiment?

A perplexing thought experiment is a hypothetical scenario that challenges our understanding of a concept or theory. It often involves a complex problem or paradox that requires critical thinking and creativity to solve.

## What is a DC electrical circuit?

A DC (direct current) electrical circuit is a closed loop of electrical components that allows current to flow in only one direction. It typically consists of a power source, such as a battery, and various components such as resistors, capacitors, and inductors.

## What are some common challenges when solving a perplexing thought experiment involving a DC electrical circuit?

Some common challenges include understanding the principles of electricity and circuit analysis, identifying the relevant variables and parameters, and determining the appropriate mathematical equations to use in the solution.

## What is the importance of solving a perplexing thought experiment involving a DC electrical circuit?

Solving a perplexing thought experiment involving a DC electrical circuit can help us better understand the principles of electricity and circuit analysis. It also allows us to develop critical thinking and problem-solving skills that are applicable in many areas of science and engineering.

## What are some strategies for solving a perplexing thought experiment involving a DC electrical circuit?

Some strategies include breaking down the problem into smaller, more manageable parts, drawing diagrams or schematics to visualize the circuit, and using mathematical equations and principles to analyze the circuit. It can also be helpful to collaborate with others and think outside the box to find creative solutions.

Replies
4
Views
2K
Replies
36
Views
4K
Replies
16
Views
2K
Replies
37
Views
3K
Replies
2
Views
10K
Replies
4
Views
1K
Replies
18
Views
2K
Replies
12
Views
3K
Replies
14
Views
3K
Replies
5
Views
1K