A basic light circuit.

[Gweedz]

Imagine a very basic DC circuit consisting of a battery, a bulb, and 2 wires connecting them.

The wires are each 1 light year long (yes, a huge circuit).

 

You connect everything together, then attach the battery...

 

How long would it take for the light to illuminate (ignoring wire resistance, battery capacity, etc)? 1 year for the electricity to get to the bulb, or 2 years for the circuit to be completed?

 

Depending on the reply I have follow up questions.

Thanks!

Edited June 5, 2015 by Gweedz

[swansont]

It's going to light up once the electrons go through the bulb. That will happen some time after 1 light year, because the signal propagates slower than c. It'll be at c/n, where n is the index of the wire.

 

The conundrum here is how the circuit "knows" it's a completed circuit. The wire is going to acquire the potential of the source, propagating at c/n. If the wire is attached to the bulb but not to the ground of the source, I think initially it won't matter — you'll get light. But the return leg of the wire would accumulate charge, and its potential would rise and the light would dim as the potential rose. I think it would behave somewhat like a charging capacitor.

[fiveworlds]

Imagine a very basic DC circuit consisting of a battery, a bulb, and 2 wires connecting them.

The wires are each 1 light year long (yes, a huge circuit).

 

Wouldn't work because the resistance would be astronomical. We have tons of stations to step up the voltage to allow electricity to flow in the grid.

Edited June 5, 2015 by fiveworlds

[studiot]

You could make the question more interesting by making one leg of the connetion 1 light year long, but the return leg 10 light years, by a more wiggly route.

[swansont]

Wouldn't work because the resistance would be astronomical. We have tons of stations to step up the voltage to allow electricity to flow in the grid.

 

I think this is safely in the realm of a thought experiment, where one is allowed to say the resistance is small.

[BR-549]

A most interesting thought. Being that the lamp requires current, it probably will take a while.

An EM pulse in parallel with the conductors would take a little over 1 year. But with the conductors, the connection is already established.

What if we replace the lamp with an electroscope?

Would that require a year?

Does the electric field have to propagate or just line up?

Would a newton's cradle a light-year long, take a year?

However, if you must use the lamp, I believe it would take just over a year.

The current would flow at the same time in both conductors.

In other words, you wouldn't have to wait for the current to return.

One conductor is pushing, while the other is pulling. Bi propagation.

Both conductors are identical except for polarity and direction.

If the conductors have no impedance, it would lamp instantly.

Replacing the return leg with 10 LY........makes me dizzy. 9 years to light.

Only after 9 years can current be established.

Great question Gweedz.

[pzkpfw]

...

Would that require a year?

...

Yes.

 

...

Would a newton's cradle a light-year long, take a year?

...

Longer. It'd involve the speed of sound, in the material the balls were made of.

 

...

If the conductors have no impedance, it would lamp instantly.

...

No. If that were true, you'd have a signal/information travelling faster than light.

Edited June 6, 2015 by pzkpfw

[BR-549]

I did not mean to hijack the post. I wasn't looking for answers, just trying to stir some thought.

Pardon me.

[swansont]

 

The current would flow at the same time in both conductors.

In other words, you wouldn't have to wait for the current to return.

 

No, that violates relativity. Instant communication.

I did not mean to hijack the post.

 

Nobody suggested that you did.

[John Cuthber]

Wouldn't work because the resistance would be astronomical. We have tons of stations to step up the voltage to allow electricity to flow in the grid.

Which part of " (ignoring wire resistance, battery capacity, etc)?" did you not understand?

[BR-549]

Ok, real good.

Lets say that the electroscope will take a little over a year to react and the lamp to light.

And lets say that even with zero impedance it still takes a year because of relativity.

The currents are still equal and opposite without violation.

The battery must receive an electron in order to give one.

And a little over a year(1 LY conductors) the current is established and we have light.

Before I proceed, can we all agree up to this point?

[studiot]

 

Great question Gweedz

 

Agreed, I have given Gweedz + in recognition.

 

As a matter of interest the length of the return wire is almost immaterial. It is only the length of the supply wire that counts.

 

There are electric supplies on this planet that are single wire and earth, used in some remote areas.

 

I calculate that the mass of a typical 1mm² lighting copper cable of length 10 light years to be about 10¹⁵ kg.

This would be a sizeable enough chunk to regard as a pretty good earth, certainly enough to sink the supply current to a light bulb, arriving from 1 light year distant.

Edited June 6, 2015 by studiot

[swansont]

The battery must receive an electron in order to give one.

 

Are you sure about that? Not even a single electron?

 

Regardless, you can look at this without current flow — how long does it take for the input to the device to achieve a potential of +V?

I calculate that the mass of a typical 1mm² lighting copper cable of length 10 light years to be about 10¹⁵ kg.

This would be a sizeable enough chunk to regard as a pretty good earth, certainly enough to sink the supply current to a light bulb, arriving from 1 light year distant.

 

That was my thinking as well — the return line is effectively ground.

[BR-549]

If only one electron left the battery, the voltage would drop. Only by adding an electron to the opposite pole can the voltage be maintained.

Can we agree on that?

[studiot]

 

If only one electron left the battery, the voltage would drop. Only by adding an electron to the opposite pole can the voltage be maintained.

Can we agree on that?

 

 

The beauty of a ground is that you can take an electron from it and stick it into the other battery terminal (to which it is connected) as many times as you like without reference to something 10mm away, let alone 10 light years.

Edited June 6, 2015 by studiot

[BR-549]

This is a DC circuit. The only source and sink is the battery terminals.

I might be mistaken, but I believe Edison powered his DC grid with 2 conductors and did not use a ground.

If I connect ammeters to both terminals, will I see one move without the other?

A water pump provides a pressure differential to the existing media......the water.

The pump does not add water.

The battery supplies a voltage difference to the existing media.....free electrons on conductor.

The battery does not source, make or add electrons, it circulates them.

This is why the current is always the same going into and out of a power source.

This is why with 2, 1 LY conductors we will see maximum current in one year at the same time on both ammeters. We will see maximum current in the 10 LY setup in 9 years on both meters.

Does anyone agree with this?

[studiot]

 

The only source and sink is the battery terminals

 

Apart from this, I basically agree with what you have said in post#16.

 

But that does not prevent one of the connections acting as a ground, which was the main point made in posts 12 and 13.

[BR-549]

There is no ground, only the current path.

The current is the same on both conductors if the distance from the terminals is equal.

When we turn the switch on, we will see both ammeters at the terminals start to rise.

It will take one year to reach maximum.

Take a 100 ft. loop of wire. Insert the lamp anywhere on the loop.

Close the switch and note the time of lamplight. Now move the lamp at different positions on the loop. That lamplight time does not change. Because the length remained constant.

Do the same with a 10,000 ft. loop. Measure lamptime. Try different positions.....same lamptime.

Not the same as 100 ft. ....its longer, but no change on same loop.

The lamptime is constant with length.

Can anyone agree with this?

The battery terminals propagate in equal amplitudes and opposite directions.

Therefore the propagation time will be one half the current path length.

And so I will change my propagation time for the 11 LY setup to 5.5 years.

Even if you put the lamp at the battery, it will take 5.5 years.

Can anyone buy that?

This is assuming the lamp only shines at full current.

[swansont]

 

If only one electron left the battery, the voltage would drop. Only by adding an electron to the opposite pole can the voltage be maintained.

Can we agree on that?

 

 

Cathode ray tubes don't seem to mind electrons taking their sweet time completing the circuit. Besides, how much drop will you get from one electron out of Avogadro's number?

 

Still moot, if one simply looks at the voltage.

 

There is no ground, only the current path.

The current is the same on both conductors if the distance from the terminals is equal.

When we turn the switch on, we will see both ammeters at the terminals start to rise.

It will take one year to reach maximum.

Take a 100 ft. loop of wire. Insert the lamp anywhere on the loop.

Close the switch and note the time of lamplight. Now move the lamp at different positions on the loop. That lamplight time does not change. Because the length remained constant.

Do the same with a 10,000 ft. loop. Measure lamptime. Try different positions.....same lamptime.

Not the same as 100 ft. ....its longer, but no change on same loop.

The lamptime is constant with length.

Can anyone agree with this?

 

You state this as if the experiment has actually been conducted. Do you have a citation for it?

[BR-549]

This is why this is such a great question.

It has been many years since this experiment for me. Any electrician, electronics tech or electrical engineering student should know this. And I would think still prove this in class.

It would be under Ohms law and Kirchhoff's current law. Both laws state the current is equal at every point in the circuit, in this setup.

Both of these laws come from the conservation of charge law.

Another way to look at it......both terminals do work. The work is equal and happens at the same time. The positive terminal works on the positive conductor and the negative terminal works on the negative conductor.

Both terminals do equal work at same time.

[swansont]

 

This is why this is such a great question.

It has been many years since this experiment for me. Any electrician, electronics tech or electrical engineering student should know this. And I would think still prove this in class.

 

200 ft of wire takes about 300 ns for the signal to traverse. I seriously doubt any electrician or EE student has done the experiment.

[Strange]

Propagation delays in wires are one of the limiting factors in the speed of electronic circuits (even on-chip wiring where the distances are pretty small).

[BR-549]

Excuse me. Not this particular experiment, true.........for you students out there........you need to think about this........great experiment for class.

If you try this, don't use coils for conductor length, keep reactance to a minimum.

What I meant was that all students confirm Ohm's and Kirchhoff's current law in class.

At least they used to.

Most propagation times in electricity are done with AC signals.(RF cabling)(AC power)

They call it velocity factor. It's usually not considered in DC circuits.

But the law that says that the current is equal at all parts of the circuit(in this setup) has been confirmed by many students for many years.

The real question for some one who is familiar with this is.......whether the current will follow the voltage in real time.....or does the electric field have to be established completely around the circuit before current starts to flow?

If the latter is true.....then 1 year for electric field.......then current starts......then another year for max current.........so 2 years for light.

So when we flip the switch......do we see current start then.......or have to wait a year to see current start?

One could get famous with this experiment. Can you give more than 1 point to a poster?

My experience tells me the current starts right away...............But maybe what I've been taught only applies to short circuits.

Does voltage have to go clear THRU(and around) the circuit before current flows?.............OR does voltage only have to be ACROSS the circuit for current to flow?

[Danijel Gorupec]

 

200 ft of wire takes about 300 ns for the signal to traverse. I seriously doubt any electrician or EE student has done the experiment.

Not sure why are you saying so. 300ns is nothing that special. We did signal propagation experiments in our classes using pulse generators, oscilloscopes and long coaxial cables. (It was fascinating to see that signal pulse arrived in the same shape as it started - then reflected, inverted or non-inverted, deepening on how the coax was terminated)

[swansont]

Not sure why are you saying so. 300ns is nothing that special. We did signal propagation experiments in our classes using pulse generators, oscilloscopes and long coaxial cables. (It was fascinating to see that signal pulse arrived in the same shape as it started - then reflected, inverted or non-inverted, deepening on how the coax was terminated)

 

I was thinking absolute timing, and not relative timing, because you need to know when the lamp lit up without propagation delays for the signal. How do you measure that with an oscilloscope? I'm open to the possibility you can do this, but I can't think of how off the top of my head.

 

You could possibly do it at the extremes of the range (i.e. one leg much different from the other) with a photodiode, since the return signal PD will be much different from the legs of the circuit, but how do you do it with equal leg lengths?