# A basic light circuit.

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As I said RF engineers do this all the time.....but in a different context.
A battery does not change amplitude. A pulse does. A pulse changes amplitudes in 2 directions also.
The conductor reacts differently to them.
This is why the question is so good.

swansont,
The propagation delay or velocity factor is a proportion of the speed of light.
It's caused by impedance. If we take the impedance out, like we did the resistance, the propagation speed is back to normal(speed of light). ok?
Edited by BR-549

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@swansont... are you talking about propagation delay of the oscilloscope measurement signal? The oscilloscope can be placed very close to the switch/battery - wires do not have to go straight.

Anyway, about the OP experiment: If the switch switches both wires simultaneously: The current will start flowing immediately through battery (possibly stronger than the steady state current). One wire will start charging, the other one discharging. A 'voltage disturbance' will travel from switch through both wires at great speed. Once one of these 'disturbances' (I am sorry, I don't know better English word) reaches a lamp, the lamp will start lighting. Closer lamps will turn on sooner than farther lamps. If a lamp is very close to any battery pole it will start almost immediately... Note: if the lamp is not exactly in the middle of the circuit, then, at first, the lamp light intensity does not have to be the same as the steady state intensity (if the wire has zero resistance, its intensity might change in two steps).

If the switch only switches one wire, while the other wire is permanently connected, then the very much same thing happens - again the 'disturbance' moves away from battery through both wires. (In both cases I am placing the switch very close to the battery)

An analogy with a water in a long circular trench helps!

However, in real circuit there are effect of EM radiation and inductance that are not taken in account in the above picture. The safest thing is to use a coax cable - then you don't have to worry about EM radiation and inductance.

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I'll toss in my hat and say 4 years.

I believe it will take 2 years for the electrons to respond to the electric field from the positive end of the terminal and 1 year to travel towards the resistor. Once an electron hits the resistance, it should produce light, which will take an additional year to travel back towards the observer.

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I would like to change my answer.
I believe it will take 1 year to lite.
1 year for voltage field(potential) to form(2 fields), and then very quickly full current.
When the electrons move in a conductor, they do not move very far. This might surprise many.
Current is local and caused by the local electric field. That's why we have to wait on the electric field first.
So I believe the lite will lite, in a couple of microseconds after the voltage field.
~ 1 year.

No current will flow until ~ year. i.e. until entire field forms.

Sorry....also, you could put the lamp anywhere on the loop......and it would still take a year. Even if lamp at battery terminal.

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...

Current is local and caused by the local electric field. That's why we have to wait on the electric field first.

...

No current will flow until ~ year. i.e. until entire field forms.

...

... and it would still take a year. Even if lamp at battery terminal.

These bits seem contradictory. Current starts flowing out of the battery, or not. Surely a lamp nearer the battery will light as soon as current flows through it?

(And doesn't this show no need for the circuit to be completed before any lamp is lit? (Yes, a complete circuit would be needed for the current to keep flowing, and for the lamp to stay lit.))

Edited by pzkpfw
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Current is local and caused by the local electric field. That's why we have to wait on the electric field first.

But how do you think the electric field is created inside the wire if not by repositioning electrons (that is, by current)? It is not that electrons just stand still while electric field is being generated - it is electrons that generate the electric field by positional rearrangement.

Current flows through battery as the electric field is being generated. Only if you consider wires without capacitance (zero diameter, I suppose), then there would be no current.

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Daniel +1 by way of encouragement.

You are showing substantial appreciation of electric circuits.

As a matter of interest, the small differences in signal paths between the various bits on a parallel data bus are sufficient to cause computer 'glitches' at the speeds of modern computer circuitry.

That is why design has moved to serial data transfer and away from parallel, for example with hard drives.

Look up 'race conditions'

http://en.wikipedia.org/wiki/Race_condition

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Hello pzkpfw,

That was confusing wasn't it.

Being that the current is local........that means for Circuit flow, all the little locals have to flow at the same time.

That means we must have the field established thru-out the circuit.

Was that any better?

Hello Danijel,

When one connects an open conductor to a terminal, as you say, it polarizes.....the electrons align.

This alignment(NOT FLOW) is and causes the potential. This potential is the electric field.

But current, actual electron flow(charge flow) does not happen until a complete path(circuit).

Current does not come from the source.

Only the potential. The current is always there.

A battery does store electrons and feed them out.

A battery can only CIRCULATE the amp hr. ,, not make it.
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A battery can only CIRCULATE the amp hr. ,, not make it.

That was Mama Cass not making it, not the battery.

Though you were quite right to note that transit time is important in modern circuitry.

Edited by studiot
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But current, actual electron flow(charge flow) does not happen until a complete path(circuit).

Let me see if I correctly understand your picture: You say that at first, when the switch is switched on, only the electric field starts spreading from the battery through the wire. This electric field front, you say, is generated by electron re-alignment (not by re-positioning). Only after the electric field is generated in the whole circuit (the electric field closes), electrons start their flow.

But such a view, as I see it, breaks the principle of locality. How does an electron here knows the moment the electric field is fully established (that is, when two fronts meet at the far side of the circuit) so that it knows it is time to start flowing?

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Danijel,

Think of the electric field as a telescopic cylinder surrounding the conductor.

The cylinders maximum diameter will be at the battery terminals, representing maximum field strength. Let's just pick 6" for talk.

As the cylinders move down the conductor, they rotate, we will ignore that for now.

Think of the cylinder coming down in steps, aligning the local electrons along the way.

The electron doesn't need to know when the circuit is complete. All it has to be is lined up and ready to go. Once all the locals are lined up, flow is automatic.

Being that in this setup, there is no resistance, the cylinder diameter will be the same at the input of lamp as at the battery terminal. If there was resistance, the cylinder would taper in proportion to the resistance.

Now at the lamp, the 6" cylinder tapers down like a cone to a zero point at the negative conductor.

This is the voltage drop across the lamp. The negative field is the same, but inverted. The zero point on its cone is the positive conductor.

This is a DC circuit and the fields are steady at these shapes as long as current flows.

Did I thoroughly confuse everybody?

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@swansont... are you talking about propagation delay of the oscilloscope measurement signal? The oscilloscope can be placed very close to the switch/battery - wires do not have to go straight.

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Wow, I wasn’t sure what direction this would go but some very interesting points here. I’ve been thinking about this on and off for quite a while now, but my paths always fall into one of the following scenarios:

SCENARIO 1

The lamp illuminates one year later - as soon as current flows through it.

At this time it doesn’t know (or seem to care) if the circuit it complete or not. We have a fully functioning open circuit!? Imagine the possibilities (oh and “circuits” would be a misnomer ).

SCENARIO 2

The lamp illuminates in 2 years when current has made a complete circuit back to the battery.

How does the lamp “know” it took the return current 1 year to reach the battery? Is some other “info” sent back to the lamp saying “circuit it good, you can light now”? If so, then it would take 1 year for this info to travel, so the lamp would really take 3 years to light, not 2. Yet it would take only 1 year for it to turn off after the battery is disconnected. Something doesn’t add up.

SCENARIO 3

Current flows from both terminals simultaneously (pushing and pulling as someone here described).

It would take 1 year for the lamp to receive both “currents” knowing that the circuit it complete – it will illuminate.

But as someone here hinted what if the lamp position was not equidistant from the terminals? Now the lamp is receiving one current before the other – which is basically identical to Scenario 1.

Or, if it received one current but does not illuminate until both currents join, then we’re back to Scenario 2.

And I'm back to being confused.

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Scenario 1 is the nearest to common sense.

Scenario 2 basically says that the current leaves the battery, arrives at the lamp and passes through it, but the lamp does not light until 1 year later, when the current has arrived back at the battery.

That is ridiculous.

Scenario 3 is even more against electrical theory. What are the agents of the two currents - they can't be the same can they?

You switch on the battery and half a year later I come and switch if off again. Will the lamp light briefly?

Or even worse suppose that scenario 2 was actually true and the lamp will not light until the current returns to the battery after 2 years.

Now for scenario 5 suppose I come along and switch off the battery after 1.5 years, ie after the current has left the battery, arrived at the lamp and is on its way back to the battery.

So where does the current or energy go now since it is not dissipated in the light and does not return to the battery?

Edited by studiot
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I think it is a combination of 1 and 3 as Danijel Gorupec says in post #27, (+1).

If we replace electricity with a water system such that we have a pump, a looong hose and a water wheel at the other end then it's obvious that when the pump starts it will push out water and cause a pressure to build up at the outlet and draw in new water which causes a lower pressure at the inlet.

These two pressure fronts will move out through the system towards each other and eventually meet somewhere in the circuit. Between the pump and the pressure fronts the water will be flowing and independently of where the water wheel is located it will start to turn when one of the pressure fronts is passing and the water starts to flow locally.

If we turn off the pump or shut a valve in the circuit then the already flowing water will continue to flow and move the pressure fronts for a short duration until friction has decreased its momentum and the pressure has evened out.

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I think it is a combination of 1 and 3 as Danijel Gorupec says in post #27, (+1).

If we replace electricity with a water system such that we have a pump, a looong hose and a water wheel at the other end then it's obvious that when the pump starts it will push out water and cause a pressure to build up at the outlet and draw in new water which causes a lower pressure at the inlet.

These two pressure fronts will move out through the system towards each other and eventually meet somewhere in the circuit. Between the pump and the pressure fronts the water will be flowing and independently of where the water wheel is located it will start to turn when one of the pressure fronts is passing and the water starts to flow locally.

If we turn off the pump or shut a valve in the circuit then the already flowing water will continue to flow and move the pressure fronts for a short duration until friction has decreased its momentum and the pressure has evened out.

Just as a matter of interest, how many pump and hose systems do you know that require a return hose to the reservoir?

Also what happens at the cut end if you suddenly cut a hose that has water being pumped through it, does the flow from the pump stop?

Electricity in cables is not (very) like water is a hosepipe.

Edited by studiot
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scenario 4,

After the switch is opened, I think that the established field would dissipate, or fall apart.

It would return to the un-organized field of the conductor before it was lined up.

I don't think there would be any induction, because there is no field storing or charge storing device or apparatus in the circuit.

I haven't thought about it, but I think the field would dissipate from the ends, towards the center.

Like I said.....great question. Great experiment. It would only take a year. Ha Ha

Excuse me, two years. One to turn on and another to see.

One thing puzzles me.

Danijel and Gweedz, why does an electron or a lamp need to know anything?

I do not recall this premise.

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Danijel and Gweedz, why does an electron or a lamp need to know anything?

When I say that electron 'knows' something, I mean that it feels a force caused by some locally existing field (created by other charged particles that might exist nearby). An electron can only behave according to its surrounding, it cannot act according to something that is happening very far away. An electron in a wire will start moving as soon as it 'feels' a local electric field - it will not wait until the filed is established across the whole wire loop.

...

Stop, Hammer time!

If one considers an experiment as described in the OP, one must make distinction between two cases:

A) The wide loop - an extreme example would be a circular loop of 2ly circumference. In this case the current will create considerable magnetic field and wires will provide considerable inductance (after all, it is a wire loop).

B) The narrow loop where two 1ly long wires run parallel, close to each other. An extreme example would be a coaxial cable. Here we could be able to ignore the inductance and energy loss due to EM radiation.

I don't know, but I have a gut feeling that these two cases would behave considerably different. For me, the A case is more complex and I don't even know if its large size makes any qualitative difference in comparison to small wire loops - not sure if the same math can be used (I suppose, yes). Anyway, i am only considering the B1 and B2 cases in the following text.

As most know, two parallel (or even coaxial) wires make a transmission line (google 'transmission line'). Basically you need to know that a transmission line has some impedance, some signal propagation speed, and some capacitance (per unit of length). Just look at the transmission line - two long parallel wires make a physically long capacitor....

It also needs to be noted here I am talking about uniform transmission line (if not uniform throughout its length, various nasty reflections would happen that could complicate our life).

THE B1 CASE

At one side of a very long cable there is a battery and a switch. At the other side, there is a lamp (a resistive load). When we close the switch:

- current immediately starts flowing from the battery and starts charging the transmission line (imagine zillions of little capacitors between two wires as they are getting charged one after the other as the voltage disturbance front moves at, say, 0.8c)

- the strength of this 'charging' current depends, among else, on the impedance of the transmission line

- when the front reaches the lamp, say after 1.25 years, the lamp will turn on

The story only ends here if the lamp resistance is equivalent to the transmission line characteristic impedance. If not, there will be reflections:

- if lamp resistance is smaller that cable impedance, then the lamp will not generate the full light at first. Reflected front of somewhat decreased voltage will be returned toward battery - when it reaches the battery, after 1.25 years, the battery will sense the decreased voltage and will increase its current to compensate. The front will again travel toward lamp, and after additional 1.25 years the lamp intensity will increase for one more step. The reflections will continue, and lamp will increase its intensity, in smaller and smaller steps every 2.5 years.

- if lamp resistance is greater than cable impedance, the lamp will reflect front of somewhat increased voltage (and will start shining more luminous than expected - up to twice as much if it has infinite resistance). After every 2.5 years their luminosity will decrease in smaller and smaller steps toward steady state luminosity.

THE B2 CASE

Here you can see that the far side of the cable is short circuited. This means 100% reflections (-100% actually, as the reflection is inverted).

- when circuit is switched on, the lamp immediately starts to shine. Its luminosity depends on its resistance and transmission line impedance. If the lamp has a low resistance, its shine might be barely visible.

- when the disturbance front reaches the far side, a near-zero-volt front gets reflected back to the lamp. When this front reaches the lamp, the battery will detect the voltage drop and will increase its current. Lamp increases its light intensity in accordance.

- I suppose, if battery+lamp resistance is not equal to cable impedance then reflections will continue further until system reaches some steady state. That is, again the lamp intensity is readjusted in steps every 2.5 years.

I must say that I did not investigate much, and made most above statements directly from my head - so be aware of possible errors.

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Good afternoon Danijel.

You refer to transmission line antics by the bulb.

Now I have assumed the bulb to be resistive and therefore it will only respond to the real part of the disturbance.

Do you not consider this will have a bearing on your case for a light oscillator based on a battery, bulb and transmission line?

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I have seen similar effects on pure resistances in practice, but I must admit that I never knew how much of the observed effect was from the transmission line effects as depicted in my previous post, and how much was from (parasitic) inductance of it (as it would be in the A case).

In real life, I suppose, the effects I describe would not be as clear (a step-by-step increase in light intensity would probably be smothered somehow), but still I believe that such oscillatory effects are, in general, real even in the pure-resistance case.

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Just as a matter of interest, how many pump and hose systems do you know that require a return hose to the reservoir?

Also what happens at the cut end if you suddenly cut a hose that has water being pumped through it, does the flow from the pump stop?

Electricity in cables is not (very) like water is a hosepipe.

Yes, I made an analogy to explain my thinking, but I never claimed it to hold in every case you can think of.

I think there will be two potential surges travelling outward in both cables, from both sides of the battery when the switch gets closed.

If we place three lamps in the circuit, the first after a half light years, the second at the furthest point a light years away, and the last at a half light years from the battery in the other wire at the opposite side as the first lamp. Like this:

┌──────────────────────────(L1)───────────────────────────┐
\ S                                                       │
+                                                         │
[B]                                                       (L2)
-                                                         │
I                                                         │
└──────────────────────────(L3)───────────────────────────┘
Then both the lamp (L1) and the lamp (L3) will light up at the same time, around half the time before the lamp (L2) will light up.
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┌──────────────────────────(L1)───────────────────────────┐
\ S                                                       │
+                                                         │
[B]                                                       (L2)
-                                                         │
I                                                         │
└──────────────────────────(L3)───────────────────────────┘
Then both the lamp (L1) and the lamp (L3) will light up at the same time, around half the time before the lamp (L2) will light up.

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

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

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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.

Increase the straw diameter so that you can put 5 strings of electrons in.

The electrons are repulsive, so they will not be in strings......all electrons will be equidistance from each other.

Can you move one electron, without moving another? Can you move 20 electrons without moving the others? No. Same reason.

Current is a system. It is comprised of a confined loop, of a set of connected (connected thru repulsiveness) charge.

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.

The force and law that is the physical cause for this action is charge repulsion......the law of charge.

That charge repulsion has a magnitude or energy level. It's always there. It's a primary property of charge. There is another equal primary property of charge. Each charge has an equal magnitude magnetic moment. A north and south pole.

This means that an alignment action, with the same magnitude is taking place, while the charge repulsion action is taking place. So when charge moves, it lines up. Once they all line up, they all want to move in the same direction. The force of repulsion with the force of alignment is the physical cause for this unifying directional movement action. You can and should repeat the experiments of Ampere and Weber to see this for yourself.

If we take a loop or ring of pure copper, A certain proportion of the free electrons on the surface will line up and make a charge path around the ring. With just background induction, several small currents can be found circulating around the ring. You can see this with a loop of copper and a oscilloscope. Try loops of different sizes and see what you get. This is all at room temps.

Why does only a portion of the free electron bank line up and move? Why not all the free electrons?

The conductor surface is a patch-quilt like pattern of fields. This appears to the free electron as a heavy thick jungle. This is THE resistance to free electron flow. The electron loses energy with all the turning thru and around the jungle. Heat loss.

If the conductor surface had a even field structure, all the free electrons would flow.

When we buy a conductor (wire), is has a current rating. This rating is usually based on a temp rise from some ambient reference. This is for safety.

Skin the wire and put a potential across it and bring the current up to the current rating.

Now slowly bring the current up until the wire starts to glow. Ionization.

Turn the lights off and back off on the current until the glow just goes away. This amperage is the true max flow (for that temp and conductor).

You can be pretty sure that all of the free electrons are flowing here.

Now if we remove all resistance, What will the flow be?

It will be the same. The maximum current remains the same. We don't want ionization. The wire remains cold at max current. Only now I don't need that high potential. Where it took a high potential to push thru the jungle, now a small breeze will push the flow thru. The load will still have resistance and ohm's law still applies.

But none of the potential will be expended across the conductor in wasted heat, only the load, which will still require potential.

The problem is that to have a even conductor field surface, we need an ordered structure material. Not grainy like conductors. Most ordered structures are insulators.

It's hard to get a free electron with an ordered structure.

It is early in the study of super cooled conductors. Perhaps one of the solid states at those temps. is acquiring these conditions.

If they are achieving super conductivity in those frozen rings, even with a small free electron count, they should see a very strong magnetic field around the ring. This is because without resistance, we should see a multi magnitude increase in electron velocity. This should be much faster than in conventional circuits. This increase in velocity is more impactful than the number (magnitude) of free electrons (charge). Velocity always has strong dominance over magnitude in nature.

This should give the ring a strong resonance and that resonance will not like being disturbed.

I could understand an atomic structure or configuration, that needs two free electrons in a certain area or position for flow without ionization, but I really doubt that electrons pair up.

It’s the refusal to pair up, that makes all of our circuits and electricity work.

In the meantime play with a loop. Depending on the size and where you are, some currents are large enough to rectify directly and listen to. With just a loop. People used to do this all the time. Free music, no battery required.

Brought to you by the self movement of confined charge.

If you doubt my word and think that a static magnetic field does not move, or can not cause movement, throw a small bar magnet in a shallow bowl of ferrofluid. You will see magnetic lines in action.

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I can't spay electrons out across the room, like I can with water.

Funny, my parent's TV could do just that back in the 1950s and I understand the LHC at CERN could now do it across a continent.

And somewhere between the two I hear they have developed a techology called electron beam welding.

Edited by studiot
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Your parents tv needs adjusting. If you did see an electron from the tv......it would be emission, not conduction. (current)

Cern is one of the few places we use positive charge. When the current beams at cern collide, the circuit is complete and the current becomes fragment emission.

In the welding......the beam will still need to be confined. If you pull the emitting head far from the work......you will lose confinement and you will lose the current.

One can cloud the conversation with the strict definition of current.

In that case, any charge movement at all could be called current.

But to me charge flow implies a system, a complete path, of moving charge.

A single charge is just emission or just a single charge. Current is a collective charge behavior.

I guess it's how you're taught, and then with your own experience, how you understand it.

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