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Consider a simple network of a bulb whose two terminals are connected to two wires with open ends A and B respectively

 A o--------💡--------o B

Now if a DC current source is attached across A and B, say terminal B is at +5 potential and terminal A is at 0 potential, then exactly what happens is that "negative charge flows to terminal at higher potential, i.e. terminal B" and what that means, as I understand it, is that : -

Because of electromagnetic forces, all of the electrons in the wire are displaced towards A with a certain velocity causing a positive current towards B. This drift of electrons heats up the highly resistive filament wire in the bulb and makes it glow.

So something like the following is a good representation of the journey of a single electron e in the wire for a DC circuit:

 A o--------💡-----e--o B  (+5)
 A o--------💡---e----o B  
 A o--------💡-e------o B  
 A o-------e💡--------o B  
 A o-----e--💡--------o B  
 A o---e----💡--------o B  
 A o-e------💡--------o B  

Assuming that my understanding of how a wire carries DC current is correct, I would like to understand how a wire carries alternating current.

I think that since AC current periodically changes direction, maybe the electrons move back and forth, (maybe oscillate along the length of the wire about their mean position?), which will also warm up the filament of the bulb and thus light it up. But I don't understand why they are able to move back and forth. Especially if the length of the wire was large, say 3 * 10^8 meters, then would the movement of electrons on one end of the wire be "in sync" with the movement of electrons on the other end?

Bonus question: how would the electron flow in DC circuits work if a bulb and a 5V voltage source kept 3*10^8 meters apart were connected by two straight 3*10^8 meter long wires? Assume that there's a switch halfway between one of the two wires and it has just been flicked to the "on" position.

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    $\begingroup$ I have often found that the word "slosh" is the right word to keep in mind when thinking about the motions of electrons on long wires and when thinking about AC. $\endgroup$ – Eric Towers Nov 22 '15 at 19:50
  • $\begingroup$ You might want, for interest's sake, to read a little about the history of the early Transatlantic telegraph cables. Once you get beyond the mere mechanics of laying a cable, it's all fun and games with DC, switches, and how long wires stop being "just wires" and start being "transmission lines". $\endgroup$ – Stan Rogers Nov 23 '15 at 3:25
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    $\begingroup$ The electrons in AC and DC move very slowly. Roughly around 2.5mm per second. So in an average room (roughly 10mx10m) it would take an electron one hour to travel from the switch to the light bulb (assuming DC). But the light bulb lights instantly. What happens is that while electrons move slowly, the holes that they leave behind move at almost light speed. So while charge moves slowly through the wire, current is very, very fast. $\endgroup$ – slebetman Nov 23 '15 at 6:36
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    $\begingroup$ I'm not sure if you live in a country that have traffic jams but if you've ever been stuck in a car in a traffic jams you can see how the holes left behind by cars moving forward in the jam move much faster than the cars (the holes move backwards through the jam - just like current moving in opposite direction to electrons). The hole started by a car at the head of the jam reaches your car long before you reach the end of the jam. $\endgroup$ – slebetman Nov 23 '15 at 6:38
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    $\begingroup$ @slebetman - But, the next "hole" will appear in front of you at "exactly" the same rate that you are moving. So, if traffic is moving at 3 meters each minute, and the "holes" are typically about 3 meters, then you will see 1 hole pass through each minute. $\endgroup$ – Kevin Fegan Nov 23 '15 at 9:23
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Because of electromagnetic forces, all of the electrons in the wire are displaced towards A with a certain velocity causing a positive current towards B.

The electrons have a small drift velocity, not moving much.

Although your light turns on very quickly when you flip the switch, and you find it impossible to flip off the light and get in bed before the room goes dark, the actual drift velocity of electrons through copper wires is very slow. It is the change or "signal" which propagates along wires at essentially the speed of light.

A single electron does not go from A to B. Think of it as each electron pushing the next one, and the signal travels with the velocity of light ,maximum, down the wire.

This drift of electrons heats up the highly resistive filament wire in the bulb and makes it glow.

True.

But I don't understand why they are able to move back and forth. Specially if length of the wire was large, say 3 * 10^8 meters, then would the movement of electrons on one end of the wire be "in sync" with movement of electrons on the other end?

Why not? When the field changes at A and B the change propagates by electrons moving back and forth over an average position. Similar to water waves, the atoms do not move much from their position, the energy is transferred atom by atom. In the case of electric fields and electrons the field is built up at the microscopic scale by the motion in situ of the electrons, in a sinusoidal way.

Very long wires enter the realm of special relativity and the limit of the velocity of light in transferring effects of fields.

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    $\begingroup$ @guntbert: For wires of reasonable lengths, we treat everything as simultaneous across the entire length of the wire; changing the voltage drop (potential difference) between the two ends of the wire has an instantaneous effect on the electric field throughout the wire, an instantaneous effect on the current throughout the wire, etc. For very long wires -- longer than {speed of light} x {time interval we're willing to fudge} -- this simultaneity can't hold, due to special relativity. $\endgroup$ – ruakh Nov 23 '15 at 0:42
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    $\begingroup$ Special relativity is not required. Maxwell's laws say changes in electromagnetic fields travel at the speed of light. In fact, investigating this is what led Einstein to special relativity. $\endgroup$ – Austin Nov 23 '15 at 4:50
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    $\begingroup$ @Austin yes, but Max equ. do not deal with atomic physics, know nothing of electrons. $\endgroup$ – anna v Nov 23 '15 at 4:52
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    $\begingroup$ @Austin the "signal" , the impulse, the energy transfer is what is limited by c , and is dependent on the microstructure i.e. drift velocities etc, the real action at the microscopic level. $\endgroup$ – anna v Nov 23 '15 at 5:23
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    $\begingroup$ @KevinFegan The simplified image is this: Even when switched off, the wire and the lamp are already full of electrons. Imagine a long tube filled with baseballs. Now you begin pushing in baseballs on one end and almost immediately there will be baseballs falling out the other end, without any single baseball travelling the whole way. $\endgroup$ – Falco Nov 23 '15 at 15:26
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As the other answers point out, there are a vast number of electrons in a piece of wire, and no single electron must traverse the whole circuit for a current to flow. You can think of an ac current as more of a sea of electrons sloshing back and forth.

I'll focus on your second question:

How would the electron flow in DC circuits work if a bulb and a 5V voltage source kept 3*10^8 meters apart were connected by two straight 3*10^8 meter long wires? Assume that there's a switch halfway between one of the two wires and it has just been flicked to the "on" position.

When the length of the wires gets longer than about 1/10th the wavelength corresponding to the frequencies in the current signal, you need to start thinking of your wires as transmission lines instead of ideal wires.

That means that you must consider that no signal propagates faster than the speed of light, and likely even slower to account for the geometry of the line and the dielectric effect of the medium between the two wires.

For the kind of situation you're asking about, the critical distance will be much less than the $10^8$ m you mentioned. For 60 Hz ac transmission, this distance would be only a 100 km or so. If you really had a fast closing switch and a high speed light source (like a laser diode) as the load, you might see transmission line effects with only a few cm between the switch and the load.

The basic result is that when you close a switch on a long line, the current signal will only propagate toward the load at about c. And when it reaches the load there will most likely be a reflected signal generated back toward the source, causing a "ring" in the signal. The details of what the propagating signal looks like depend on the details of the line geometry and the material between the signal and return lines.

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  • $\begingroup$ OPs second question was about "DC" circuits, and your answer is talking (nearly) exclusively about "AC" circuits. $\endgroup$ – Kevin Fegan Nov 23 '15 at 9:37
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    $\begingroup$ @KevinFegan, s/he used the term "DC", but the circuit he described contains a switch that changes state. This circuit can't be analyzed as a DC circuit. $\endgroup$ – The Photon Nov 23 '15 at 16:19
  • $\begingroup$ I read it (second question) as a pure "DC" question: 1) a bulb, 2) a 5V "DC" power source, 3) 2 (very) long wires of equal length connecting the 5VDC to the bulb, 4) a simple On/Off switch at the midpoint of one of the wires, and 5) the switch moved one time from the Off position to the On position. $\endgroup$ – Kevin Fegan Nov 24 '15 at 2:18
  • $\begingroup$ @KevinFegan, that's how I read it too. And that circuit can not be analyzed as a dc circuit. $\endgroup$ – The Photon Nov 24 '15 at 4:39
  • $\begingroup$ @ThePhoton I did mean DC. I don't understand why it can't be analyzed as a DC circuit. Can't there be long DC circuits? DC circuits can have switches $\endgroup$ – Peeyush Kushwaha Nov 24 '15 at 13:56
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Specially if length of the wire was large, say 3 * 10^8 meters, then would the movement of electrons on one end of the wire be "in sync" with movement of electrons on the other end?

No, they wouldn't and this fact is crucial for understanding antenna operation.

Note that even short conductors become electrically long if the frequency is high enough.

Essentially, if the length of the conductor is very small compared to the wavelength of an EM wave at the operating frequency, then the voltage and current along the wire at any instant of time is essentially uniform, i.e., we can treat the conductor as a lumped element.

However, for conductor lengths comparable (or much longer) to the wavelength, the voltage and current along the conductor will vary in space and time. For example, look at this distribution, at one instant of time, of the voltage and current along an antenna:

enter image description here

Bonus question: how would the electron flow in DC circuits work if a bulb and a 5V voltage source kept 3*10^8 meters apart were connected by two straight 3*10^8 meter long wires? Assume that there's a switch halfway between one of the two wires and it has just been flicked to the "on" position.

This is not too far from a standard problem when studying transmission line theory.

Essentially, the two wires form a guide for electromagnetic waves and have an associated characteristic impedance. You might, for example, think of your two conductors as something like 300 Ohm twin lead transmission line.

A transient analysis of a switched transmission line problem will give solutions that look like propagating step functions of voltage and current that involve reflection at the ends (due to impedance mismatch) and dissipation. Eventually, the system settles into steady state.

enter image description here

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Think of AC as something that starts out as a positive DC voltage. Then it starts going in the opposite direction. Then back again. And continues doing that over and over. Then smooth the current change out and make it sinusoidal. Now you have AC.

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  • $\begingroup$ This helps. Could you also answer the part of my question about the electron movements at two ends being in sync if conductor is very long? $\endgroup$ – Peeyush Kushwaha Nov 22 '15 at 16:50
  • $\begingroup$ Yes. In the DC case, when the voltage source(battery) is introduced into the open circuit, the wire connected to the negative terminal(1.5 units long) will start filling up with electrons till it gains a sufficient negative charge(still quite weak). The wire connected to the positive terminal(0.5 units long) will start losing electrons till it gains a sufficient positive charge(still quite weak). The charge density of the shorter wire is 3x that of the longer wire. Now the switch is flicked and current starts flowing again. After 0.5 seconds the current will reach the (to be continued...) $\endgroup$ – Laff70 Nov 22 '15 at 17:56
  • $\begingroup$ light bulb and turn it on. The current will also reach the positive end of the battery now. Since the battery is starting to become negatively charged and the negatively charged terminal isn't as negatively charged as it'd normally(like if the battery was neutrally charged and not connected to anything) be, the negatively charged terminal starts emitting electrons again. After 0.5 more seconds the part of the circuit without any current will start to have current flowing through it. (AC will be discussed in the next comment) $\endgroup$ – Laff70 Nov 22 '15 at 18:04
  • $\begingroup$ In the AC case, it depends on how fast the current is alternating. In this case lets say the current is alternating at 1 time a second(so it takes 1 seconds for the current to go from positive to negative and back). Lets say were following the propagation of current through the wire. At 0 seconds the AC current generator is emitting a current of +1 and the current we're following is in the same place and has a current of +1. 0.5 seconds later, the +1 current is on the opposite side of the wire and the AC current generator is emitting a current of -1. Thus, they are out of sync/phase. $\endgroup$ – Laff70 Nov 22 '15 at 18:25
  • $\begingroup$ @Laff70 You can edit your answer, which is easier to read than a multi-part comment. One way to handle this is to add update after comment to the end of your original question, and then add your response there. That preserves your original answer so that people can better follow the discussion. $\endgroup$ – garyp Nov 22 '15 at 22:31
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It is due to the electric field that is set up that will cause the electrons to move. The drift velocity of the electrons is much slower. There will be a delay in switching on the bulb, and it is equal to approximately $l/c$, $l$ being the length of the wire. Your diagram is not exactly right, as it shows as if the electrons are being produced at one end and being received by another, electrons will always be present (in abundance), so the end electron and the start election will be out of sync by $l/c$, the same works for AC and DC.

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  • $\begingroup$ Isn't there a net flow of charge between two terminals in DC? I understand that the conducting wire is full of electrons, and it's not like one end is producing electrons and other end consuming them, in my diagram I've shown the journey of a single electron (surrounded by many more neighboring electrons traveling with it). $\endgroup$ – Peeyush Kushwaha Nov 22 '15 at 16:46
  • $\begingroup$ There is a Net flow of charge in DC, but in AC the as you said there is a movement but the mean position remains the same $\endgroup$ – Oswald Nov 22 '15 at 16:52
  • $\begingroup$ ah okay, my diagram was about the DC circuit. Made edits to make it clearer now. $\endgroup$ – Peeyush Kushwaha Nov 22 '15 at 16:56

protected by Qmechanic Nov 23 '15 at 15:47

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