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This question already has an answer here:

I was looking at this textbook example that was posted to reddit:

Image of circuit with high travel time

The accompanying comments basically thrashed it saying everything was wrong.

I had a specific question about it, though, and this is probably due to my not understanding how electric fields work.

If the switch is placed immediately before the light bulb (say, 6" away), will the light turn on almost immediately?

I fully understand that electricity doesn't generally behave instantaneously. My concern has to do with whether there will be some kind of charge buildup on the wire that if the switch is moved closer to the lightbulb, it will wind up lighting within the amount of time it takes light to travel between the switch and the bulb as opposed to the source and the bulb.

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marked as duplicate by John Rennie, Danu, user36790, Sebastian Riese, Kyle Kanos Nov 14 '15 at 11:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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  • $\begingroup$ I don't think so. Wherever the switch is placed, the electric field travelling at $c$ will take time = $\frac{x}{c}$ where $x$ is the length of wire between battery and light including the turns on earth. $\endgroup$ – SchrodingersCat Nov 13 '15 at 17:04
  • $\begingroup$ Relevant earlier question / answer: physics.stackexchange.com/q/215166/26969 $\endgroup$ – Floris Nov 13 '15 at 17:09
  • $\begingroup$ @Floris: That one has the second switch, but is conflated by having both switches opened and closed at the same time. I'm wondering if the wire would have some kind of capacitive effect contributing to turning on the light sooner than 1 second. $\endgroup$ – JustLoren Nov 13 '15 at 17:19
  • $\begingroup$ @JohnRennie: Both of those don't seem to answer the question I'm asking, which is about switch placement with regards to the source and length of wire. Maybe I'm naive and it doesn't matter. If that's the case, I'm open to that being the answer. $\endgroup$ – JustLoren Nov 13 '15 at 17:20
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Let's think about the circuit you drew. It contains a battery, switch, bulb, and a very large inductor. In fact, the inductance of a wire that goes 10 times around the Earth can be calculated (I am going to assume an air core - in fact there is a piece of iron in the middle of the Earth which makes the resulting inductance greater).

$$L\approx N^2 R \mu_0 \mu_r \left(\log\left(\frac{8R}{a}\right)-2\right)$$

For a wire with 1 mm diameter, I calculate about 18 kH (kilo Henry) inductance for an air core - more because of the iron core but I don't know how much.

Now let's assume we have a 12 V battery and an 1.0 W light bulb - corresponding to a resistance of 144 Ohm. For a circuit with an inductor and a resistor in series, when you turn on the circuit it responds with a very slow time constant $\tau = \frac{L}{R} \approx 100 s$. A simple circuit simulation shows the current does indeed increase very gradually:

enter image description here

In a situation like this, you really have to think about the real physics - the "parasitics" in the circuit are not parasitic, they are essential to understand the behavior.

With that said - the location of the switch does not matter.

UPDATE a couple of comments said: "what if you arranged the wire so the inductance would be zero?". There is a way to almost do that: it is called "bifilar winding", where a wire is first wound one way, and then the other. in its simplest form, it would be a wire that is folded onto itself - it could be wound or straight, but because the "out" current and the "return" current are right next to each other, the magnetic field far away decreases very rapidly (dipole field) and it will be confined to the (small) region between the wires. This is in fact the principle of the transmission line.

If you use a transmission line, then it will indeed look "magically" that the electricity is traveling faster than the speed of light. Let me explain with a little diagram.

enter image description here

You see a battery, a lamp, a switch, and a long wire that returns on itself. This wire is a "shorted transmission line". Transmission lines have a "characteristic impedance" given by the capacitance and resistance per unit length,

$$Z = \sqrt\frac{L}{C}$$

Now even ordinary wire has a certain capacitance per unit length because the outgoing wire and its return are close together, and it also has a certain inductance because of the finite diameter of the wire. If you put the wires closer together the inductance goes down and the capacitance goes up, so the characteristic impedance goes down.

Key about transmission lines is that they "look just like a resistor" until the input voltage reaches the other end, and is reflected. So if you have a $3\cdot 10^8\;\rm{m}$ cable with a characteristic impedance of 50 Ohm and short circuited at the end, then it will look like a 50 Ohm resistor until the signal travels all the way to the short circuit (1 second) and the information "there is a short circuit" travels all the way back again (another second).

For such a transmission line, the light would turn on immediately - but not at full brightness: it would glow as though there was a 50 Ohm resistor in series. After a full two seconds (incidentally, the time it would take light to travel all the way to the end of the wire and back down the other leg), the bulb would become fully bright as the wire would now look like a short circuit.

The whole analysis is a little bit more complex, but I hope you get the idea. Note that different geometries of wire will give different behavior, and that the speed of propagation of a signal through a transmission line is not always the speed of light: that depends on the characteristics of the medium between the wires. So analyzing a particular situation properly will be quite tricky. At any rate, the capacitance between the wires plays a very significant role in determining how things behave. You can think of the situation right after you turn on the switch as follows:

enter image description here

The red capacitor is "in series" with the circuit the moment the switch is closed, and briefly supplies the current to keep the lamp lit; a short moment later, the next capacitor "sees" the circuit and provides a current path, then the next one, etc.

You can Google "Telegrapher's equations" to learn much more about this topic.

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    $\begingroup$ Thank you for taking the time to answer! It seems that the coiling effect is posing some form of complication that I think neither the cited textbook nor myself understood existed. Does the problem change if you straighten the wire out? An imaginary 1 light second long circle with the switch close to the lightbulb? $\endgroup$ – JustLoren Nov 13 '15 at 17:45
  • $\begingroup$ Can't you arrange the wire so that the inductance is zero? Then doesn't the question make sense? $\endgroup$ – Mike Dunlavey Nov 13 '15 at 17:53
  • $\begingroup$ @MikeDunlavey yes you can; thanks for the suggestion, I have expanded my answer accordingly $\endgroup$ – Floris Nov 13 '15 at 22:02
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Electricity actually flows faster than the speed of light. If you think of electrons as each having a little engine, it makes it very easy to understand.

I hope this helps.

Edit:

There seems to be some confusion as to my statement so I'm going to clarify.

Okay. Then. Imagine I've got a block. I pull the block. It moves. Now, imagine I've got two blocks, stacked. I pull the bottom block, it moves, and then the top block moves. With me so far? Okay, so, instead I pull the top block. It does not move, because the bottom block is in the way. Now, let's say I have 3 blocks. I pull the bottom block. It moves. Then, I pull the second block. It moves. Now I pull the third block. It moves. Now, I start with the second block, in my three block stack If I pull that block, it does not move. The bottom block is in the way. Now, I pull the third block. It cannot move, both blocks below are in the way. Okay, so, what if I pull the second block, and then remove the first block? The second block will suddenly move. The third block will not. But, what if I was pull on the second block and the third block? Neither will move. If I keep pulling, however, and then move the first block, both the second and third blocks will move at the same time. Because they both already have force acting on them, but their force does not exceed the force preventing the movement. As soon as the negating force is removed, all blocks which have a force enacting on them will suddenly move.

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    $\begingroup$ Were I in your place I'd have checked Wikipedia before posting. $\endgroup$ – John Rennie Nov 13 '15 at 17:08
  • $\begingroup$ What fuels that engine? $\endgroup$ – 299792458 Nov 13 '15 at 17:27
  • $\begingroup$ Hi Guy. The top block doesn't start moving the instant you move the bottom block. When you start moving the bottom block it creates a shear wave that propagates at (roughly) the speed of sound and the top block only starts to move when the shear wave reaches it. For a few small blocks the delay is too short to notice, but imagine large blocks made of something soft like jelly and it should be clearer why the top block lags the bottom block. $\endgroup$ – John Rennie Nov 13 '15 at 17:47
  • $\begingroup$ @The Dark Side electromagnetism $\endgroup$ – Guy Nov 13 '15 at 17:52
  • $\begingroup$ @John Rennie - I hate jelly. Can you perhaps give me another example that doesn't involve a disgusting food? Thanks! Also, Jelly isn't as firmly bound as the electrons in a wire. Your example clearly doesn't apply here. $\endgroup$ – Guy Nov 13 '15 at 17:53

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