Basically, how long does it take for electricity to determine there is a closed circuit and how does it know that the circuit exists? I'm curious to know how it knows there is a closed circuit at any length


For clarification of the thought experiment, picture an electrical source of some type with a 1 light year wire coming from both terminals and connected in the middle. The wires are of zero resistance and capacitance.

If electrons were to start flowing immediately then information (one end of the circuit knowing the circuit is now closed at the opposite end) has obviously traveled faster than light, which is impossible.


It has been pointed out that at least some capacitance is required. So if required, consider that as a possibility in the hypothetical setup. I'm not quite sure how that affects this situation due to my lack of knowledge. I'm trying to not get to bogged down in the setup and trying to focus more on "How does the circuit know its complete & how long does it take to determine it (particularly when the distance is long)"

  • $\begingroup$ related: physics.stackexchange.com/questions/335/… $\endgroup$ – Mark Eichenlaub Mar 11 '11 at 10:07
  • $\begingroup$ @wayne: About your edit: Ahhh... You are wrong... only electrons at the very begining of the wire end will start immediately. $\endgroup$ – GJ. Mar 11 '11 at 12:03
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    $\begingroup$ It is not possible for a wire to have zero capacitance. If you make this assumption you will run into paradoxes. This is analogous to the question about what happens when you push a long perfectly rigid rod at one end. The question is nonsensical because relativity places limits on how rigid a rod can be. $\endgroup$ – Philip Gibbs - inactive Mar 11 '11 at 12:04
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    $\begingroup$ Such a wire has some capacitance and inductance per length. Nobody can cange this values for a "interstellar" wire. This is just a one turn-coil with a huge diameter, not more, not less. Any old radio ham will laugh at this thread. $\endgroup$ – Georg Mar 11 '11 at 12:54
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    $\begingroup$ Who is downvoting here all answers in this incompetent manner? $\endgroup$ – Georg Mar 11 '11 at 17:59

You need to analyze the circuit as a transmission line to understand how signals propagate along the wires to other parts of the circuit:


The velocity factor is the ratio of the velocity of a voltage signal traveling along the transmission line, to the velocity of light. A typical value for twin lead is 82%:


On connecting the source to the transmission line, electrons will start to flow from the negative terminal of the source to the wire it's connected to, and from the other wire to the positive terminal of the source. This process propagates down each wire at 82% the speed of light in the case of twin lead. So after 1/82% = 100/82 years, curent will start to flow along that part of the circut 1 light year away from the source, using your example.

  • $\begingroup$ I think that this is the best answer; it's a transmission line. And the speed assumes the geometry of the wire, but "cylinder" is natural. $\endgroup$ – Carl Brannen Mar 12 '11 at 3:07

There is absolutely nothing in any electrical circuits which is nonlocal. Everything is perfectly local and complies with all the local laws of physics. An element of an electrical circuit only "knows" about its immediate vicinity and it can affect or can be affected by the conditions in its immediate vicinity. About your specific question, let me say that if you close a circuit at some point then it takes finite nonzero time for an element of the circuit to "feel" it. The effect of closing at a particular point in the circuit will cause its immediate neighborhood to come to the same potential. That means an elementary length in both sides of the joint now comes to the same potential which in turn implies that current will flow in that elementary length (but not throughout the conductor for a very long conductor). Immediately the next neighborhood elements also come to the same potential. So more parts of the conductor will come to the same potential. All these takes nonzero time. For a one light year long conductor the whole of the conductor will take at least one year to come to the same potential in all of its parts. There is no way you can get an FTL effect for sure.

  • $\begingroup$ OK answer, here you have +1. $\endgroup$ – Luboš Motl Mar 11 '11 at 17:17
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    $\begingroup$ This answer is partly wrong because it omits the consequences of the inductance of that "antenna". There will be an oscillation. $\endgroup$ – Georg Mar 11 '11 at 19:56
  • $\begingroup$ I'm not sure which answer to mark as right! $\endgroup$ – user2482 Mar 14 '11 at 10:06
  • $\begingroup$ Yes, this answer applies to wire with significant resistance and capacitance, for example the first few underwater telegraph cables of 1850-60s. In it there are no actual EM waves, and the propagation speed is way below the speed of light. A better explanation includes inductance and the concept of Z, line impedance. $\endgroup$ – wbeaty Jul 19 '14 at 20:30

John McVirgo is correct -- it's a transmission line.

Let's say you have a 1 light year long pair of wires. As soon as you connect a battery to one end of the pair, energy starts racing towards the other end of the pair.

That energy is mostly in the space between the two wires -- electrical energy does not normally flow inside metal objects. Poynting-flow diagrams show where the energy flows, although they are a bit difficult to draw.

Various popular cables have a wave propagation speed that ranges from 42% to 95% of the speed of light. Twin lead has a wave propagation speed about 80% of the speed of light. If your cable happens to have the same wave propagation speed, then that pulse of energy will hit the end of the 1 light-year-long cable about 1/0.80 years = 15 months later.

Up until the time that pulse hits at time t=+15 months, it makes no difference whether that far end of the cable is shorted together, open and disconnected, or connected to a lightbulb.

If the battery is connected for 10 seconds and then disconnected, then a light bulb at the far end of the cable will remain dark for about 15 months, then glow for 10 seconds, and then go dark again.

  • "How long would it take electrical energy to travel 1 light year along a pair of wires?" As we just discussed, roughly 15 months.
  • "How long would it take for electrons to begin flowing out the - terminal of a battery after you hook it to a 1 light-year-long cable?" That happens pretty much instantaneously when you connect the battery.
  • "How long would it take an electron to travel 1 light year along a long wire?" That's going to be orders of magnitude longer than a year -- see Speed of "Electricity" by Bill Beaty.
  • "How long would it take for electrons to begin flowing into the + terminal of a battery after you hook it to a 1 light-year-long cable?" That happens pretty much instantaneously when you connect the battery.

Think of it like a straw: you start sucking on the straw (current) whatever is in the straw is pulled up (capacitance) when what your sucking on pulls on whatever is on the other end of the straw (voltage) begins the journey thru the straw (induction).

~15 months for twin-lead, as pointed out earlier.


The fastest response time on the other side of superconductor with no capacity and no inductivity is light speed. And this is valid only for DC (direct current). If we are using alternative current than we have resonant frequency, which is at ideal superconductor equal superconductor length divide c(light speed) . Depend of current source, which supply the circuit, electrical impedance (which is composed from capacity, inductivity and resistance) and length of real conductor (wire) we can calculate response time. Current will start to flow immediately if electric potentials are different on both sides of conductors before you do short-circuit.

  • $\begingroup$ It cannot start flowing immediately (see my clarification above, information violating FTL). $\endgroup$ – user2482 Mar 11 '11 at 11:20
  • $\begingroup$ A conductor (super is not necessary) without capacity and inductance will conduct immediately! Thus it is nonsense. Capacity and inductance of conductors are a property of the space surrounding the conductor, the same property that causes light speed being what it is $\endgroup$ – Georg Mar 11 '11 at 11:28
  • $\begingroup$ @wayne: Disagree, the only reason why current start flow is electric potential. So if electric potential (voltage) before we short-circuit both conductor ends isn't equal then current will start to flow and it takes some time to get potential (informations) to the other side of conductor! $\endgroup$ – GJ. Mar 11 '11 at 11:44
  • $\begingroup$ @Georg: I'm talking about electric (voltage) potentials. So current start to flow immediately and it takes some time to get potential (informations) to the other side of conductor!. $\endgroup$ – GJ. Mar 11 '11 at 11:48
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    $\begingroup$ @GJ You are talking about nonstationary conduction without knowing the simplest basics. $\endgroup$ – Georg Mar 11 '11 at 12:26

protected by Qmechanic Mar 22 '13 at 10:40

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