If I arranged an experiment where light raced electricity what would be the results? Let's say a red laser is fired at the same time a switch is closed that applies 110 volts to a 12 gauge loop of copper wire with a meter at a distance of ten meters. Also, does the speed of the electricity depend on the voltage applied or the resistance of the conductor? For this test let's say the distance is ten meters through air. I'm not looking for an exact answer. An approximation is fine.

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    $\begingroup$ A standard demonstration is sending an electrical pulse into a cable a few hundred metres long. The incident and reflected pulses can be visualised on an oscilloscope, seperated by about a microsecond, strongly dependent upon cable length, allowing one to determine the speed of signal propagation in the cable. $\endgroup$ Commented Sep 24, 2017 at 12:13
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    $\begingroup$ You used to be able to do this experiment with a television, before the days of digital TV. A television tuned to a local channel on a poorly shielded cable TV line could sometimes get a ghost image slightly out of sync from the main image. It was caused by the over-the-air local broadcast signal arriving slightly sooner than the stronger over-wire signal. I'd imagine that with the right equipment you could measure the difference in latency to a pretty high degree of precision. $\endgroup$ Commented Sep 24, 2017 at 13:53
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    $\begingroup$ Relevant: ibiblio.org/harris/500milemail.html $\endgroup$ Commented Sep 25, 2017 at 12:52
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    $\begingroup$ Victor Storm's answer, which you chose as "best", talks about "good cables" and "excellent cables." But, Peter Cordes's answer is the only one that actually talks about the physics of transmission lines. Transmission line theory enables you to actually calculate what is a "good" cable, and what is a "best" cable for any given application, and it lets you actually calculate the speed with which a signal will propagate. $\endgroup$ Commented Sep 25, 2017 at 13:43
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    $\begingroup$ I thought analog TV ghosting was caused by multi-path effects. $\endgroup$
    – JEB
    Commented Sep 25, 2017 at 15:12

5 Answers 5


The speed of electricity is conceptually the speed of the electromagnetic signal in the wire, which is somewhat similar to the concept of the speed of light in a transparent medium. So it is normally lower, but not too much lower than the speed of light in the vacuum. The speed also depends on the cable construction. The cable geometry and the insulation both reduce the speed. Good cables achieve 80% of the speed of light; excellent cables achieve 90%. The speed does not directly depend on the voltage or resistance. However, different frequencies have different attenuation. In your example, the very moment of switching on represents a high frequency front that will be attenuated. While at the input the voltage would increase very fast, at the output it would increase gradually, as if with a delay. It is not really a delay per se, because the initial low level signal would get there almost with the speed of light, but its amplitude would only gradually increase and reach the full voltage with a substantial delay that would depend on the cable and circuit impedance (mostly on the cable inductance). If you use a high speed coaxial cable (like a 3GHz satellite TV cable) instead of a wire, the delay would be much shorter (80-90% of the speed of light to the full voltage). Hope this helps.

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    $\begingroup$ Light in a glas fiber is also slower than the vacuum speed of light. $\endgroup$
    – Uwe
    Commented Sep 24, 2017 at 12:36
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    $\begingroup$ I'm curious about the "excellent coax" so I've asked Are there types of standard coaxial cable with a propagation velocity of 0.9c? What would be the application?. $\endgroup$
    – uhoh
    Commented Sep 24, 2017 at 16:09
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    $\begingroup$ @uhoh: This doc shows standard velocities up to 91% cdn.shopify.com/s/files/1/0986/4308/files/Cable-Delay-FAQ.pdf - Also there are methods to further increase them as I've successfully tested (you can change the dielectric effect by applying a high voltage bias). The need for high velocities is high not for the velocity value per se, but because cables with higher velocities also have a higher bandwidth and lower jitter that is critical in telecom and digital audio applications. $\endgroup$
    – safesphere
    Commented Sep 24, 2017 at 16:57
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    $\begingroup$ It's a tradeoff between cost, practicality, and performance. If you want a coax with propagation speed close to c, then the dielectric will have to be mostly air. However air does not hold the center conductor in the center, which is of very high importance in a coax. So some dielectric supports are required. The dielectric can be PTFE foam, for example, among others. $\endgroup$
    – bobflux
    Commented Sep 24, 2017 at 19:15
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    $\begingroup$ However, almost no-one is interested in the fastest possible propagation speed in a coax. The reason "mostly-air" dielectrics are used is that they have very low losses, and this is important if the power being transmitted is huge (so the losses would melt the dielectric) or the distance is very long... $\endgroup$
    – bobflux
    Commented Sep 24, 2017 at 19:17

Also, does the speed of the electricity depend on the voltage applied or the resistance of the conductor?

Not just the resistance of the conductors, but the inductance. And also the capacitance to ground and/or to the other conductor.

Remember that an electric circuit requires a complete loop, unlike a laser. Wiring to carry electricity normally includes 2 conductors (and sometimes a 3rd ground conductor). This is the case for household wiring.

A transmission line can be modeled as a "ladder" of resistive and inductive elements with capacitors to the other conductor. (Image from the linked wikipedia article). This is one "block" of a transmission line. A real transmission line can be modeled by repeating this, and taking the limit as the number goes to infinity while the resistance / inductance / capacitance goes to zero. (You can usually ignore Gdx, the resistance of the insulator separating the conductors.)

telegrapher's equations

This model of a transmission line is called the telegrapher's equations. It assumes the transmission line is uniform over its length. Different frequencies in the same wire "see" different $R$ and $L$ values, mainly due to the skin effect (higher resistance at higher frequency) and proximity effect. This is unfortunate for us, because an impulse from flipping a switch is effectively a square wave, which in theory has components at infinitely high frequencies.

Wikipedia's transmission line article derives this equation for the phase shift of an AC signal in a transmission line of length $x$. (They point out that an advance in phase by $-\omega \delta$ is equivalent to a time delay by $\delta$.)

$V_out(x,t) \approx V_in(t - \sqrt{LC} x) e^{-1/2 \sqrt{LC} (R/L + G/C) x}$

The end result of all this is that electrical signals propagate at some fraction of the speed of light. This makes sense, because the electromagnetic force is carried by (virtual) photons (https://en.wikipedia.org/wiki/Force_carrier).

Further reading:

  • https://practicalee.com/transmission-lines/ shows practical vs. ideal (lossless) and shows the $t_{PD} = \sqrt{L_0 \cdot C_0}$ propagation-delay formula and $\displaystyle Z_{0}={\sqrt {\frac {L_0}{C_0}}}$ characteristic impedance, and some stuff about geometry of traces on a printed circuit board.

I haven't had much luck finding numbers for the transmission-line characteristics of household wiring. They're unsuitable for sending high-frequency signals, so it's not something most people bother to measure.

Ethernet wiring (like Cat5e) twists the conductors together, and has tight restrictions on the uniformity of twists per meter (and other characteristics). This is important for carrying high-frequency signals, because variations in the wiring change the characteristic impedance (for AC signals) and cause signal reflections. (https://en.wikipedia.org/wiki/Impedance_matching). AC power cables usually don't twist the wires at all, so high frequency signals will lose energy to RF emissions.

Even though the power switch is only in one conductor, flipping the switch applies a voltage difference across one end of the transmission line. What we want to know is when (and in what shape) that pulse will appear at the other end.

Household power is 50 or 60 Hz AC, so if you happen to throw the switch while the voltage difference is (almost) zero, your meter won't measure anything for transmission delay + the fraction of a second for the phase to change past the meter's sensitivity threshold. It's easier if you assume that doesn't happen and just model it as a DC spike (since the power phase changes much slower than the transmission line delay over 10m of wire.)

Thus, the transmission-line characteristics of the wire are what determines the time delay from a power switch being flipped to power "appearing" at the far end of a wire.

If anyone wants to argue about relativity / simultaneity, then do the experiment with a mirror and a transmission line that puts the detector physically next to the switch, but still electrically separated by 10 meters of wiring.

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    $\begingroup$ Chiming in from electronics stackexchange: I upvote this because you're the only one mentioning transmission lines, which are the real answer to the question... so here's a high five. $\endgroup$
    – bobflux
    Commented Sep 24, 2017 at 19:13

Consider by analogy, water in a pipe, with a valve at one end.

If the pipe is empty, when you open the valve, the molecules of water have to travel the entire length of the pipe before you see any water emerge at the far end. The time taken represents the velocity of the water in the pipe.

On the other hand, if the pipe is already charged with water, as soon as you open the valve, water begins to flow out of the far end. This much shorter period of time represents the speed at which the information (opening of the valve) traveled down the pipe - essentially the speed of sound in water.

Lining up the analogy between water and electricity:

The first case corresponds to the velocity of the electrons themselves (or electron drift); the second case corresponds to the propagation of electromagnetic waves.

In the case of an electrical circuit, the correct water analogy would be the pipe already filled with water. The electrons carrying the energy along the wire are always present; the switch simply applies or removes the potential to push them along. Measuring the "speed" of electricity by the time taken for the closing of a switch to have an effect somewhere along the conductor is measuring the speed of electromagnetic waves in the medium (electrical conductor) which is comparable to (almost) the speed of light in a vacuum.


It would depend all on the circumstances of the medium the light is traveling through, and the type of wire electricity is passing. However, if both are able to be neglected, the speed of light will be faster. The reason for this is because light is an electromagnetic wave, meaning it has no mass, since photons don't have mass. On the other hand, electricity is a flow of electrons, which have a mass, and although little, it will affect the overall speed. However when in this case, we are talking about the speed of the electrons. If talking of the speed of the energy flowing will always equal the speed of light no matter what. Usually though, the speed is calculated of the energy going through the wire, which is then slower than that of light. A clearer explanation is shown here:


hope this helps!


In a strict sense there is no $"$speed of electricity$"$. It must be distinguished among charge and EM-field. Speed of electricity can either be the drift speed of electrons (amounting a few mm/sec,) or the speed of the EM-field surrounding the cable, close to c. Electric power is exclusively transmitted by the EM-field as indicated by the Poynting-vector $S = E \times H $. (E and S are zero within a perfect conductor). For DC, the rule simply is: a) Within a conductor, there is charge transmission (current), but no power transmission. b) Within an isolator, there is power transmission, but no charge transmission.


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