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Q&A

How long does it take for energy to propagate in a circuit?

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The premise

In a recent video by the pop-sci channel Veritasium, the concept of the flow of electricity and energy transmission in a circuit was discussed. In that video a thought experiment is presented:

Thought experiment

The video concludes that, after the switch is flipped, the lightbulb will turn on after 1/c seconds (1 meter divided by the speed of light), since the lamp and the battery are 1 meter apart. This is explained by showing that, in an electrical circuit, energy is transmitted through an EM field, which propagates at the speed of light, and not through movement of particles in a conductor.

This explanation did not sit right with me, for mainly two reasons:

  1. Clearly, the conductor plays a role in the transmission of energy. This is actually pointed out in the video as well, by saying that after a connection in a circuit is made, the EM field will propagate along the conductor at the speed of light. To me, this would mean that the field would take 1s to propagate through the cables in the thought experiment, and the lamp will turn on after 1s.
  2. For the lamp to turn on, current must flow through it, heating the filament. Simply being in an EM field generated by the battery would not work, there needs to be some potential difference at the lamp's terminals. The signal doesn't "know" what happens at the load until it reaches it. This is a big part of reflection in signal and transmission lines, yet seems to be ignored here. Therefore, the voltage will take 1s to arrive at the lamp in this case.

The question

Is the answer of 1/c seconds correct, or should the answer be something else? If the answer is not 1/c, where does the mistake in the video's reasoning lie?

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1 comment thread

A really poor thought experiment (1 comment)

2 answers

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So here's the thing. Energy doesn't "propagate" as the video suggests. Energy is used. A voltage potential may propagate, but a voltage potential is not energy.

Considering the video

With the diagram shown in the video, the electric potential is greatest between the switch and its contacts at t=0. after t=0 the switch is connected, and the high density of free electrons between the left side of the switch and the battery will exert a force on the lower density of free electrons everywhere else. As they do this, the electrons move, creating a current. this current moves the voltage potential from the switch, and spreads it out proportionally among the resistances in the circuit. As stated in the video, the wire has resistance 0, and as such this voltage potential need only spread out to the lamp itself, after which current can flow normally. The question then becomes, how fast does this happen.

Propagation
Propagation in electronics is how fast a voltage potential (aka charge density) can travel from one location to another. Propagation is limited, but often not determined by, indeed, the speed of light 'c'. Light itself in fact is the same phenomena I just described; A voltage potential moving from one point to another. As a matter of fact, if we tuned the natural parameters (R,L,C, and to some extent the shape) of the wires correctly, and could make the switch flip fast enough, then this moving voltage potential could indeed send radio waves, and sometimes very effectively. Radio waves aside, you may have noticed that I said 'not often determined by' the speed of light, and that is because of a natural quality of all things called magnetic permeability. Magnetic permeability is something that everything has (including empty space!), and there is a universal minimum for this number. It is this quality specifically which limits how fast an electric potential may charge. For engineering purposes, you must know that this defines inductance, and inductance defines the rate at which current can change. For a more abstract approach, you can attempt to derive this equation from Maxwell's Equations:

Image alt text

Where mu 0 is the magnetic permittivity of a vacuum.

conclusion

This answer really only holds true if the wires were made of free space. The inductance of a wire cannot be zero, and that by default, adds time to lighting the bulb. This is admitted in the video.

The video however, assumes that the switch and the battery are one in the same, which they are not, and this also adds some propagation delay.

Going from the initial question to the final answer, there's a lot of goal posting, and under some theoretical conditions 1/c could be the right answer. However, there is no condition in real life where these conditions could come close to the theoretical time, and further there's no way to derive an accurate guess for this question based on the info given before they ask for an answer.

So what conditions is the answer correct

The closest way that we could arrive at a 1/C time, would be to ask the battery to be an RF source, and the wire itself to be a receiver. There are a lot more parameters which could be specified, but in general, this is the answer that the video really asks you to come to.

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0 comment threads

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Is the answer of 1/c seconds correct

It can't possibly be. The question is looking for a time value. "C" is a speed, which has units of distance/time. "1/c" therefore has units of time/distance.


In this case, 1 is 1 meter

"1" is never one meter. "1 m" or "1 meter" is one meter.

Added: I guess a discussion of the physics has been delayed long enough so that there have been enough consequences for the sloppy use of units.


One answer discussed propagation of current down the wire and radio propagation from the switch to the light bulb. There is another possibility, which is the transmission line effect.

If each out-and-back segment at left and right were a transmission line, then current would be induced in the far conductor very quickly. Another way to think of the same thing is that the near wire is the primary of a transformer, and the far wire the secondary. When current changes in the primary, there will be an induced voltage in the secondary.

In this case, with the near and far wires being 1 m apart, that coupling is rather weak. Most of the magnetic field produced by the current in the near wire won't wrap around the far wire. The energy transfer between the two will therefore be minimal. Another way to put this is that the transformer has large leakage inductance, and small coupled inductance.

If the two wires were right next to each other, then this effect could be significant. With perfect coupling and 0 wire resistance, any current in the primary wire would immediately cause an equal and opposite current in the secondary wire, which causes the magnetic field to stay at 0. In that case, the light bulb would light almost instantly. In the case actually presented, a tiny voltage could be detected after the magnetic field propagated from the front to the back wire. The bulb lighting would have to wait until the current propagates in the wire to the bulb.

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2 comment threads

I think that you have the right answer Olin. Of course, the stronger the transmission line effect, th... (1 comment)
In this case, 1 is 1 meter, the distance between the lamp and the battery. I will edit the question t... (3 comments)

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