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

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.

posted about 3 years ago

CC BY-SA 4.0

3y ago

Olin Lathrop‭

8059 reputation 18 332 961 469

Try this out for size: -

Hope it's clear. Maybe a little simulation of 10 km 600 Ω line might help paint a better picture: -

The voltage (V1) that feeds the t-line via the 100 Ω lamp is a 1 volt step. The end of the t-line is shorted: -

Points to note: -

• Immediately the t-line acts like a 600 Ω line and immediately some current (1.429 mA) is passing through the lamp. This is important because it basically tells you that the daft video is er... daft.
• After some time has passed, the current eventually rises in steps to a value of 10 mA (1 volt / 100 Ω).

posted about 3 years ago

CC BY-SA 4.0

3y ago

Andy aka‭

2319 reputation 19 75 268 282

I add my answer here because I think the other answers are incorrect or irrelevant: namely, the answer of Andy is nice after he has redrawn the schematic and changed the problem, answering to another (actually more interesting) question.

The answer of Olin seems to me a bit misleading: even if the two lines were very close, inducing a strong coupling, the lamp will certainly not glow continuously, but very briefly, until a continuous regime is reached. If the power supply were not a battery, as indicated by the schematic, but an AC supply of sufficient frequency, things would be different and far more complex, as we would have to take into account both the direct coupling and the continuous regime. This could be a very interesting high level problem.

As for the original question, the answer is: the lamp could have a very brief almost instantaneous glowing, if the current provided by the battery is huge, because of the tiny inductive and capacitive coupling of the wires (assuming they are not coaxial wires). But only after 1s, the time needed by the speed of light to reach the lamp via the wire, will the lamp glow more or less continuously, until a constant regime is reached after several reflections; then it will glow uniformly.

Actually, that's not right: the electric wave do not propagate at the speed of light in electrical wires, even not in coaxial one; this is obvious for ordinary wires which own some inductance, but is also true for coaxial wires where the inductance is locally canceled by the capacitance at high frequencies. The telegrapher equation says that the velocity of the electric signal is a fraction of the speed of light (e.g. 90%). So, the lamp will begin to glow continuously somewhat after 1 s.

My impression is that this question was created to illustrate one aspect of the electrical propagation, and was very badly thought and asked.

posted almost 3 years ago

CC BY-SA 4.0

3y ago

coquelicot‭

704 reputation 17 33 90 142

This is someone trying to rehash a Light experiment with circuits, however I do not think it holds up. Inside a medium, charged particles are subject to Drift Velocity. When you work through the calculations I do believe the particles move at the rate of Centimetres per second through the wire. The energy would propagate at the speed of light except that there is inherent Capacitance and Inductance to take into account which has the effect of building up energy within it, slowing down its propagation.

If the experiment were talking about a Radio Wave transmitted or some such event it may work. For the sake of engineering method, it is assumed that the conditions are ideal.

The other thing to consider is that within a wire not all the electrons are pushing the same direction. You can not control the direction of travel of an electron. You build up potential at one end and those moving the oppostite way tend to propagate. The others collide creating heat. So not all the energy is even moving the right way, in terms of this experiment.

posted about 3 years ago

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Meat5000‭

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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:

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.

posted about 3 years ago

CC0

3y ago

Enginidiot‭

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