Does electricity have a propagation delay?

[astrodummy]

Suppose you have a battery wired to a light bulb and switch via a long pair of conductors, say 1 light second long. Assume for now that electricity travels at c.

The switch is thrown. How long does it take the bulb to light? Does the electricity take 1 second to get from the battery to the bulb, or is it instantaneous? If it takes one second how does the electricity 'know' the switch was thrown?

 

[Drakkith]

The speed at which a 'signal' travels through a conductor varies greatly, going from roughly 0.5 to 0.99 c. When I say 'signal' I mean any change in the voltage or current in a conductor, such as what happens when you open or close a switch.

One source: https://en.wikipedia.org/wiki/Velocity_factor#Typical_velocity_factors

 

[astrodummy]

What I am getting at is, in the case of the switch being very close to the bulb does this 'signal' move from the switch to the bulb when the switch is thrown, of from the battery to the bulb? If from the battery, what 'signal' traveled from the switch to the battery to get the current moving?

 

[Dale]

The signal starts at the switch and moves away. In your configuration it will reach the light bulb first and the battery later. The light bulb will light before the signal reaches the battery.

 

[Drakkith]

The signal starts at the switch and moves away. In your configuration it will reach the light bulb first and the battery later. The light bulb will light before the signal reaches the battery.

If we modify our circuit and place the switch between two light bulbs, then both bulbs will light before the signal reaches the batter, correct?

 

[Dale]

If we modify our circuit and place the switch between two light bulbs, then both bulbs will light before the signal reaches the batter, correct?

Yes. Remember, there is a voltage across the switch while it is open, and bringing the two sides to the same potential requires a movement of charge, or current.

 

[astrodummy]

Yes. Remember, there is a voltage across the switch while it is open, and bringing the two sides to the same potential requires a movement of charge, or current.

Soooo, the electric charge is waiting AT THE SWITCH even through it is open. That makes sense to me, thank you.

 

[davenn]

Soooo, the electric charge is waiting AT THE SWITCH even through it is open

Careful there, as that comment could lead to a misunderstanding.
There is no build-up of charge waiting at the switch ready to move as soon as the switch closes.
The measured charge measured anywhere along the wire, either side of the switch, at the battery terminals,
within the filament of the globe will be the same.
When you close the switch, there will be a pulse in the electric field surrounding the wire that will propagate
around the circuit at near light speed ( refer to @Drakkith 's earlier comment about velocity factors)
The electric field sets the electrons ( charges) into motion. and they start moving around the circuit.
For every charge that enters the circuit from the battery ( other power supply) one leaves the other end
and back into the power supply.
A final comment, The energy is carried in the electric field, not in the wire. The charges in the conductor
are generating the electric field. And to take it a little deeper, a moving electric field generates a moving magnetic field
and so there you end up with an electromagnetic field around the conductor
This is why then the switch is turned on, battery connected etc, the globe etc, comes on nearly immediately
( taking the velocity factor into account)

 

[Dale]

There is no build-up of charge waiting at the switch ready to move as soon as the switch closes.

There is a build up of charges at a switch. Typically it is small because the capacitance is small and the voltage is low. But it is there.

 

[davenn]

There is a build up of charges at a switch. Typically it is small because the capacitance is small and the voltage is low. But it is there.

hmmmmm ok

 

[Drakkith]

There is no build-up of charge waiting at the switch ready to move as soon as the switch closes.

As Dale said, there is a build up of charges. Indeed there must be in a material like a conductor, as there must be a voltage opposing the voltage source otherwise you'd have current flowing. It's just like a parallel plate capacitor. The buildup of electrons on one side and the reduction on the other sets up an electric field that opposes the battery and stops current flow to/from the switch contacts.

And it is exactly this buildup of charges waiting to move that generates the 'signal' we're talking about when the switch closes. It's like short-circuiting the plates in a capacitor.

 

[davenn]

As Dale said, there is a build up of charges.

well it must be freakin tiny

 

[Drakkith]

well it must be freakin tiny

I don't know what the capacitance across a switch's contacts are, but I imagine it is quite small.

 

[davenn]

I don't know what the capacitance across a switch's contacts are, but I imagine it is quite small.

yeah, so in the big scheme of things it's insignificant.
I suspect, going by the OP's comment, that he thought that there was this BIG lot of charge sitting there
waiting for the switch to close, which isn't true and hence why I made the comment I did, which under the circumstances think is valid

 

[sophiecentaur]

Careful there, as that comment could lead to a misunderstanding.

Oh yes - as could many of the other comments here. The fact is that even in the simplest circuit, you may have to consider what goes on in terms of EM Waves, propagating through it. To be sure of getting the best understanding out of this exercise I think we need to approach it in steps. (This is all optional, of course, and is almost never considered in small, simple circuits; in any case, most mechanical switches take a significant time to connect and disconnect.)

The most straightforward involves a 'double pole switch', right next to the battery and the distant bulb is fed with a pair of parallel wires. With this switch, both sides of the battery can be connected and disconnected at the same time. With the switch 'off', the Potential ``Difference across the wires is zero and the battery has a PD of, say 6V This is 'easy' to comprehend because of the symmetry. When the switch is turned on, the PD of the battery is transferred to the first section of the pair of wires and there is an Electric Field between them , which quickly propagates along the pair (as a step pulse and a tiny current flows from the battery) until it reaches the bulb, causing a current in the bulb which will light up. It is only then that the current through the bulb starts to affect the wires and a step pulse of current travels back along the wires and the battery starts to pass current into the wires. All this is almost instantaneous but the speed of propagation is just below c.
The battery doesn't 'know' about the presence of the bulb until the Voltage pulse has got to the bulb and then the Current pulse has got back to it. It then 'sees' the resistance of the bulb.

It's not hard to go to the next step of having a single pole switch (on just one leg of the circuit) and pulses travel 'there and back', with a similar delay to the first case.

If, otoh, you have the bulb and battery near each other and if just one of the wires it an enormous loop (one second long) then it's much more complicated because the current wave is no longer a step function and it gets far too hard for first time through the discussion. In applications where this is a significant delay - for instance in high speed computer circuits where 1cm delay is significant for circuits working at 10GHz - signals are always carried by the shortest routes and over a 'ground plane' so the pulses keep their shapes. It becomes an RF problem as much as a circuit problem.

 

[vanhees71]

The speed at which a 'signal' travels through a conductor varies greatly, going from roughly 0.5 to 0.99 c. When I say 'signal' I mean any change in the voltage or current in a conductor, such as what happens when you open or close a switch.

One source: https://en.wikipedia.org/wiki/Velocity_factor#Typical_velocity_factors

The Wikipedia article is not very accurate. E.g., in the standard classical model of dispersion the wave front moves with the vacuum speed of light, not with $c/n$ (also $n$ is a function of the frequency of the em. wave, i.e., there is dispersion); $c/n$ is the phase velocity of the em. wave in the medium at the given frequency.

Rather have a look at the Feynman lectures (here the section on waveguides as an example for em. signal propagation):

https://www.feynmanlectures.caltech.edu/II_24.html

 

[sophiecentaur]

I don't know what the capacitance across a switch's contacts are, but I imagine it is quite small.

A pair of twisted wires is about 20pF per foot. A good start when trying that sort of estimation.

 

[vanhees71]

Look for "telegrapher's equation", which is a simplified version of the Maxwell equations for wires/landlines and the like using effective quantities like resistance, capacitance, and inductivity per unit length:

https://en.wikipedia.org/wiki/Telegrapher's_equations

 

[tech99]

I think the safe assumption, if you are interested in the initial switch-on impulse or AC transmission, is to treat all circuits and wires as transmission lines. When the switch closes, it applies the supply voltage to the characteristic impedance of the line, and that dictates the initial current by Ohm's Law. Characteristic impedance is called surge impedance by power line engineers.