An interesting question from Veritasium on YouTube

[nsaspook]

What's the big deal? 25:30 on the video
The results are the same using CA Transmission theory and Field theory.

As someone here said once.

Circuit theory describes how much energy is transferred, but it never makes any claim about where energy is located within a circuit element nor where energy crosses a lumped element's boundary. There is no conflict here because circuit theory makes no claim on the question.

 

[Dale]

Normal circuit theory isn’t applicable. The circuit violates the small circuit approximation.

 

[nsaspook]

Normal circuit theory isn’t applicable. The circuit violates the small circuit approximation.

Sure, that's why a engineering transmission line theory (telegrapher's equations ) lumped model is used to solve the problem. The physical interpretation (how energy is transferred) of these solutions seems to be the issue here for some reason.

 

[Interested_observer]

The physical interpretation (how energy is transferred) of these solutions seems to be the issue here for some reason.

Isn't that the WHOLE reason for science?

 

[Baluncore]

The physical interpretation (how energy is transferred) of these solutions seems to be the issue here for some reason.

That is because the problem is deliberately presented to trap the unwary and inexperienced.
1. The transmission line components are drawn as normal wires on the diagram.
2. Some people do not yet understand transmission lines.
3. Two-terminal transmission-line inputs are twice used as series components in the bulb circuit.
4. Instead of “bulb will glow” it should ask when “a circuit current will flow in the load”.
5. The circuit has no relative values, so every analysis can be different. What is the;
a. Wire diameter, or Zo of the transmission line?
b. Battery voltage? Globe voltage?
c. Globe power, or cold/hot resistance relative to Zo?

 

[nsaspook]

Isn't that the WHOLE reason for science?

Sure, but how energy is transferred has been an answered question in engineering and science for more than a century.

 

[Swamp Thing]

With engineers and physicists, it becomes second nature to pick a model that suits the regime that one is currently looking at, knowing all the while that any contradictions between those models are not something to worry about.

When addressing non-specialists, it's a kind of ethical responsibility not to sow confusion at first, only to then seek credit for rescuing the audience from that confusion. Maybe Veritasium has sailed a bit close to the wind in this video.

 

[nsaspook]

With engineers and physicists, it becomes second nature to pick a model that suits the regime that one is currently looking at, knowing all the while that any contradictions between those models are not something to worry about.

When addressing non-specialists, it's a kind of ethical responsibility not to sow confusion at first, only to then seek credit for rescuing the audience from that confusion. Maybe Veritasium has sailed a bit close to the wind in this video.

The Veritasium video IMO is fine for the media audience it's directed to. The electrical misconceptions in the general public from grade-school level simplifications are not Veritasium's fault.

https://www.abc.net.au/science/articles/2014/02/05/3937083.htm

 

[OCR]

Lol. . . Derek Muller's Veritasium video about electric energy not traveling inside wires raised some questions. . .

.

 

[cmb]

You do not understand transmission lines.

I have been thinking more about this and I now completely agree with you about your transmission line model, and your previous circuit. Apologies if it sounded like I doubted you personally, was not the intent and I now see the light.

I still feel that there are different relevant models that could be considered (like it being a dipole or transformer) but it's irrelevant as all the answers should come out the same.

In regards suggesting I don't understand transmission lines, you're partially right because the one thing I did not understand in the first instance was a relationship regarding transmission line capacitance that I could see this would imply, that initially I had been very reluctant to believe, but now accept it.

The implication is this;
As you say, prior to the conductors being charged (forming an electric field that can carry the power 'directly'), the battery will see a load of Z+ B + Z, where Z is the transmission line impedance and B the bulb impedance.

The problem I struggled with was the capacitance of the transmission line. For simplicity I will alter the question and talk here only about what happens when a battery is first placed across both lines of a transmission line (this gets rid of the bulb and splitting the voltage of the battery).

Taking a transmission line, let's now say the propagation velocity of transmission is W (m/s) and battery voltage V (Volts), and capacitance per unit length C' being units of F/m.

The power required to charge up the transmission line is therefore (1/2)*C'*W*V^2, while the power out of the battery is I^2*Z.

So (1/2).C'.W.V^2 = I^2.Z
therefore (1/2)*C'.W.Z = 1, or W = 2/(C'.Z)

This is the thing I could see coming but struggled to believe in the first instance. I felt sure the voltage should come into it, capacitance being a squared term of voltage and power being linear, I couldn't see in my head how those would square up if I wrote out equations.

I now accept that the propagation velocity of the electrical energy (charging up the transmission line), the transmission line's capacitance and its impedance, are in a strict relationship.

Now that I think more about that it seems sort of inevitable in hindsight. I just didn't buy it the first time around. So in that respect, yes, there was indeed something more I needed to understand about transmission lines.

Sorry for taking my time getting there. I don't know if that is a 'known' relationship, I expect it is, but I was not aware of it.

It means the maximum capacitance of a 50 ohm coax is 133pF/m, I would not have suspected there to be such limits on such coax. Make it smaller and the capacitance keeps going up, is what I assumed? I guess not?

 

[OCR]

I now see the light.

Lol. . . .
.

 

[idea2000]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?
2. If someone broke the wire at the point closest to the Sun, how long would the signal take to get to the light bulb?
3. If there were two light bulbs, one at the point closest to the Sun, and one at the spot which is one meter away from the battery, which one would light up first?
4. If the wires were shielded, would the energy take longer to get to the bulb? So, would the bulb take longer to light up?
5. How long does the bulb take to ramp up to full brightness, after 1/c?

 

[Baluncore]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?

Insulated or shielded how? So long as there were still two transmission lines it would still work. The lines could even be a twisted pair, inside a conductive pipe. A solid insulation will cost more than space. The lights in your house still work with insulated wire, so long as the wire was striped where there are connections.

2. If someone broke the wire at the point closest to the Sun, how long would the signal take to get to the light bulb?

1/c. The battery and bulb would not 'know' the line was cut until a reflection got back from the cut. That reflected counter current would cancel the initial current, then turn off the bulb suddenly after maybe 15 minutes.

3. If there were two light bulbs, one at the point closest to the Sun, and one at the spot which is one meter away from the battery, which one would light up first?

The one that is one metre from the battery.

4. If the wires were shielded, would the energy take longer to get to the bulb? So, would the bulb take longer to light up?

No. A metal shield would lower the line impedance, but it would not slow the signal. On the other hand, insulation would slow down the signal because the dielectric constant is greater than space.

5. How long does the bulb take to ramp up to full brightness, after 1/c?

The voltage and current through the bulb take 1/c to stabilise. Then the thermal mass of the filament gradually heats up to the point where it radiates light energy. Maybe 0.1 second.
You can experiment with that. Turn off a filament bulb. Face the bulb. Then shut your eyes just as you switch it on. If you get the timing right, you will see the filament start to glow. You can reverse the process to see the filament dim. Use a low power filament bulb and take care.

 

[cmb]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?

Insulation - All else being equal; the lines would 'charge up' more slowly so the whole circuit would 'close' later (when the circuit finally 'completes' and passes the whole voltage to the bulb). The addition of a dielectric will increase the capacitance of the transmission line, thus slow the propagation.

The impedance of the line might change, depends on geometry and material of insulation, and if so that would affect the bulb brightness when it turns on when the switch is closed.

Given the geometry, almost imperceptibly as a thin dielectric

By way of example; the propagation velocity in a solid PE coax is around 0.6c, whereas it is 0.8c for a foamed dielectric. In a coax that was 'just vacuum' between inner and outer, it'd be at 'c'.

 

[A.T.]

1. If the wires were insulated or shielded, what would happen? How would the energy propagate to the bulb?

Insulated or shielded how? So long as there were still two transmission lines it would still work. The lines could even be a twisted pair, inside a conductive pipe.

What about a coaxial cable? The Wikipedia article seems to suggest that all the power flows between the outer & inner conductor:

https://en.wikipedia.org/wiki/Poynting_vector#Example:_Power_flow_in_a_coaxial_cable

...the power given by integrating the Poynting vector over a cross section of the coax is exactly equal to the product of voltage and current as one would have computed for the power delivered using basic laws of electricity.

 

[Baluncore]

What about a coaxial cable? The Wikipedia article seems to suggest that all the power flows between the outer & inner conductor:

That is true, what about it? It does not matter if the wave is guided by two wires through space or by two surfaces within a closed space. The inside and the outside of the braid are separate, and not connected through the holes. The external voltage wave on the outside of the coax braid has no fixed impedance, and will be quickly limited by inductance. Only the internal wave will continue because it is following a balanced ladder network.

 

[A.T.]

What about a coaxial cable? The Wikipedia article seems to suggest that all the power flows between the outer & inner conductor:
https://en.wikipedia.org/wiki/Poynting_vector#Example:_Power_flow_in_a_coaxial_cable

That is true, what about it?

Would the bulb still light up after 1m/c if instead of two wires, a single coaxial cable was used (laid out like one of the two wires in the Veritasium experiment)?

 

[Baluncore]

Would the bulb still light up after 1m/c if instead of two wires, a single coaxial cable was used (laid out like one of the two wires in the Veritasium experiment)?

The coaxial cable has two terminals at the near end. The two-wire transmission line, composed of two wires 1 m apart, also had two terminals, one on each wire.
If both terminals are connected, it will still work with coax.

 

[A.T.]

If both terminals are connected, it will still work with coax.

So it will still take only 1m/c for the bulb to light up, even when connected via one 1 light-second long coaxial cable?

Is the Wikipedia article wrong in suggesting that the Poynting vector outside of a coaxial cable is zero?
https://en.wikipedia.org/wiki/Poynting_vector#Example:_Power_flow_in_a_coaxial_cable

Outside the entire coaxial cable the magnetic field is identically zero since paths in this region enclose a net current of zero (+I in the center conductor and −I in the outer conductor), and again the electric field is zero there anyway.

Or does "Poynting vector is zero outside of a coaxial cable" apply only in steady state, and not when the current builds up?

 

[cjl]

I feel like there's poor communication going on here. At least the way I read it, Baluncore is assuming that the coax cable is hooked up using the ring connector and the center connector to give the two connections, one hooked to the lightbulb and one hooked to the switch. In this case, yes, it will act as a transmission line with its characteristic impedance.

It seems to me that A.T.'s question though is about using coax in a more conventional manner, with the ring grounded and using the center wire to connect to both the bulb and the switch (which, necessarily, means you'd see two parallel coax cables running off into the distance if you were sitting at the location of the switch. In this case, no, no current would flow through the bulb until the signal has propagated all the way around the circuit.

On a separate note, this is wrong:

The voltage and current through the bulb take 1/c to stabilise. Then the thermal mass of the filament gradually heats up to the point where it radiates light energy. Maybe 0.1 second.

The bulb won't achieve full brightness until at least the time it takes light to go out to the end of the circuit and come back, possibly longer if there's an impedance mismatch that causes several reflections before steady state. Steady state, the battery doesn't see the transmission line impedance, it only sees the DC resistance of the bulb and the wire.

When you first turn it on, the battery sees the transmission line impedance and bulb resistance in series, but you get a substantial ramp up in current after the field propagates fully around the circuit and reaches the bulb, at which point the battery no longer sees the wire as a transmission line. This is very obvious in the simulated behavior seen in this video for example (simulation results start just after 6 minutes in, but the setup is interesting too).

How much brightness you'll see originally vs after the field has fully settled to steady state depends of course on several factors, but you'll always see an increase in current after the field has propagated fully around the wire.

 

[Baluncore]

I feel like there's poor communication going on here.

The problem is very poorly specified, as are the questions.
This is certainly not the place to discuss transmission lines in general.

 

[A.T.]

The problem is very poorly specified, as are the questions.

Sorry if I didn't describe the setup clearly enough. What I had in mind is using the 2 strands of a coax cable in the same way the 2 simple wires are used in the Veritasium setup:

For example:
- Inner coax strand connects: battery positve to light bulb terminal 1
- Outer coax strand connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart.
- The switch at the battery interrupts both coax strands.

How long will it take for the light bulb to receive any power after the switch is closed?
About 1m/c?
About 1s?

 

[Baluncore]

What I had in mind is using the 2 strands of a coax in the same way the 2 simple wires are used in the Veritasium setup:

But there are four simple wires in the symmetrical Veritasium setup. Describing a circuit diagram in words is not a wise move.

Fundamentally, the Veritasium setup uses the near end of a transmission line as a local component in a local circuit. It does that twice, once on the left hand side, and once on the right hand side of the circuit diagram.

- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart.

What happens at the far end of the two coaxial cables? Are they joined to make the mid-point of one longer cable, or are they separately short circuited, while remaining 1 m apart?

Since there is no signal on the outside surface of a coaxial cable, the fact that it could be laid out like a hairpin, one metre apart seems irrelevant. It might as well be a circle with a circumference of one second. Maybe I misunderstand you.

If you connect a battery to the near end of a one second long transmission line, the connection transient will take one second to reach a globe at the far end of the line. The geographic path that the line takes is irrelevant. It does not matter how far apart the ends of the delay line are, the signal must follow the line. That is quite different to the Veritasium circuit where there is a reflection from a short circuit at the far end of each transmission line.

- The switch at the battery interrupts both coax strands.

That should not be necessary. One switch pole should be sufficient to make or break the circuit at the battery. I wonder why you make that point.
What is a “coax strand”? Maybe it is another opportunity for a miscommunication.

 

[A.T.]

The problem is very poorly specified, as are the questions.

- Inner coax strand connects: battery positve to light bulb terminal 1
- Outer coax strand connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each.
- The switch near at the battery interrupts both coax strands.

How long will it take for the light bulb to receive any power after the switch is closed?

What happens at the far end of the two coaxial cables? What is a “coax strand”?

In my coax setup there is only one coaxial cable. At the far end (1/2 lightsecond from the battery & bulb) it simply bends around. By strands I meant the two coax conductors: outer & inner.

If I understand you correctly it would take 1s in my coax setup?

That is quite different to the Veritasium circuit ...

Of course it is.

... where there is a reflection from a short circuit at the far end of each transmission line.

For the original Veritasium question it is irrelevant what happens at the far end. The bulb there will light up after ~1m/c, thus long before any information about the switch closing can even reach the far end.

 

[cmb]

In my coax setup there is only one coaxial cable. At the far end (1/2 lightsecond from the battery & bulb) it simply bends around. By strands I meant the two coax conductors: outer & inner.

TBH I have no idea what your setup is, precisely, and I am looking through all of your posts.

Why not draw up an equivalent circuit diagram in spice, like baluncore has done. It resolved, for me, the original issue I had with his explanation. Diagram is needed here, if you really want any proper discuss about it, and if you do put it into spice, then just hit the 'run' button and let spice tell you the answer!

 

[A.T.]

TBH I have no idea what your setup is, precisely, and I am looking through all of your posts.

You only need to read this:

- Inner coax conductor connects: battery positve to light bulb terminal 1
- Outer coax conductor connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart. At the far end (1/2 lightsecond from the battery & bulb) the coax cable simply bends around.
- The switch at the battery interrupts both coax conductors.

How long will it take for the light bulb to receive any power after the switch is closed?

 

[Baluncore]

How long will it take for the light bulb to receive any power after the switch is closed?

Are you asking how long a signal takes to travel along a line that you have specified as being one second long, or is this a trick question ?
How do you define "to receive any power" ? Do you mean to receive any energy ?

 

[cmb]

You only need to read this:

- Inner coax conductor connects: battery positve to light bulb terminal 1
- Outer coax conductor connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart. At the far end (1/2 lightsecond from the battery & bulb) the coax cable simply bends around.
- The switch at the battery interrupts both coax conductors.

How long will it take for the light bulb to receive any power after the switch is closed?

I'm supposed to guess what the switch is connected to?

Just do the diagram in spice and press the 'run' button.

 

[A.T.]

Are you asking how long a signal takes to travel along a line ...

No, just how long the signal takes, without additional constraints on how it gets there.

How do you define "to receive any power" ? Do you mean to receive any energy ?

What difference does it make whether I ask about "time to receive any power" or "time to receive any energy"?

 

[A.T.]

- The switch at the battery interrupts both coax conductors.

I'm supposed to guess what the switch is connected to?

The switch interrupts/connects both coax conductors simultaneously, near the battery. What is ambiguous about that?

 

[hutchphd]

I'm supposed to guess what the switch is connected to?

The description is complete. (Yikes! get a grip)

In this case the bulb will receive a very small signal through the air immediately but will heat up more than one second later, depending upon the coax propogation speed.

 

[A.T.]

- Inner coax conductor connects: battery positve to light bulb terminal 1
- Outer coax conductor connects: battery negative to light bulb terminal 2
- The battery is 1m away from the light bulb
- The coax cable is 1 light-second long, laid out in two straight parallel segments of 1/2 light-second each, 1m apart. At the far end (1/2 lightsecond from the battery & bulb) the coax cable simply bends around.
- The switch at the battery interrupts both coax conductors.

How long will it take for the light bulb to receive any power after the switch is closed?

In this case the bulb will receive a very small signal through the air immediately but will heat up more than one second later, depending upon the coax propogation speed.

Thanks for the reply. Is the initial very small signal through the air because the fields outside a coax cable are not completely zero in the transient phase, when the current builds up?

 

[Baluncore]

No, just how long the signal takes, without additional constraints on how it gets there.

1/c. The parallel external coaxial braids form a separate external two wire transmission line, shorted at the far end, where the line loops. Call that say Zo = 600 ohms, until the short circuit appears after about 0.7 seconds, because the coaxial external velocity factor will be greater than the inner. That circuit is closed on the centre conductor side by stray inductive and capacitive coupling between battery and bulb, terminals and wires.

Exactly 1 sec, when the signal on the inside of the coax gets to the globe.

Power is the rate of flow of energy. It is not energy.

 

[Baluncore]

Is the initial very small signal through the air because the fields outside a coax cable are not completely zero in the transient phase, when the current builds up?

No. It is capacitive and inductive coupling of the local wires. The outside of the folded coaxial cable is a quite separate transmission line to the inside of the same coaxial line.

 

[cmb]

The description is complete. (Yikes! get a grip)

So it is a quadruple pole switch, that is on both sides of the battery at the same time?

Somehow?

Really, a diagram would have clarified this.

OK, it's over my head now then!... I'm out.