# An interesting question from Veritasium on YouTube

Veritasium on YouTube https://www.youtube.com/c/veritasium/community has posted this problem, inviting viewers to say what would happen:

In series with the battery, switch and lamp are two transmission lines. At ##t = 0^+##, we will have twice the characteristic impedance ##2Z_0## of the line in series with the lamp. Since the wire spacing is 1 meter, ##2Z_0## would be quite large, which will limit the current and won't let the lamp glow. Once the voltage steps make one round trip (simultaneously) though the left and right lines, the series impedance will drop to zero and the lamp will light up.

But if we replace the lamp with a neon lamp or an LED, and use a high enough voltage, then the lamp would light up instantly because then the ##2Z_0## would be less than the lamp resistance.

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Demystifier

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Coupled lines, I've seen myself. I'd crimped a DB25 onto a 300 foot RS232 cable - shielded 6 strand twisted pair. Keyboard input transmitted out the transmit lead (pin 2) would result in matching characters being received as terminal output (pin 3) despite there being nothing plugged in at the far end and no physical connection between the strands.

scottdave, protonsarecool, Nugatory and 1 other person
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But if we replace the lamp with a neon lamp or an LED, and use a high enough voltage, then the lamp would light up instantly because then the 2Z0 would be less than the lamp resistance.
Wouldn't that mean that you can send a signal faster than the speed of light?

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Wouldn't that mean that you can send a signal faster than the speed of light?
One assumes that a few nanoseconds to get a field across the one meter air gap counts as "instant".

I'm not smart enough to know whether this thing works like a capacitor or a transformer, but it is clear that the connected ends of the wire 1/2 light year away are irrelevant in the near term.

protonsarecool and hutchphd
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but it is clear that the connected ends of the wire 1/2 light year away are irrelevant in the near term.
Not irrelevant. You need Maxwell's Equations to solve that problem, not circuit analysis.

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The question is a mares nest, made up of unspecified parameters;

1. The voltage of the battery. It may be 1 V or 1 kV.
2. The impedance of the transmission line. (What diameter are the wires?) 1 m = short circuit.
3. The DC resistance of the very long transmission lines.
4. The cold resistance of bulb B.
5. What is actually meant by “will light up”, a momentary flash, or a continuous dull red glow.

You can get any answer you want by setting different parameters.

Klystron, OmCheeto and Dale
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Is it safe to say that from a practical point of view, a circuit with the dimensions of several light years is impossible?

What if the circuit is of more modest proportions? Doesn't the current flow generally at something short of the speed of light in vacuum?

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Doesn't the current flow generally at something short of the speed of light in vacuum?
That sub-luminal velocity would be along the transmission line if there was a dielectric insulation rather than only vacuum. vf = √ Er .

The near end of the line will appear to be a small capacitance with ½ m long leads, in parallel with the characteristic impedance of the line. Closing the switch might light the lamp for a couple of nanoseconds.

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That sub-luminal velocity would be along the transmission line if there was a dielectric insulation rather than only vacuum. vf = √ Er .

The near end of the line will appear to be a small capacitance with ½ m long leads, in parallel with the characteristic impedance of the line. Closing the switch might light the lamp for a couple of nanoseconds.
Forgive my ignorance of EE, but isn't that a basic DC circuit diagram?

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Forgive my ignorance of EE, but isn't that a basic DC circuit diagram?
Not if it has a switch that is flipped every 100 years.

hutchphd
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Not if it has a switch that is flipped every 100 years.
What's this about every hundred years?

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The turn on transient is clearly an AC signal.
How long ago was the electrical equipment manufactured and turned on for the first time ?
DC is just very low frequency AC.

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The turn on transient is clearly an AC signal.
How long ago was the electrical equipment manufactured and turned on for the first time ?
DC is just very low frequency AC.
I find that post less than enlightening!

malawi_glenn, OmCheeto and hutchphd
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I find that post less than enlightening!
If you wait for a year, the reflection from the far end of the two short-circuited transmission lines will get back, then the status of the filament in "bulb B" will be illuminating.

pbuk
Staff Emeritus
The question smokes out two principles.
1. One of the base assumptions of CA (Circuit Analysis) is "The time scales of interest in CA are much larger than the end-to-end propagation delay of electromagnetic waves in the conductors.
Source https://www.physicsforums.com/insights/circuit-analysis-assumptions/"

Typical EM propagation speeds in wires is of the order 0.7-0.8 c.

2. Forget electricity and circuits. Basic relativity tells us that we can't propagate information faster than light. So you can't have a switch that triggers a bulb 0.5 light years away with less than 0.5 years delay. Drawing circuitry in this question is a diversion. The mechanism is irrelevant. Signals can't travel faster than light.

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So you can't have a switch that triggers a bulb 0.5 light years away with less than 0.5 years delay.
The bulb is only 1m away.

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The bulb is only 1m away.
Not following the path of propagation. Not unless we say that free space radio transmissions cross the 1m to light the bulb.

Astronuc and alantheastronomer
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Drawing circuitry in this question is a diversion. The mechanism is irrelevant. Signals can't travel faster than light.
The input impedance of a transmission line is immediate and not related to the length of the line.

tech99
Gold Member
You can get any answer you want by setting different parameters.
This - so not a very interesting question.

There is obviously some capacitance > 0 between the wires, and the more ideal we make all the components the more likely it is that enough current flows for long enough to light the bulb.

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Not following the path of propagation. Not unless we say that free space radio transmissions cross the 1m to light the bulb.
Your aerial [edit: antenna to some] pair is my capacitor

jbriggs444
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Did no one read #2 by @jbriggs444 ????
When I first read it, I just assumed he'd been drinking. Now, I think I understand it.

NTL2009, Mondayman, hutchphd and 1 other person
Lol. This thread is an unexpected gold mine of humorous nuggets. Do carry on.

hutchphd and PeroK
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Here is a simulation of lines matched to the global load. For time, treat the seconds as years.
The two 1G0 resistors are needed to satisfy the initial DC bias of the transmission line model.

When V1 first rises to 100 V, the 100 ohm globe receives half current because the circuit resistance is 50+100+50=200 ohms. When the reflection of the short circuit finally gets back, the resistance falls to 0+100+0=100 ohms, so the full current flows from then on.

Lowering the line impedance will brighten the globe initially, but reflections will go on forever.

Dale
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Here it is with 10 ohm transmission lines and immediate visible light.
Will a filament bulb survive the second year with more than 10% over-voltage ?

hutchphd
Gold Member
How long ago was the electrical equipment manufactured and turned on for the first time ?
I’m wondering about this too. How is the switch relevant? What I mean is: Does the pulse of current that happens when the switch is closed depend on how long the battery has been hooked up? That is, when you set up this circuit (with the switch open), does the fact that it takes a non-negligible amount of time for the fields to equilibrate globally affect the behavior of the circuit when the switch closes? Or as @jbriggs444 noted, does the local coupling between the wires settle into a steady state relatively quickly?

Another question: does the position of the switch in the circuit matter? If the switch is right beside the battery (and presumably ~1m from the light bulb), then closing it would obviously have a different effect than if the switch was 0.5 light years away.

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How is the switch relevant?
The state, and the position of the switch in the circuit is relevant because it controls the initial distribution of the electric field that originates at the battery. When the switch is closed the electric field must re-locate, so capacitance must be discharged here while being charged there, with the inductance of the circuit in between limiting the rate of propagation.

TeethWhitener

Spinnor, hutchphd and PeroK
Does anyone have a nice online reference about the precise physics of how energy is transferred in electrical circuits?

cmb
Veritasium on YouTube https://www.youtube.com/c/veritasium/community has posted this problem, inviting viewers to say what would happen:
View attachment 292546

In series with the battery, switch and lamp are two transmission lines. At ##t = 0^+##, we will have twice the characteristic impedance ##2Z_0## of the line in series with the lamp. Since the wire spacing is 1 meter, ##2Z_0## would be quite large, which will limit the current and won't let the lamp glow. Once the voltage steps make one round trip (simultaneously) though the left and right lines, the series impedance will drop to zero and the lamp will light up.

But if we replace the lamp with a neon lamp or an LED, and use a high enough voltage, then the lamp would light up instantly because then the ##2Z_0## would be less than the lamp resistance.
I spotted this video the other day and did not get further than him asking this question. I will find time for it later in the week, maybe.

I was (will be) wondering how he deals with it because it is a bit of a can of worms, but what I can say is that the best and most plain-language explanation I have ever seen, which also serves to offer an ideal primer on the impedance of pair/coax lines, is on the following website.

I recommend a read of this link before posing too many follow up questions, because to my mind this answers ALL the questions around this sort of scenario, albeit the reader might have to make one more step to address their particular question/problem but it should put anyone into the right thought process to answer these questions for themselves;-

It is a very good text IMHO. HTH.

At the end of the day, if you want to get into bed with electronics at a deeper level of understanding, you should work towards understanding that there is actually no such thing as 'DC'. There is 'quasi-DC' at short timescales and there is what looks like DC on longer timescales when all reactances have found an equilibrium and all resonances damped, and it makes perfect sense to talk in terms of DC for basic real-world electronics at slow switching speeds.

But once you get into 100's kHz then you need to start understanding more stuff on top of what you know about DC. And what you know about DC circuits is virtually useless to any electrical engineering over 2MHz switching rates.

Dale
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The input impedance of a transmission line is ... not related to the length of the line.
?
Maybe characteristic impedance? This xmsn line is terminated.

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"The input impedance of a transmission line is ... not related to the length of the line."
Why do you deliberately misquote me by cutting out the middle of the sentence?
The input impedance of a transmission line is immediate and not related to the length of the line.
It is immediate, now. Until the applied signal can get to the termination and back, the input impedance will appear to be the characteristic impedance of the line. Once the reflection returns, the input impedance will appear to change from the characteristic impedance, (unless the line was perfectly matched at the far end, so there is no reflection).
The characteristic impedance is fixed by the line construction.

This xmsn line is terminated.
It is mismatched at the far end. But you cannot determine the termination mismatch without transmitting another signal to the far end, then waiting to get the reflection back.

DaveE
Interested_observer
Did no one read #2 by @jbriggs444 ????
Yes. EM coupling. But he didn't specify the time lapse between send and receive.

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But he didn't specify the time lapse between send and receive.
A 300 foot line is less than 1 usec there and back. At 9600 bits per second, which is about 100 usec per serial bit, you might notice the edge distortion with an oscilloscope, but not a character time delay of less than 1 usec.

RS-232 risetime was very slow, several microseconds, so it would have been capacitive crosstalk between the parallel wires. Being open circuit, inductive coupling would have been cancelled, while the reflection would double the voltage, making capacitive coupling more of a problem.

It is also possible with RS-232 that the Rx and Tx data signals might have used two wires from the one twisted pair. There was no telling how the mates of the pair would be grounded in RS-232. It was not until RS-485 that the signals were balanced.

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Interested_observer