An interesting question from Veritasium on YouTube

[cmb]

I've now watched the video and read the posts.

I need to mention something that is missing in the conversation.

It does not contradict anything in the video but there is a missing piece. He sort of hinted at it.

The issue is that there is no real DC, it is all RF. DC is just zero frequency RF, really, and one has to consider the impedance of all parts of the system.

The term 'impedance' doesn't simply mean resistance because it is both reactance but also it applies to sources as well as loads.

In this case, the load is the wire circuit and the bulb, and the impedance of those is down to the combined interaction of those parts, not taken in isolation.

But the most important part of this circuit is the impedance of the battery, which has not been mentioned.

To match power into a circuit one must consider how the impedances match. It would be theoretically possible to create a battery with an internal impedance that is wholly mismatched to the impedance of the load, and in that case no power would flow.

He did mention this, about how much voltage would be seen straight away, and it is that. But he skipped the impedance of the source itself.

I'll give an example.

Let's say the battery is capable of delivering 10W and the bulb is also 10W. Is that enough information to tell us what will happen? If the battery were a 10mV 1000A battery with a 1uOhm internal resistance, as the contact was closed what would happen is that the full 1000A would try to flow through the battery, shifting charge from one arm of the conductor to the other. As the wires are virtually unlimited in length, the whole 1000A could flow. But also the current would not ramp in an instant but dictated by the reactance, in this case the capacitances to the other side of the line where the bulb is, and the inductance of the cables to the battery.

You can then view the voltage across the bulb as being either from the capacitance shifting on the wires to which the bulb is connected, or as a magnetic induction, being two parts of a transformer with two half-turns as primary and secondary. It doesn't matter which way you view.

But the outcome is that a voltage would be induced onto the other bulb. However, what that voltage is would depend on its own inductance, the current might ramp slowly in which case the voltage is high, or low and if the inductance is the same as the 'primary' side then the voltage ramps would match.

Meanwhile, the 10W bulb on the other side might be a 1000V 10mA bulb, in which case the 10mV voltage across it wouldn't light it up.

If the reactance of the bulb was perfectly mismatched to the energy coming from the battery side then the bulb might not light at all. Likewise, if the battery was perfectly mismatched to the circuit then the battery may not be able to release any energy.

So, yes, it is all correct but one must also look at the impedance matching between the battery and the circuit too.

Also, the matter of 1/c is OK as the question is posed because no dielectric insulation was mentioned, but the precise answer will depend on the dielectric properties of the materials inside the loop. If it was a loop that wound itself right around the earth, for example, then the bulb would take k*{earth-diameter}/c, where k is the relative permittivity of the Earth. Further, the permeability of any materials inside the loop would affect the loop's inductance, or to look at it another way, it'd affect that Poynting vector equation as Veritasium mentioned, and again the bulb will take longer to respond.

So my answer would have been; about 1/c, discounting the impedance of the space inside the loop, and necessitating good impedance matching between the battery, loop circuit and bulb.

 

[cmb]

There is a further thought I have had about the description offered by Veritasium.

If we consider his diagram where the 'energy' is flowing out of the battery and to the bulb along Poynting vectors, in the closed circuit, as he demonstrated very nicely the energy flow can be seen to be flowing out of the battery and the blub has energy incoming.

In the very-large-loop example, when the switch is thrown the bulb remains, initially, effectively isolated. One can construct a diagram showing incoming energy for the bulb, but likewise any similar circuit in the vicinity would also respond likewise.

The load becomes arbitrary to the source, and the source has to match into free space and the collection of electrically active elements within it. It is not exclusively tied to the bulb unless and until the electrical charge on the cables reaches a steady equilibrium.

In other words, activating the battery results in a radiating source, and the bulb is an arbitrary receiver, it is no different to any other similar receiver in the area, there is nothing 'special' about the bulb receiving the battery's electrical power exclusively until all the connecting lines are also filled with charge and forming electric and magnetic fields.

Once the very long lines are 'charged', the circuit impedance may change depending on the other parameters in the circuit, but one can be reasonably sure that it'd undergo an oscillation as the lines over-charge, then under-charge, etc, due to their inductance and their 'resistance' to rates of change of current. Tis is called 'ringing' and all circuits do it but usually limited to ee's attention for small circuits at high switching rates.

This oscillation on a long line due to switched DC is the problem the early long distance telegraphers suffered. The oscillation should damp, but I think that it might remain driven by the battery, or, if the Q of the circuit is high enough, even resonate and slowly increasing the voltages and currents in the circuit. Not entirely sure about that but this sort of thing can happen with reactive sources, so would need more investigation on that.

 

[haushofer]

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

I found two:

Energy transfer in electrical circuits: A qualitative account
Igal Galili and Elisabetta Goihbarg
Am. J. Phys. 73, 141 (2005); doi: 10.1119/1.1819932

and

Understanding Electricity and Circuits: What the Text Books Don’t Tell You
Ian M. Sefton
Science Teachers’ Workshop 2002

 

[Hornbein]

The answer is intended to amaze so only 1/c qualifies.

In college I learned to not even try to give correct answers. Give the answer Teacher wants.

 

[Dale]

Another question: does the position of the switch in the circuit matter?

Yes. The signal originates from the switch, not the battery.

 

[Dale]

activating the battery results in a radiating source, and the bulb is an arbitrary receiver, it is no different to any other similar receiver in the area

This is my issue with the claim. The bulb is different from other receivers like antennas. In order to light up a bulb must have a substantial amount of current through it. It is not designed to do that in response to a small RF signal propagating through free space. Although such a signal exists and would produce some minuscule current, the bulb would not “light up”.

 

[Baluncore]

In other words, activating the battery results in a radiating source, and the bulb is an arbitrary receiver, it is no different to any other similar receiver in the area, there is nothing 'special' about the bulb receiving the battery's electrical power exclusively until all the connecting lines are also filled with charge and forming electric and magnetic fields.

The battery was activated to the circuit when connected to the open switch and return conductor.

The bulb is not an "arbitrary receiver" it has energy guided to it's filament by the conductors that lead the electric field there, and the magnetic field which is guided by the EM "mirror like" surfaces of the conductors, causing a current to flow on the conductors.

You are indirectly measuring the strength of the conductor-guided electric and magnetic fields when you measure their proxies, the differential voltage and the current in at least one of the conductors.

 

[cmb]

The battery was activated to the circuit when connected to the open switch and return conductor.

The bulb is not an "arbitrary receiver" it has energy guided to it's filament by the conductors that lead the electric field there, and the magnetic field which is guided by the EM "mirror like" surfaces of the conductors, causing a current to flow on the conductors.

You are indirectly measuring the strength of the conductor-guided electric and magnetic fields when you measure their proxies, the differential voltage and the current in at least one of the conductors.

This is only the case once the current across the battery has caused the conductors to charge up (whichever polarity).

But this is not that situation. Here, the battery shifts charge from one side of it to the other and it is a significant time before that happens, so for the period in which the legs of the conductor are charging (capacitively, in fact) then the bulb remains essentially isolated from a 'direct current'.

The direct current is, in fact, the capacitive charging of the conductors in the circuit, which I think will actually take quite a long time.

 

[Baluncore]

This is only the case once the current across the battery has caused the conductors to charge up (whichever polarity).

What is only the case?
Which conductors? Everything in the diagram is conductors, except the battery electrolyte.

But this is not that situation. Here, the battery shifts charge from one side of it to the other and it is a significant time before that happens, so for the period in which the legs of the conductor are charging (capacitively, in fact) then the bulb remains essentially isolated from a 'direct current'.

What takes a significant time to shift charge from one side to the other?
The nearest ends of the transmission lines take less than 3 nsec to appear as resistors of value Zo, connecting the battery to the bulb. See the simulation in post #24.
Later, when the reflection gets back from the far end of the line, the near end changes to look like a short circuit, so the bulb current will then step up again.

The direct current is, in fact, the capacitive charging of the conductors in the circuit, which I think will actually take quite a long time.

To me, "quite a long time" is longer than a piece of string, and shorter than a parliamentary term.
I do not believe in a "direct current" concept. I believe the concept of DC is all in your imagination.
See the overshoot simulation in post #25. When and where does DC current flow in that?
Are you referring to the time it takes to charge the local wires, or to charge the entire length of the transmission line capacitance?

 

[cmb]

What is only the case?
Which conductors? Everything in the diagram is conductors, except the battery electrolyte.

The case that the battery is electrically connected to the bulb.

It is only electrically connected once the wires to it have charged up to the potentials of the battery terminal.

The 'electrical connection' we normally associate with physical wiring is no longer applicable here. It is when those wires have become charged and offer an electric field between the conductors.

Until the two wires to the bulb actually have a differential voltage between them for their entire length then there is no electric field between them that can sustain a transmission path, as the Poynting vector would disclose, between source and load.

The only transmission path is as a propagating wave across the 1m gap.

Until the battery has charged up the wires, there is simply no circuit. It is a virtual disconnection.

Consider this view instead. Let the system be in equilibrium and therefore the two lines totaling 10^16m are at the same potential (equalised via the bulb). Consider them as forming a capacitor; assume 10pF/m for an isolated wire, so we have there a 100kF capacitor.

The limit in your model, I believe, is that the bulk capacitance of the transmission line is being ignored, and is not included in the model because normally this isn't that important.

Let's say the internal resistance of the battery is 10mOhms.

How long does it take to charge a 100kF capacitor through a 10mOhm resistor?

That is one thing (and also to mention that it would take 7.2MJ to 'charge' this capacitor to 12V, thus a 160Ah 12V battery would be completely flattened just charging these wires up).

But the reason the circuit wiring would not, probably never (given that energy requirement), charge up is because as far as segments of the wire are concerned, they are a perfect match for each other, so when you attach a wire end to the terminal of the battery then an 'infinite' current will flow along the wire, thus totally neutralising the voltage on it to zero, like the rest of the wire is at the far end of this propagating wave of charge, which propagates down the wire at some function of c (according to the permittivity of its material and local space around it).

So I have now arrived at a new interpretation of what will happen;-

What would happen if you connect a battery to a 100kF capacitor? It would basically appear as a short, and after all the sparks and stuff, melted connections and all the rest, the battery will over heat and die.

It'd have to have at least 160Ah to charge up the OPs circuit to 12V, and its limiting current would be dictated by its internal impedance, for a typical cranking battery that would be around 10mOhm, so that would be an internal power dissipation of 15kW.

It'd have to do this with an RC constant of 1,000s, so would dissipate about 15kWh of thermal heat into the battery (so would have to be a lot bigger than 160Ah just to cover its own internal impedance losses).

And this is all before the circuit wiring has actually reached an equilibrium charge, capable of then dissipating through the bulb.

So my new, possibly surprising answer, is that the bulb would probably flash on quite violently (depending on its own impedance) as it is on the secondary of an air core transformer with two half-turns, for as long as the battery can sustain any voltage output at all, which wouldn't be for very long as it would 'blow up'. The IEC short circuit test on a battery of the type Veritasium is for 20 second periods.

 

[Baluncore]

The limit in your model, I believe, is that the bulk capacitance of the transmission line is being ignored, and is not included in the model because normally this isn't that important.

Now I understand why it is you make your claims. Our understanding of transmission lines differ.

A transmission line does not really have a bulk capacitance, it has distributed capacitance. It is a very long ladder network of series inductors, with parallel capacitance, distributed along the line. The key parameters are inductance per unit length, L; and capacitance per unit length, C.
From that the characteristic impedance is Zo = √ L/C .
The impedance of the line determines the ratio of voltage to current. If you connect a 50 volt signal to a 50 ohm line, 1 amp will flow, until a reflection returns. If the far end of the line is connected to a 50 ohm resistor there will be no reflection, so the 1 amp will continue to flow and you will not know how long the line is.

At the moment a current starts to enter one side of the parallel transmission line, the same current is induced to flow out of the other side of the line. That is necessary since the capacitance of the first unit length is being charged through the inductance of the two parallel unit lengths of wire. The two wires are also inductively coupled so their currents are equal and opposite. That is why the bulb is actually connected to the battery from the start, about 3 nsec into the process.

 

[cmb]

Now I understand why it is you make your claims. Our understanding of transmission lines differ.

A transmission line does not really have a bulk capacitance, it has distributed capacitance. It is a very long ladder network of series inductors, with parallel capacitance, distributed along the line. The key parameters are inductance per unit length, L; and capacitance per unit length, C.
From that the characteristic impedance is Zo = √ L/C .
The impedance of the line determines the ratio of voltage to current. If you connect a 50 volt signal to a 50 ohm line, 1 amp will flow, until a reflection returns. If the far end of the line is connected to a 50 ohm resistor there will be no reflection, so the 1 amp will continue to flow and you will not know how long the line is.

At the moment a current starts to enter one side of the parallel transmission line, the same current is induced to flow out of the other side of the line. That is necessary since the capacitance of the first unit length is being charged through the inductance of the two parallel unit lengths of wire. The two wires are also inductively coupled so their currents are equal and opposite. That is why the bulb is actually connected to the battery from the start, about 3 nsec into the process.

For moderate/short lengths.

For a huge loop as we are discussing, the capacitance becomes significant. It has to.

One can look at capacitance from the point of view of 'what is voltage'. A thing 'only has' a voltage because it has a capacitance, the relationship being that for a given amount of charge on any object (conductive or otherwise), its capacitance then determines the voltage of that object, for a given quantity of charge (imbalance).

A short track in a circuit will have virtually zero capacitance so only needs to charge up with a very small amount of charge to adopt the voltage of the source. Not many electrons need to flow to sustain a given voltage on a low capacitance track.

A thing with 'zero capacitance' (in a theoretical world) would have infinite voltage for one single elemental charge imbalance. In the real world, things have to have a 'proper' finite capacitance so that 'X' amount of elemental charges equals 'Y' voltage. There is no other way to define that relationship.

A conductor simply cannot be at a finite voltage unless the conductor has capacitance (its 'capacity' to hold charge being a ratio of the voltage versus charge).

For 'short' (sub 1000km) wires, the actual proper capacitance is rarely an issue, even in 800kV lines at 50Hz, the capacitive losses are substantial but not dominant. For millions of km of wire, this is a different game altogether and one I doubt spice is programmed to deal with.

 

[Swamp Thing]

Now I understand why it is you make your claims. Our understanding of transmission lines differ.

When it comes to high frequencies, fast transients and transmission lines, "Baluncore" has to be a user name to reckon with.

 

[Baluncore]

For a huge loop as we are discussing, the capacitance becomes significant. It has to.

There is no huge loop. The two parallel wires are a single two-wire transmission line. The two wires are coupled by their distributed L and C. The signal is differential, between the wires. The signal certainly does NOT go down one wire and then come back up the other. Maybe it would be more obvious if the parallel wires were drawn closer together on the schematic.

The voltage risetime is limited by line capacitance. The infinite current "surge" you fear is limited by the line inductance. The total distributed capacitance is broken up by the distributed line inductance. So the total capacitance is never all present in one place at one time.

The LTspice model of a transmission line that I used to model the circuit in post #24, is accepted and used by engineers world wide. It obeys all the physics, Maxwell's equations, and accurately predicts what will happen with that circuit.

You really need to study transmission line theory before you continue to make predictions using an unrealistic model. Maybe it would help if you first looked up the meaning of the word mumpsimus.

 

[cmb]

There is no huge loop. The two parallel wires are a single two-wire transmission line. The two wires are coupled by their distributed L and C. The signal is differential, between the wires. The signal certainly does NOT go down one wire and then come back up the other.

The voltage risetime is limited by line capacitance. The nfinite current "surge" you fear is limited by the line inductance. The total distributed capacitance is broken up by the distributed line inductance. So the total capacitance is never all present in one place at one time.

The LTspice model of a transmission line that I used to model the circuit in post #24, is accepted and used by engineers world wide. It obeys all the physics, Maxwell's equations, and accurately predicts what will happen with that circuit.

You really need to study transmission line theory before you continue to make predictions using an unrealistic model. Maybe it would help if you first looked up the meaning of the word mumpsimus.

We disagree.

Unless and until the wires are 'charged' with the battery source voltage there is no complete circuit, only the reactive coupling between the lines.

Applying the battery across those two lines will result in the battery pushing charge from one side to the other, so that those connecting wires adopt the same relative terminal potentials as the battery terminals. But because they start out at the same relative potential, the battery has to push apart their potentials while the connecting cables try to pull the battery's terminals together.

The connecting cables will win.

I disagree that the equalisation current in the cables will be limited by inductance. Inductance can only delay the rate of change of current, but it cannot limit the maximum it may reach.

As I mentioned, each segment of cable matches perfectly with the next segment along, so the wire is as if it is a perfect connection between the part of the cable where the charge is at the original voltage and the end 'close to' the battery. The conduction path along the cable beats the conduction path out of the battery.

Imagine a cable segment a metre from the battery and another cable segment 1km from the battery. What is the current with which those to cable segments will try to equalise their potential? Now what is the current from the battery plates to that first 1m-away segment? The former trumps the latter by some OOM, so that 1m segment will be closer to the 1km segment than the battery plates.

As I propose, look at the fact that the connecting cables have a capacitance, and thus will have some charged capacitive energy once up to the battery terminal voltage. Where does that energy come from?

This isn't a conventional transmission line problem because in a conventional transmission line, actually 'charging up' a conductor to the applied voltage (whether DC, AC RF) is considered negligible, but the specifics of this problem change that.

For a cable which is several hundred to thousands of km long, I agree with you. For megametre lengths of cable, I believe the usual features we normally neglect to consider for transmission lines become relevant.

 

[Baluncore]

A circuit in which the conductors are at some given voltage will become capacitively charged, in DC, no time domain thus no transmission line effect.

For a 50 volt battery, charging a 50 ohm transmission line, only 1 amp will flow into the line. So the charging of the line will take a long time. It will charge from one end to the other, not exponentially.

What would the capacitive energy of a light-year scale circuit be, wherein the voltage on the lines is in equilibrium, and what size of battery would that energy come from?

The hypothetical battery would be sufficiently large to charge the hypothetical long line.

 

[Dale]

Reopened after cleanup

 

[DaTario]

I watched the video and I guess it seems reasonable to think that after a time interval of the order of (1 m / 3 x 10^8 m/s) the bulb starts suffering some kind of EM influence, but essentially in a transient regime. Nothing similar to the step functions we would expect to see if the cuircuit were much smaller. But the question raised in this video is very interesting in deed.

 

[jbriggs444]

order of (1/c)

Units. I think you mean 1 meter / c -- about 3 nanoseconds using Grace Hopper's rule of thumb.

 

[DaTario]

Thank you, jbriggs444. But it is implicit the option shown in OP.

 

[jbriggs444]

Thank you, jbriggs444. But it is implicit the option shown in OP.

Yes. And the OP is incorrect on that point.

 

[DaTario]

Ok. My point is that, if the circuit has the huge dimensions described in the video, the Poyinting vector pattern will not be complete (or stationary) right after t=0, as it is in the figure bellow (credits to Veritasium). But some of these yellow filaments will be arriving at the bulb, forming a scenario for a transient regime and/or behavior.

 

[cmb]

For a 50 volt battery, charging a 50 ohm transmission line, only 1 amp will flow into the line. So the charging of the line will take a long time. It will charge from one end to the other, not exponentially.

I'd like to invite you to consider my previous comments, where I sought to generalise the impedance matching between the battery and the line, because we just don't know what those are. I have mentioned that previously.

I generalised it by saying 'one has to consider the impedance of all parts of the system' because there is surely no way that the line will act as a 50 ohm line other than for one specific frequency. I can only guess/estimate what that frequency is, but the battery will attempt to discharge at the lowest impedance it can, and when it is connected to the line it'll initially act as a virtual short.

In turn, this will produce more likely than not, by my guess, a half-Gaussian pulse with a very wide spectral content, because the inductance of the system will mercilessly squash any attempt to feed a square wave into it.

I would expect the resonant frequency of that circuit to be in the order of microHertz.

The Gaussian pulse will have a wide spread of spectral content, and while the frequency at which your 50ohm calculations are made, for sure that spectral content will be conducted at 1A for a 50V source. But that will be just at one particular frequency, I think this will be in the order of a few hundred nanoHertz for a 1 light year dipole?

The battery has an impedance wherein it will happily discharge at a much faster rate than the months-long period of the frequency at which 50 ohms will apply.

It'd help my understanding if you could please explain why you are picturing a model with, and performing the calculations with, a 50 ohm transmission line impedance.

This is effectively a centre-fed dipole of unlimited length (for at least half a year after it is powered on), with no monochromatic signal driving it but for a milliOhm-impedance DC source, so the 50 ohm impedance seems an odd choice of value to pick for this system.

 

[Baluncore]

I generalised it by saying 'one has to consider the impedance of all parts of the system' because there is surely no way that the line will act as a 50 ohm line other than for one specific frequency.

You do not understand transmission lines. They are broad-band up to a certain cutoff frequency in the GHz range. Frequency is unimportant. Square waves propagate along lines, even though they are broad-band signals.

...but the battery will attempt to discharge at the lowest impedance it can, and when it is connected to the line it'll initially act as a virtual short.

That is completely wrong. Lines simply do not work like that. You clearly do not understand transmission line theory, and have no real experience with transmission lines.

In turn, this will produce more likely than not, by my guess, a half-Gaussian pulse with a very wide spectral content, because the inductance of the system will mercilessly squash any attempt to feed a square wave into it.

Your Gaussian pulse would actually be a half-sine pulse if all the L and C parameters were local, but they are not, they are distributed, it is a transmission line.

I would expect the resonant frequency of that circuit to be in the order of microHertz.

The transmission lines have a reflection time of 1 year = 31,557,600 sec.
The fundamental would be; 1/year = 31.688 nHz.

It'd help my understanding if you could please explain why you are picturing a model with, and performing the calculations with, a 50 ohm transmission line impedance.

A two-wire line with 1 m between the wires will have an impedance between about 100 ohm and 1000 ohm, depending on the diameter of the wires. My model used 100 ohms for the lines and for the bulb.

I use 50 ohm in explanations because 50 ohm coaxial cable is used for interconnection between instrumentation such as pulse generators and reflectometers.
https://en.wikipedia.org/wiki/Time-domain_reflectometer

This is effectively a centre-fed dipole of unlimited length (for at least half a year after it is powered on), with no monochromatic signal driving it but for a milliOhm-impedance DC source, so the 50 ohm impedance seems an odd choice of value to pick for this system.

The current in the two wires is always equal and opposite, so the magnetic and electric fields will cancel away from the lines. It is not a dipole antenna, it will not radiate. That is why efficient open-wire transmission lines are balanced, so they will not radiate.

 

[Dale]

A two-wire line with 1 m between the wires will have an impedance between about 100 ohm and 1000 ohm, depending on the diameter of the wires.

Is there a simple formula for that?

 

[Baluncore]

Is there a simple formula for that?

Two wire, open line; Zo = ( 120 / Er ) · LOGe( 2 · D / d )
Where; D is separation of centres; d is diameter of the wires; Er = 1, relative permittivity for air.

https://en.wikipedia.org/wiki/Transmission_line#Twin-lead
https://en.wikipedia.org/wiki/Twin-lead#Characteristic_impedance

 

[cmb]

You do not understand transmission lines. They are broad-band up to a certain cutoff frequency in the ...

A two-wire line with 1 m between the wires will have an impedance between about 100 ohm and 1000 ohm, depending on the diameter of the wires. My model used 100 ohms for the lines and for the bulb.

You are assuming that this is a transmission line and I cannot comprehend that.

If you are arguing that two diverging cables running 180 degrees in opposite direction away from an electrical source is a transmission line, then for sure I don't understand that.

It is two lines diverging away from a dipole feed, it is a dipole antenna.

There is only one singular point on each leg of the connections which is equidistant from the source and 1m from each other. The rest are 'something else'.

The electrical source from the battery would NOT propagate down the lines on which the bulb is situated until a propagating EM wave has bridged the gap. The electric potentials on the other side would then be shorted together across the bulb, again wholly dependent on the impedance of the bulb. If the bulb was a dead short (let's arbitrarily use a 1megaAmp bulb) then clearly the leg of the circuit running in parallel 1m away would be a metal rod for all intent and purpose.

IF the battery were located at one narrow end of the circuit and the bulb on the other narrow end, with two parallel lines running between them, then I'd go with a description of a transmission line.

But it is specifically not that.

You call the circuit a transmission line and I say it is clearly not and I am really struggling to understand why you call it that. One feed wire goes one way, the other 180 degrees in the opposite direction. Transmission lines are characterised by conductors running in parallel away from the source.

There is no consistent distance between the one side of the battery to the other. The cable starts out 'very close' and diverges to 1 light year away. I'm unable to see a transmission line here. I say it will behave as a dipole antenna with a dipole receiver 1 m away, and I think it is self-evident it is 'that'. But I am not going to argue with you, I am just going to say again that we disagree and would prefer to leave it at that, for others to form an opinion/solution.

A transmission line consists of a source, a line of constant impedance and a load at the far end of it. This is clearly not 'that'.

Transmission lines work by the energy propagating between and parallel to the conductors. The only conductors that the energy propagates 'between' and parallel to (in the first few months) are two 1 metre legs of wire separated by one light year.

 

[Baluncore]

It is two lines diverging away from a point, it is a dipole antenna.

You call the circuit a transmission line and I say it is clearly not and I am really struggling to understand why you call it that. One feed wire goes one way, the other 180 degrees in the opposite direction. Transmission lines are characterised by conductors running in parallel away from the source.

Two long wires, 1 metre apart, make one two-wire transmission line.
One two-wire transmission line goes one way.
The other transmission line goes in the opposite direction.

 

[cmb]

Two long wires, 1 metre apart, make one two-wire transmission line.
One two-wire transmission line goes one way.
The other transmission line goes in the opposite direction.

What is the load these transmission lines go to?

What is the source placed across these two-wire transmission lines?

 

[Baluncore]

What is the load these transmission lines go to?

The short circuit at the far end of each.

 

[Dale]

Oh, that is what makes it reflect back down the line?

 

[ergospherical]

Is there a simple formula for that?

As a crude analysis from the transmission line equations, one finds solutions $V = V_0 \mathrm{exp}[-i(kz-\omega t)]$ and $I = I_0 \mathrm{exp}[-i(kz-\omega t)]$ with $kV = \omega LI$, and the characteristic impedance of a transmission line is defined $Z \equiv V/I = \sqrt{L/C}$. It remains to determine the capacitance and inductance (per unit length) of two parallel wires of radius $R$ separated by a distance $d \gg R$. For the inductance: the magnetic field of the first wire wire is $B = \dfrac{\mu_0 I}{2\pi r}$ therefore the flux per unit current through the relevant rectangular plane is\begin{align*}
\dfrac{\Phi}{I} = \dfrac{\mu_0 }{2\pi} \int_R^{d-R} \dfrac{dr}{r} = \dfrac{\mu_0 }{2\pi} \mathrm{log}\left( \dfrac{d}{R} \right)
\end{align*}The self-inductance is double this value (each wire contributes a flux), i.e. $L = \dfrac{\mu_0 }{\pi} \mathrm{ln}\left( \dfrac{d}{R} \right)$. For the capacitance: consider two line charges of density $\pm \lambda$, then the field between them is\begin{align*}
E = \dfrac{\lambda}{2\pi \epsilon_0 r} + \dfrac{\lambda}{2\pi \epsilon_0 (d-r)}

\implies V = \int_R^{d-R} E dr &= \dfrac{\lambda}{2\pi \epsilon_0} \int_R^{d-R} \dfrac{1}{r} + \dfrac{1}{d-r} \\
&= \dfrac{\lambda}{\pi \epsilon_0} \mathrm{log}\left( \dfrac{d-R}{R} \right)
\end{align*}which means that $C = \dfrac{\lambda}{V} = \dfrac{\pi \epsilon_0}{\mathrm{log}((d-R)/R)}$. Finally you have for the impedance $Z \sim \dfrac{Z_0 \mathrm{log}(d/R)}{\pi}$ so long as $d \gg R$ and where $Z_0 \sim 377 \ \Omega$ is the impedance of free space.

 

[cmb]

The short circuit at the far end of each.

What's the significance of the bulb then? What part does that play, how will its impedance affect the impedance the source experiences?

 

[Baluncore]

For the first year the impedance of the input end of the transmission lines will appear to be Zo. So the impedance presented to the battery when the switch closes is; Z = Zo + Rbulb + Zo.
Ohms law gives us the current through the bulb. I = V / Z.

After one year the lines suddenly look like short circuits. I = V / Rbulb.

 

[Baluncore]

Oh, that is what makes it reflect back down the line?

I don't need to explain why conservation of energy is important. It is all here.
https://en.wikipedia.org/wiki/Time-domain_reflectometer#Explanation