I An interesting question from Veritasium on YouTube

[Swamp Thing]

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

Is the following answer correct?

Spoiler

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.

 

[jbriggs444]

Is the following answer correct?

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.

 

[anorlunda]

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?

 

[jbriggs444]

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.

 

[anorlunda]

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.

 

[Baluncore]

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.

 

[PeroK]

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?

 

[Baluncore]

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.

 

[PeroK]

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?

 

[Baluncore]

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.

 

[PeroK]

Not if it has a switch that is flipped every 100 years.

What's this about every hundred years?

 

[Baluncore]

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.

 

[PeroK]

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!

 

[Baluncore]

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.

 

[anorlunda]

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.

 

[pbuk]

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.

 

[anorlunda]

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.

 

[Baluncore]

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.

 

[pbuk]

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.

 

[pbuk]

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

 

[hutchphd]

Did no one read #2 by @jbriggs444 ?

 

[PeroK]

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.

 

[Bandersnatch]

Lol. This thread is an unexpected gold mine of humorous nuggets. Do carry on.

 

[Baluncore]

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.

 

[Baluncore]

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 ?

 

[TeethWhitener]

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.

 

[Baluncore]

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.

 

[A.T.]

 

[haushofer]

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

Is the following answer correct?

Spoiler

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;-

https://www.allaboutcircuits.com/textbook/alternating-current/chpt-14/characteristic-impedance/

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.

 

[rude man]

The input impedance of a transmission line is ... not related to the length of the line.

?
Maybe characteristic impedance? This xmsn line is terminated.

 

[Baluncore]

"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.

 

[Interested_observer]

Did no one read #2 by @jbriggs444 ?

Yes. EM coupling. But he didn't specify the time lapse between send and receive.

 

[Baluncore]

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.

 

[jbriggs444]

It is also possible with RS-232 that the Rx and Tx data signals might have used two wires from the one twisted pair.

Yes. I was an utter newb at wiring the things and had no prior inkling that putting transmit and receive on strands in separate pairs could have been a wise move. I may well have put pins 2 and 3 on a single pair.

 

[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.