"Veritasium" (YT) - "The Big Misconception About Electricity" ?

[T3sl4co1l]

Just watched it.  Excellent presentation of the situation, and yes indeed it is about energy flowing in fields.   (And, as for skepticism, this one seems pretty straightforward.  And indeed, the sponsor uses the mechanism that is the subject of the video, in a fairly direct way, so it would be strange for them to object to its content.  Not that that's saying much, as it's a pretty general topic, with respect to anything electronic at all. )

And yes, he did have to make a sneaky definition, which is, although the lightbulb has some given resistance, for argument's sake we're going to say it's very high so that it responds to any significant change (i.e. to a change in voltage comparable to the battery), so that the round trip delay is not required/incurred, and instantaneous operation results.

Another catch: the type of lamp he had there, may be a one of those "backup" LEDs with the internal battery, and DC leakage detector: when no AC is applied, and a DC path exists (such as through the pole/pad transformer), it lights up.  This condition will take some 10s of ms to detect, skewing the timing measurement (but, given the ratio between prompt and reflected waves, this is still more than good enough to discriminate the cases in the thought experiment.)  Else, if it's an ordinary LED bulb, it will still be at least a few ms to charge up its power supply (filter caps, and stabilize whatever kind of converter it's using, if any).  Now, it's also possible that it's the simplest type of LED: just a string of them, with current-limiting resistor and FWB.  This will actually light quite promptly: some nanoseconds to propagate into the lamp itself, and a few nanoseconds more for the LEDs to emit.  A few more ns and the yellow phosphors become active, and there you have it.  (LEDs have been proposed for one-way data transmission; data rates of low 10s Mbit are easily achieved.)

Other catches that frustrate a real physical experiment: the lines will radiate, even if they do not have resistance (must be superconducting); there's also a common mode applied to each line, whereas if they were paired together they could cancel out.  So they will radiate strongly in this configuration.  But we could indeed wrap up both transmission line stubs into a cable, and have equivalent behavior in a compact, lossless (and hopefully non-dispersive) medium, that, if low enough loss overall, could indeed illustrate an arbitrarily long delay.

What does the bulb actually need to be?  Well, it needs to be a high resistance, to suit the high impedance lines connecting it; and it needs to run at low voltage, since it's just a 12V battery or whatever powering the whole circuit.  A small automotive (e.g. dash) light, or just some indicator LEDs (and current limiting resistor), will do the job.

The twin-lead configuration will have a characteristic impedance on the order of 600 ohms, so applying 12V to two in series draws about 10mA, and therefore a lamp of several kohm will light noticeably.  And this will be true within the few nanoseconds claimed, because mechanical (switch) contacts are actually quite fast indeed (fractional ns), giving a measurable wavefront spanning the line separation distance.

If we allow that less propagation delay is acceptable, then we could indeed measure such a setup, say using a high speed camera and lines of some microseconds length.  An extremely fast (streak or equivalent-time sampling) camera would however be needed to observe the direct propagation.  Else, if we permit an oscilloscope, it can all be measured at once.


The one thing I don't like / get, is introducing undersea cables at the very end; why they failed to perform, is not explained.  One might indeed assume from the video, that the Poynting vector works fine on them, as well as anything else, so what's the deal?  The real problem is something more subtle, and so it's no accident it's omitted, but it's a shame not to mention it at all, really?  I at least would prefer hinting at a deeper mystery, than not mentioning it at all.

(The real reason is that, not only does the line have inductance and capacitance, but resistance and conductance as well, and both must be balanced in order to have low dispersion, that is, to preserve time-domain wave shape.  This is not a nonlinear distortion, but a time distortion which is objectionable to our time-domain signaling systems -- Morse code.  The solution was, inserting loading coils periodically, in series.  This can be modeled as effectively tuning out the capacitance of the intervening TL sections, but also at the same time, balancing the resistive (lossy) and reactive (propagating) effects, giving a flat frequency response.  It turns out, in most cables, the dielectric has significantly less loss than the conductor, and to balance this, the inductivity of the line must be increased.  Hence the loading coils.  Mind, this particular approach only flattens the bandwidth to a point: the periodic structure happens to exhibit a bandgap or lowpass response, so it's flatter below the cutoff, but completely and utterly useless above it.  The chosen values and spacing, of course, were more than adequate for 10s of wpm telegraph.)

(Later in history, multi-channel (T1) voice cables were developed and laid.  These offered a bit over a MHz of analog bandwidth, with low dispersion and excellent flatness.  (Analog trunk lines worked by multiplexing voice channels into frequency bands, much as ADSL and DOCSIS work today for transmitting digital data.  A ~1MHz trunk can therefore carry 30 voice channels or so.)  Attenuation was also kept very low, by using not just loading coils, but repeater amplifiers -- these were in turn powered by DC send along the cable, the amplifiers being wired in series along it, so each station had a +/- some thousand volt supply "charging" the cable.  Even as late as 1980, these used vacuum tubes of the latest technology and refinement -- with an extremely high purity cathode, liberal use of gold for the grids, and transient protection to deal with inevitable dielectric discharges, these operated for several machine-centuries without failure, until being superseded more recently with fiber optic repeater cables.  Truly marvels of reliable engineering.)

Tim

[HendriXML]

Do mind it's worth bringing some skepticism to Derek's videos these days -- three reasons:
1. Just because, of course; try not to take things at face value, but understand what relationships or motivations might underlie a claim.
2. YouTube revenue.  He's quite open about this, tuning everything from content to thumbnail to optimize viewership.  This isn't necessarily a bad thing -- greater viewership and a good explanation introduce more people to a technical subject.  But it does affect how the subject is presented, more sensationalized perhaps, creating drama from academic disagreement, etc.  (And also not that this has specifically happened -- just that it's something to beware of.)  And of course, the major downside of popular science presentation, the explanations can be oversimplified, and the content very shallow, so it may not even be all that useful if you want to get into the subject.  (But that's an audience problem -- it's an introductory video, you're simply looking in the wrong place if you want depth.  Can't have everything, unfortunately.)
3. Corporate sponsorship, when applicable.  The criticism of his recent driverless car video is particularly apt.  Look for similar patterns in, well, anything you consume, of course: we can especially place blame in this case when the channel's byline is "an element of truth", but in general, anywhere you see noncritical presentation or acceptance of facts, especially when the presenter may have a vested interest in the subject (sponsorship is a fine example!), keep your guard up.  Let alone possible omitted facts -- these can be hard to spot without broad knowledge in a subject, and so are an common strategy.

Tim

Spot on!

If this would be a troll video. I'm gonna unsubscribe.

Just kidding.. Some seem to be desperate to excersice (even very little) power.

However I already read in the comments,  that some physics teachers now believe that the bulb shines in 1/c   time. (That can only mean at least some 90٪, not some initial low energy transfer). If it turns out to be bs, then his scientific reputation would suffer. Even though he is largely a video producer.

Personally I've searched and seen more videos about this topic. Not one was completely satisfying, so I hope someone makes a better one. 

[bdunham7]

The twin-lead configuration will have a characteristic impedance on the order of 600 ohms,

Could you elaborate on that?  Are we talking about two copper wires 1 meter apart in space or did I miss something?

Edit:  So actually doing a little math, the characteristic impedance by the simplified formula using the impedance of free space, 377 ohms and Z₀ = 377/pi * ln (2D/r), with D being the 1 meter and r being the radius of the wire I'm getting characteristic impedances of 1K or so, so not far off depending on the wire size.  However, that formula doesn't hold down to DC, you need the full one for that, and that has frequency in it.

[mdubinko]

I think @Simon311 is on to something. Unlike the case with most circuits we'd ever find ourselves looking at, with this one you need to take into account the pre-conditions. The circuit diagram shows the positive side of the battery permanently connected to 1ls (light-second) of wire, then the bulb, then another ls of wire, then a switch, and a trivial amount of wire back to the negative terminal. We are to assume zero-resistance wire.

Even with the switch open, when arranging the circuit, the moment you connected the positive terminal, something strange would happen. The positive terminal has a deficit of electrons, while the wire is thoroughly neutral. So a current would flow as the entire circuit [excepting the open side of the switch and the short piece of wire from there to the negative terminal] stabilized at a positive charge relative to the negative end of the battery. (The current would trend to 0, so the voltage drop across the bulb would also reach 0 in the limit)

When closing the switch, there would be an immediate difference in potential, so it's certainly reasonable to predict that *something* would happen within nanoseconds.

What would happen if you had a double-pole switch fencing both sides of the battery? That forces these preconditions into post-conditions.

But here's what I don't get. Say there was a second light bulb, the same distance from the battery as the first, but with no wires whatsoever. Still on the double-pole switch scenario, how long after throwing the switch would it take for there to be a noticeably different electric field between the two bulbs? It would seem that since the only difference is the wires, it would require the impulse to travel the length of the wires (after all, there could be a break in the wire half a light-second out). Until that point, there's no electric field at the bulb, so long as E is zero, E x B is also zero. No Poynting vector, no energy, bulb doesn't light.

OK, now tell me all I've gotten wrong.

[rfeecs]

I agree that there is at least theoretically an initial response, the question is the magnitude and whether that can justify the assertions in the video.  I think the previous poster who suggested considering the initial response as if the ends were not terminated has the best idea for analyzing the initial conditions.  And I'm not sure that the initial currents will be equal and opposite, could you explain that one?

OK, off the cuff:

Say initially the voltage at one end of the switch is zero, and the voltage at the other end is +V.

Close the switch.  Current will start to flow through the switch.

This current creates a magnetic field rotating around the switch and wires attached to it in the vicinity of the switch.

This changing magnetic field will create an electric field rotating around the magnetic field.

This electromagnetic field will propagate across the distance to the other wire at approximately the speed of light.

When the electric field hits the wire, charges on the surface of the wire are going to move to try to maintain no field tangential to the wire, and no changing magnetic field inside the wire, producing a current.

So that is where the 1m/c time delay comes from.

Are currents in the wires initially exactly equal and opposite?  Maybe not.  I was thinking based on ideal transmission line theory.

(Edit:)
If the two wires form an ideal transmission line of say 600 ohms, then initially they will look like simply a 600 ohm resistor.  So any current going in to one end of the resistor will come out of the other.  This results in equal and opposite current on the two wires.

[bdunham7]

This current creates a magnetic field rotating around the switch and wires attached to it in the vicinity of the switch.

This changing magnetic field will create an electric field rotating around the magnetic field.

This electromagnetic field will propagate across the distance to the other wire at approximately the speed of light.

When the electric field hits the wire, charges on the surface of the wire are going to move to try to maintain no field tangential to the wire, and no changing magnetic field inside the wire, producing a current.

So that is where the 1m/c time delay comes from.

Are the wires initially exactly equal and opposite?  Maybe not.  I was thinking based on ideal transmission line theory.

OK, but the permittivity of the free space is much, much lower than the conductor--that's why conductors conduct after all.  So the current going down the side connected to the battery will only be limited by the characteristics of the wire--self inductance and resistance.  The reaction of the other wire is going to be only very loosely coupled and much less current will flow.  I suppose the resistance of such a long wire will be so high as to make the whole experiment pointless.

I think T3sl4co1l, who apparently suffered through the whole video ads and all, has the right take.  And since it is essentially a clickbait sort of situation, if he is 'called out' on it he will probably set up something that he claims is equivalent with an even lower characteristic impedance, light up a special type of light bulb and declare victory.  Perhaps he's looking for a sucker to bet $10,000.

[sleemanj]

I think I sort of understand, am I right to say that when the switch was closed, LS1 would "light" (for some definition thereof) ~immediately, LS2 ~0.5 seconds later, and LS3 ~0.5 seconds later again, and that if the switch was then opened again that LS1 would extinguish (for some definition thereof) ~immediately, then ~0.5 seconds later LS2 goes out, and ~0.5 seconds later LS3 goes out.

[T3sl4co1l]

Correct.

We can avoid using fields for a bit, on these sorts of problems, by modeling the transmission line as a delay between two ports, where local circuits around the ports obey nice "DC" rules like Thevenin.

Note that Thevenin is violated for the whole system, because there's clearly one piece of wire between the battery and LS3, and the currents on it don't match at any given instant (assuming we're switching it on and off enough that it doesn't come to equilibrium, anyway).

When the speed of light matters, when waves are propagating, we have to use a more restricted form, so that speed of light again does not matter, for the parts of the system we're using the analysis on.

So, the circuit can be changed to look like this:



As long as the local loops, and wavelengths / wavefronts, are small enough that the current is reasonably equal around the loop (and the voltages add up, same idea), then we can break it up this way.  Read the rectangles as a particular kind of dependent source: namely, that has some impedance (Zo), and the voltage at the port is the sum of applied and transmitted voltages, the transmission being delayed from the port it's paired with.  Put another way, an ideal isolation transformer, plus an ideal delay, and it has impedance.

There doesn't need to be any common ground here; ideal ports are two terminals, perfectly floating from anything else, no continuity between them.

We would, of course, need to assign grounds for purposes of SPICE simulation, say.  And, real transmission lines do have common mode impedance, so we need to model that, to the extent it's relevant to the analysis.  (Assuming we have a ground to measure CM with respect to, we can model that as its own transmission line, with the ports common-grounded to that reference plane.  That's another common structure: having single-terminal ports over ground, such as anything with an array of coax connectors coming out of it.  RF and EMC setups are typically designed this way.)

Note that, if we want to better emulate this IRL, we can increase the CM impedance, say by loading the line with magnetic cores (with ferrite beads, or windings on a core, as a transformer / CMC) -- this allows us to maintain the fiction of port isolation, with respect to some minimum CM impedance (i.e. Zcm > Zcm(min)), over some modest bandwidth (extending to ever lower frequencies by increasing core and turns, though never to DC).

As it happens, calling a transformer a transmission line, really isn't an accident; transformers do indeed have characteristic impedances and cutoff frequencies, corresponding to winding geometry and wire length.  This is an extremely useful way to design transformers, at least when you can afford to -- typically the case at RF, sometimes difficult at switching frequencies, and not very common below that.

Tim

[Peeps]

I watched the video and read through the thread, but isn't the answer to this fairly simple?

Since he talks about EM fields, he implies the existence of inductance. So his 1/2 light second long cables inherently have an incredible amount of inductance to them. You can simulate the circuit, but we know intuitively that anytime you have a great big inductor, it will take some amount of time for the circuit to reach steady-state due to the energy required to build or collapse the EM fields.

So in practice, the bulb would not turn on instantly but light up extremely slowly as the fields built up, and if you dared to flick the switch off, you would get a great big arc across the contacts as the fields collapsed.

[T3sl4co1l]

I watched the video and read through the thread, but isn't the answer to this fairly simple?

Since he talks about EM fields, he implies the existence of inductance. So his 1/2 light second long cables inherently have an incredible amount of inductance to them. You can simulate the circuit, but we know intuitively that anytime you have a great big inductor, it will take some amount of time for the circuit to reach steady-state due to the energy required to build or collapse the EM fields.

So in practice, the bulb would not turn on instantly but light up extremely slowly as the fields built up, and if you dared to flick the switch off, you would get a great big arc across the contacts as the fields collapsed.

That's a way to approach it -- but it is just an approximation.

Indeed, the inductance will be on the order of 754 henries.  But here is the key question:

At what frequency (and any other conditions if relevant) do we measure this approximation?

The catch is, the LF approximation only holds once standing waves have mostly decayed.  If the lamp has matched impedance (about a kohm, per above), the reflected wave will be absorbed, and no apparent inductance or capacitance remains in the system -- something weird's happened in that first second, but after that, it's essentially steady state.  If it were an ideal inductor, you'd expect it to continue exponentially, with measurable change after several seconds.

If the lamp were say a headlight, so the equilibrium current draw is a few amperes, well we know the first-pass current will be about 10mA, so it'll take around a hundred passes for the current to finally ramp up to its final value -- and here we do see the effective inductance, if we smooth over the standing waves (or allow that the line has just enough (dielectric) loss that the waves die out sooner, while still having zero DC resistance so current still reaches the desired final value).  And this will indeed take some minutes, and so we get some idea of what frequency it's reasonable to measure the LF equivalent at -- some fraction of the electrical length of the line, at the very least!  (So, some 100s mHz in this case.)

We can often skip this depth of analysis in practical circuits, by using such low frequencies and impedances (or high impedances, when the equivalent capacitance is relevant), and such short lengths (e.g. ~cm PCB traces in a digital circuit with 10s ns edges), and enough loss or dispersion, or slow enough excitation (or scopes to read it), that we wouldn't (or couldn't) notice the standing waves, even if present -- but we do need to be careful when dealing with modest length transmission lines (where the standing waves would be significant).  And this situation is especially contrived to violate all those expectations, because there's simply no such thing as a transmission line with that kind of length, and (low) lossiness.

Tim

[sandalcandal]

At what frequency (and any other conditions if relevant) do we measure this approximation?

The (ideal) switch turning on will be a Heaviside step function, taking the Fourier transform of that is a Dirac delta at 0 Hz (0 rad/s) plus an imaginary rectangular hyperbola.
https://mathworld.wolfram.com/FourierTransformHeavisideStepFunction.html

More "realistically" you could use a logistic function https://en.wikipedia.org/wiki/Logistic_function the Fourier transform is a bit more messy as you'd expect but akin to "flattening out" the 0 Hz (0 rad/s) Dirac delta as you might expect https://arxiv.org/pdf/1502.07182.pdf

Without having done the complete mathematical modelling, I believe Derek is "theoretically correct" even with the dispersion caused by the incredibly bad 1 light year [second] wide transmission line, if you had an ideal bulb that only requires >0A (>0W) to detect the EM field change then with a finite rise time source step [or any frequency source for that matter] you'd be able to detect the change after 1/c. I think the math/physics might screw up a bit if you with use an actual Heaviside step with infinite rate of rise or you try use QED where any known physically possible detector cannot interact with infinitely small energy but even then I think the answer would come pretty close to 1/c.

Could be an interesting exercise to try derive the dispersion function of this transmission line. We'd need more info on the source and load characteristics to get a complete "picture" of how the "bulb" would actually light up.

Edit: Using the Poynting vector analysis might be easier than trying to work this problem into transmission line analysis. We should probably watch the expert interviews and analysis Derek put up.
Edit2: Analysis slides using transmission line models linked by Derek: https://ve42.co/bigcircuit [Would light up at 1/c but would take ~2 sec (for the 1 light second total width case) to reach peak]

Edit3: Another interesting paper I saw posted on this forum a while back about energy transfer within a [toroidal] transformer using Poynting vectors. https://www.researchgate.net/publication/43483876_Power_flow_in_transformers_via_the_poynting_vector

Edit4: IMO the "energy goes through the wires" picture is alright for basic, non-relativistic modelling for electrical systems. This Poynting vector stuff only practically matter when you get into RF territory (even then you can get around it to an extent using transmission lines in your lumped model analysis) and even then Derek's model has limitations because it fails to account for quantum behaviours which requires stepping the model up to using QED.
Edit5: then you could go even further to using electroweak interaction which is at the edge of the current Standard Model AFAIK. My formal education only went as far as starting initial principles of QED (quantum electrodynamics)

[HendriXML]

I did a little experiment with a square wave signal from a AWG.

One short wire and gnd (about 50 cm) to channel 1. One longer one trough a roll of double wire of more than 30m to channel 3. That path should have accounted for at least a 100 ns delay, if length mattered.

The scope doesn't show much of a difference between the 2.

(Using a shorter vs longer coax cable does show a difference)

I wouldn't have expected that!

Really nice and somewhat mind blowing.

[RoGeorge]

Seems like the answer should be easy to find experimentally, at a reduced scale. 

Take a few meters of wire and a 2 channels oscilloscope and measure the delay for different layouts of the wires.  I'll expect the final distance between the source and the load to be irrelevant for the delay. 

My intuition tells me the delay is not dictated by the shortest path between the battery and bulb (that's what the video concludes, delay=1m/c, and 1/c seems the wrong conclusion to me).

I 'll say the delay is dictated by the length of the EM field traveling path, and that length in the video is enforced by the piece of wire between the switch and the light bulb, so my answer for question in the video is:

B) 1 second 


LATER EDIT:
I was wrong, now I think the correct answer is 1m/c.

[RoGeorge]

I did a little experiment with a square wave signal from a AWG.

One short wire and gnd (about 50 cm) to channel 1. One longer one trough a roll of double wire more than 30m to channel 3. That path should have accounted for at least a 100 ns delay, if length mattered.

The scope doesn't show much of a difference between the 2.

(Using a shorter vs longer coax cable does show a difference)

I wouldn't have expected that!

Really nice and somewhat mind blowing.

I've typed my previous post on a page opened long ago, and didn't see your measurement.

The result is not what I would've expected, either.   

Are you sure it is not capacitive coupling what you are measuring?  Is there any load attached?  Can you draw by hand the circuit layout, or attach a photo of the layout, please?

[HendriXML]

Are you sure it is not capacitive coupling what you are measuring?  Is there any load attached?  Can you draw by hand the circuit layout, or attach a photo of the layout, please?

No load or anything fancy 

I redid the experiment using only a single wire (of the pair), but still rolled up. It does not really matter as can be seen.

[HendriXML]

Using an 50 ohm load on the longer wire shows the inductance of that wire (at a much larger (or smaller?) timescale).

But I wonder wether inductance can be left out of the model when talking about electric and magnetic fields and energy transfer  .

[bdunham7]

One short wire and gnd (about 50 cm) to channel 1. One longer one trough a roll of double wire of more than 30m to channel 3. That path should have accounted for at least a 100 ns delay, if length mattered.

Were your wires a meter apart?

[HendriXML]

One short wire and gnd (about 50 cm) to channel 1. One longer one trough a roll of double wire of more than 30m to channel 3. That path should have accounted for at least a 100 ns delay, if length mattered.

Were your wires a meter apart?

No, it wasn't really an attempt to recreate the thought experiment.  I had to verify whether using normal wires would result in similar delay as coax cables, but it seems they don`t.

But my setup might be flawed in getting a definite answer. I think the video should have included one 

[Bud]

Not Electron flow again.   

Yes. That just reminded me of the tongue-in-cheek question I asked a little while ago, which was like "does current actually flow through anything".

Step barefoot on the wet ground and touch a live electric wire through a 1MOhm resistor to find out.

[Bud]

I'll expect the final distance between the source and the load to be irrelevant for the delay. 

If that was the case, TDR (Time Domain Reflectometry) would never work.

[bdunham7]

No, it wasn't really an attempt to recreate the thought experiment.  I had to verify whether using normal wires would result in similar delay as coax cables, but it seems they don`t.

But my setup might be flawed in getting a definite answer. I think the video should have included one 

Yes, your wires need to actually be spaced out otherwise you run into all sorts of issues.  Try this, and put a 50R load on each scope input if you can.  And make sure you have at least 2 ns of screen space for every foot of wire.

[aneevuser]

I've just watched this, and I thought the presentation was rather confusing. In particular, I misunderstood what the actual claim was. I was assuming that Derek's claim was that the bulb started delivering full power after the stated 1/c s. This may be my fault (I watched it quickly), or maybe the video wasn't clear. [Just watched it again: he says "lightbulb turning on" which I think is deliberately misleading]

However, having skimmed this thread and pondered a bit, I guess the way to think about it is that:

a) the wires close to the battery form a dipole antenna
b) the wires close to the bulb form a dipole antenna
c) shortly after turn on time, the closing of the switch allows a current to flow in the vicinity of the battery, as the E fields in the wires on either side of the switch reach equilibrium
d) due to c), we have accelerating charges in the wires near the battery, and thus an EM wave propagates outwards
e) the dipole antenna close to the bulb feels the EM wave 1/c s later, a current is induced in it, and (in principle) the bulb begins to shine (or perhaps better: a scope displays a pulse)

So near the battery and bulb we are essentially replaying Hertz's famous experiment, and the amount of energy transferred could be calculated by a Poynting vector calculation based on details of the EM wave propagating directly from the source antenna to the sink antenna, and the geometry of the wires near the bulb.

However, it's not clear to me how things work thereafter. Given that the B field around a straight wire falls off as 1/r from the wire, I would assume that the maximum energy propagation does indeed occur either in the wire, or at least very close to it, since |E x B| will be largest there, and of course, any energy losses to resistance will occur due to electron scattering processes in the wire itself, not outside it (and how could that occur if the energy was being transmitted outside the wire?)

So it seems to me that, after a fairly negiglible "antenna-to-antenna" energy transfer, we would indeed have to wait for the current to flow in the wire (propagating around initially at some speed characteristic of the transmission line properties of the circuit) before the bulb receives any significant power.

Any RF specialists able to comment?

[bdunham7]

Any RF specialists able to comment?

I'm certainly not an RF specialist, but I pretty much agree with your overall thoughts, especially about how the problem has been set up to be misleading and support wild "blow your mind" claims.  EM coupling isn't the same as a conductor.  One part I've had to rethink is that if you model it as a transmission line, then the initial conditions actually would have both currents equal because the apparent input resistance for a step is the characteristic impedance and that isn't as high as I first thought, even for wires a meter apart.  The self-inductance of the wire is the thing limiting the initial current flow, but that folds into the transmission line equations.  And of course the whole circuit as stated is impossible for other reasons, so we'll have to settle for some scaled-down version of it to argue about. 

[TimFox]

Until you actually look at the equations for characteristic impedance, it is not obvious how hard it is to achieve a high characteristic impedance (without a helical conductor to increase the inductance per unit length), even with large dimensions, due to the logarithmic dependence on spacing.
For a twin-lead parallel-wire transmission line (without dielectric between wires), with the spacing D much larger than the wire size d, the approximate formula for characteristic impedance Z₀ is:

Z₀ = (276 ) x log₁₀(2 D/d).

[bdunham7]

Z₀ = (276 ) x log₁₀(2 D/d).

Yes, I was a bit surprised upon actually doing some math. 

That applies if we can assume that we are in the transmission-line domain, right?  You need the the more complete formula that has frequency in it, and then the characteristic impedance goes to infinity as you approach DC as long as your dielectric leakage conductance is zero.