Tear This Guys Theory Apart

Merry Christmas

EC Dan

Member
Location
Florida
Occupation
E&C Manager
It's an interesting thought experiment. I am by no means saying I have this correct, so please correct me if I have any of this fundamentally wrong:

Initially, as charges begin to move around the battery, capacitive coupling occurs with the wire 1 m away. This broadly results in negative charge on the positive terminal of the bulb since the positively charged wire around the battery is closest to the wire on the positive side of the bulb. The opposite happens on the negative side of the battery and the negative side of the bulb. This actually causes localized current to flow in the reverse direction across the bulb than would would normally occur. This capacitance forms based on the time constant of the two-wire capacitor as the electric fields are formed across the 1 m gap (starts forming at t = 1/c). As the charges move along the light-second long wire away from the battery at some fraction of the speed of light, the capacitance propagates down the length of the wires, continuing to cause current to flow in the opposite direction from the negative terminal of bulb to the positive terminal. The interesting thing happens when the positive charges from the battery encounter the capacitively-formed negative charges from the opposing wire. The capacitive effects break down and the positive charge propagation from the battery along the wire overcomes the capacitively-formed negative charge in the wire and starts to result in current flow in the expected direction.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
I am thinking of this problem like a transmission line.

With conductors 1m apart and 0.001m in diameter, the characteristic impedance is about 800 ohms

So we have 2 transmission lines extending off to the left and right, a supply, and a load. Make the load 1600 ohms. You now have 3200 ohms in series. Apply 56V (56^2/3200 = 1 ), and you end up with 1/2 watt going to the bulb and 1/4 watt going down each transmission line.

Some time later that 1/4 W hits the end of the transmission line where it hits a dead short, and gets reflected back. At some point that reflected power end up hitting the starting point of the circuit, and suddenly you have a full watt going to the bulb.

At least that is my guess. This is far enough outside of my area of work that I am rather grasping at straws and don't have the necessary tools to check my answer.

However I think I could assemble the tools to test this out, not with DC and 600,000,000m of wire, but at say 1MHz and 10 feet of twin lead.

-Jon
 

K8MHZ

Senior Member
Location
Michigan. It's a beautiful peninsula, I've looked
Occupation
Electrician
The parallel conductors are transmission lines and adhere to all the math that would be applied to an antenna fed with a parallel line, which is quite popular. What if, instead of having the conductors parallel, they form a big circle? What would that do to the math?
 

rambojoe

Wireman
Location
phoenix az
Occupation
Wireman
Ah, yes, the lure of special knowledge. What they don't want you to know, what most people think is wrong, etc. I know people to whom a preface like that is all it takes to convince them that what follows is the truth. It makes them feel special.
Whats funny is, when it comes to time and space, there is the truth for the photon, the truth for the electron, then the truth for us...
Then there is the whole "speed of light" paradox..
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
The part of the answer that says the initial influence of closing the switch will hit the light 1m/c later remains unchanged if the space between battery and light remains 1m.

The response curve of how the voltage hitting the light changes over time after the switch closes will change as the geometry of the wire changes.

Jon
I think that the simplest analysis of the current versus time, starting with a step from 0 to the source DC voltage is most easily approximated by applying transmission line theory to the long loop of wire and noting that the results will be quite different depending on whether the line characteristic impedance is higher or lower than the bulb impedance.
To make it more tractable, we should assume the bulb resistance is constant (resistor-ballasted LED for example) rather then temperature dependent.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
The parallel conductors are transmission lines and adhere to all the math that would be applied to an antenna fed with a parallel line, which is quite popular. What if, instead of having the conductors parallel, they form a big circle? What would that do to the math?
It would make it much harder, among other things. You would still be faced with distributed capacitance and inductance, but they would vary with postion around the loop.
 

Ravenvalor

Senior Member
Seeking practical applications.

Is the video implying that the power is flowing directly out of the battery and going straight to the light bulb which is about a meter away?
Is there a way to test this? Perhaps increase the distance far enough away so that it can be timed?
Also, is the video implying that the power feeding the light bulb in my home is flowing directly (as the crow flies) from the source feeding the grid (coal, nuclear, solar, wind)?
Can we think of the grid as a giant battery? If so is the power flowing directly (as the crow flies) from the grid to the light bulb in my home?
Can we think of the wiring in my home as a large battery? If so is the power flowing directly (as the crow flies) from the light switch to the light bulb in my home?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
IMHO practical applications come from a better understanding of how the power flows.

And no, it doesn't flow straight as the crow flies from power plant to house. While it can be described as flowing through the field, if you look at the shape of where the field is strong, that follows the wires. In fact we do lots of design to intentionally contain the regions where the field is strong.

Jon
 

Reluctance

Member
Location
Central Jersey
This forum discussion brings in some interesting comments, and some of them focus on the part that still puzzles me; i.e., the answer to the quiz is 1m/c seconds, which means that the light bulb first glows because EM energy travels from the space surrounding the battery to the space surrounding the bulb, over the one-meter distance between them.

I don't think Muller (Veritasium) adequately explained why that's the correct answer, though I may have missed it. Here's my own step-by-step reasoning.

In the short moments after the switch is made, an E-field propagates at the speed of light (which is a misnomer, since it's really the "speed of causality") from the E-field around the battery to the E-field that is consequently set up along the two wires connected to the battery.

(Almost) immediately accompanying that E-field surrounding the two wires, is the motion of electrons in the wires. This motion (almost) immediately sets up an M-field around the two wires, which now means that we have an EM-field surrounding the two wires a short distance away from the battery. This EM-field contains a flow of energy, as Poynting shows, which is directed away from the battery in the space surrounding the battery, and towards the two connecting wires in the space surrounding them. Energy is not propagated inside the wires, nor in the direction of the wires. The wire direction is the direction of current, and current is the flow of charge, not the flow of energy.

This EM field propagates in all spatial directions, as with an antenna, not just in the direction of the wires. And in a time 1m/c seconds, that field strikes the space around the bulb and the two wires connected to it. The EM-field now surrounding the bulb and its two wires can be shown by Poynting to direct EM energy into the bulb and into its two connecting wires.

My puzzle is, why do we need wires connected to the bulb, in this explanation? I have a hunch that we don't. Even without wires connected to the bulb, the same thing would happen to the bulb. There will be a flow of energy into the bulb filament (which is really only a wire), but without connecting wires that go back to the battery, that flow won't be maintained for any appreciable time, because the electrons that flowed inside the filament, without connecting wires, have nowhere to go. Their motion thus stops after a very short time, say x/c, where x is about the length of the filament. So without wires, a teeny weensy amount of energy will flow into the filament, not enough energy to significantly increase its temperature.

Thus, the wires are necessary to provide an approach to steady state. Of course, that steady state, as Muller explains, happens after a complicated transient interaction between the establishing EM-field, the battery, wires, and bulb. During this transient, the wires act as a transmission line; there is reflection when the establishing EM-field hits the far end where the wires turn 180 degrees, for instance, and so much more in this very complicated transient. But after a couple seconds, we see steady light from the bulb.

So now having reasoned out these details carefully, I think that's getting to a more full answer to Muller's quiz. Interesting how individual humans can reason, step by step, but in large groups, we often fall on our faces and eventually destroy most everything such careful reasoning has produced.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
This forum discussion brings in some interesting comments, and some of them focus on the part that still puzzles me; i.e., the answer to the quiz is 1m/c seconds, which means that the light bulb first glows because EM energy travels from the space surrounding the battery to the space surrounding the bulb, over the one-meter distance between them.

I don't think Muller (Veritasium) adequately explained why that's the correct answer, though I may have missed it. Here's my own step-by-step reasoning.

In the short moments after the switch is made, an E-field propagates at the speed of light (which is a misnomer, since it's really the "speed of causality") from the E-field around the battery to the E-field that is consequently set up along the two wires connected to the battery.

(Almost) immediately accompanying that E-field surrounding the two wires, is the motion of electrons in the wires. This motion (almost) immediately sets up an M-field around the two wires, which now means that we have an EM-field surrounding the two wires a short distance away from the battery. This EM-field contains a flow of energy, as Poynting shows, which is directed away from the battery in the space surrounding the battery, and towards the two connecting wires in the space surrounding them. Energy is not propagated inside the wires, nor in the direction of the wires. The wire direction is the direction of current, and current is the flow of charge, not the flow of energy.

This EM field propagates in all spatial directions, as with an antenna, not just in the direction of the wires. And in a time 1m/c seconds, that field strikes the space around the bulb and the two wires connected to it. The EM-field now surrounding the bulb and its two wires can be shown by Poynting to direct EM energy into the bulb and into its two connecting wires.

My puzzle is, why do we need wires connected to the bulb, in this explanation? I have a hunch that we don't. Even without wires connected to the bulb, the same thing would happen to the bulb. There will be a flow of energy into the bulb filament (which is really only a wire), but without connecting wires that go back to the battery, that flow won't be maintained for any appreciable time, because the electrons that flowed inside the filament, without connecting wires, have nowhere to go. Their motion thus stops after a very short time, say x/c, where x is about the length of the filament. So without wires, a teeny weensy amount of energy will flow into the filament, not enough energy to significantly increase its temperature.

Thus, the wires are necessary to provide an approach to steady state. Of course, that steady state, as Muller explains, happens after a complicated transient interaction between the establishing EM-field, the battery, wires, and bulb. During this transient, the wires act as a transmission line; there is reflection when the establishing EM-field hits the far end where the wires turn 180 degrees, for instance, and so much more in this very complicated transient. But after a couple seconds, we see steady light from the bulb.

So now having reasoned out these details carefully, I think that's getting to a more full answer to Muller's quiz. Interesting how individual humans can reason, step by step, but in large groups, we often fall on our faces and eventually destroy most everything such careful reasoning has produced.
You need the wires attached to the bulb. You can understand this simply by realizing that the filament will not be heated without actual current flow. regardless of what the E and M fields are in the vicinity of the bulb.
The Poynting vector field in the immediate vicinity of the battery shows in which direction energy from the battery is flowing, but it does not show where that energy stops. For that, you need to look at the Poynting vector field in the immediate vicinity of the bulb. Without the wires attached to the filament and current flowing in those wires you simply have energy from the battery passing unimpeded past the location of the bulb.

So far, I find the transmission line analysis approach to be by far the most satisfying way to understand this problem and predict the exact outcome.
Without that you are making hard to quantify hand waving statements about the effects of the capacitance associated with the wires.
 
Last edited:

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
IMHO you don't need wires connected from the battery to the bulb, but you do need wires somewhere.

Go back to the transmission line analysis, but this time put open circuits at the far end.

Close the switch and 1m/c later the bulb lights (* see note). Meanwhile the wavefront travels down the line, hits the open circuit, and gets reflected.

About 1 second after the switch is closed the bulb goes out again.

* usually when transmission lines are analyzed, the approximation used considers the speed of light along the line, but ignores the speed of light across the line. Clearly that approach is incorrect to analyze the initial lighting of the bulb. The initial hint that the switch is closed will take 1m/c to get to the bulb, but full voltage will take somewhat longer.

Jon
 

Russs57

Senior Member
Location
Miami, Florida, USA
Occupation
Maintenance Engineer
I guess I just never really thought about it yet knew it on some level. Energy and charge flow in different ways. Electrons aren't photons.

I know all this but I never really equated it to speed of current being slow in a wire. It is the watts that are going fast.

In many ways old naming conventions are a big part of the problem.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
I guess I just never really thought about it yet knew it on some level. Energy and charge flow in different ways. Electrons aren't photons.

I know all this but I never really equated it to speed of current being slow in a wire. It is the watts that are going fast.

In many ways old naming conventions are a big part of the problem.
What problem? Whether you envision current flow as electron movement in a wire or energy flowing outside a wire in response to something the electrons are doing, it's still just a mathematical abstraction contrived to explain what we are observing. It doesn't matter which is "correct" because a) neither one is in an absolute sense, and b) they both are as long as they can be used predictively to achieve desired results.
 

Carultch

Senior Member
Location
Massachusetts
I guess I just never really thought about it yet knew it on some level. Energy and charge flow in different ways. Electrons aren't photons.

I know all this but I never really equated it to speed of current being slow in a wire. It is the watts that are going fast.

In many ways old naming conventions are a big part of the problem.

It takes a LONG TIME for the electrons to make it from source to load. And in an AC circuit, the electrons never make it from source to load.

It is the propagation of the fields that make it from source to load in order to transmit the energy and information. It is due to the fact that there are so many (perhaps octillions) of electrons involved in the circuit, that make the slow electrons able to generate the fields moving quickly despite moving slowly themselves. The upper limit for how fast the fields propagate, is the speed of light. It depends on the electrical properties that get in the way, and there is a velocity factor that reduces the speed from the speed of light for a practical circuit's ability to transmit a signal.
 

Carultch

Senior Member
Location
Massachusetts
Does it make Tables 8 and 9 or the 310.15 Tables incorrect? If not, I really don't care.
Derek's thought experiment assumes ideal conductors with zero resistance. If such a material existed with strictly zero resistance, it wouldn't matter how thick the wires are, to construct his thought experiment.

The tables you mentioned are for considering the practical implications of conductors in the real world having a resistance that generates heat, how to size the conductors for mitigating the temperature generated due to this heat, and for how to size the conductors to mitigate the parasitic loss of voltage due to generating this heat.
 

Carultch

Senior Member
Location
Massachusetts
My puzzle is, why do we need wires connected to the bulb, in this explanation? I have a hunch that we don't. Even without wires connected to the bulb, the same thing would happen to the bulb. There will be a flow of energy into the bulb filament (which is really only a wire), but without connecting wires that go back to the battery, that flow won't be maintained for any appreciable time, because the electrons that flowed inside the filament, without connecting wires, have nowhere to go. Their motion thus stops after a very short time, say x/c, where x is about the length of the filament. So without wires, a teeny weensy amount of energy will flow into the filament, not enough energy to significantly increase its temperature.

You can transmit wireless energy across two conductive wires that are parallel to one another, without closing the circuit on the two ends. And this is not just in theory, there are practical examples of this working.

The problem is, that if you don't close the two ends, you cannot sustain the circuit with DC. Eventually, the current will stagnate and just form a static charge at both ends of the source antenna. Lack of motion of charges in the primary antenna means a lack of energy transmitted to the secondary antenna. The fact that the video I linked is using AC instead of DC is why this can be sustained, as AC allows the EM waves to be sustained.

The need to close the circuit on both ends, is to stop the EM waves from radiating beyond the circuit, and concentrate the power transmission from source to load. That way, the path of the circuit doesn't impact the efficiency of power transmission, from a poor field coupling as you would have with unconnected antennas transmitting the waves. You would be hindered by the inverse square law to wirelessly transmit power over long distances, which is why we don't transmit power without wires in practice.
 
Top