Tear This Guys Theory Apart

ggunn

PE (Electrical), NABCEP certified
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.
Precisely. Unless this revelatory concept somehow makes it necessary to change the way we design systems using these tables, I really don't care if the power flows inside or outside the wires. It is just a mathematical abstraction, anyway.

ggunn

PE (Electrical), NABCEP certified
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.
I thought it was to stop all the electrons from spilling out onto the floor. Carultch

Senior Member
I thought it was to stop all the electrons from spilling out onto the floor. That's what insulators and the air do. You need an electric field strong enough to break the dielectric strength of air or of the insulator to get electrons to move out of the network of metal.

The best electrical insulator is a perfect vacuum, so the surfaces of our conductors automatically do that.

Precisely. Unless this revelatory concept somehow makes it necessary to change the way we design systems using these tables, I really don't care if the power flows inside or outside the wires. It is just a mathematical abstraction, anyway.

Who it would matter to in practice, is the people who design our data/communication systems. The radio frequency applications of electrical engineering. That is where it is important to model how the field propagates both inside and outside the metal of the cable, to know how to the unintended capacitance and inductance in a data communication circuit will do to your signal. This impacts the speed at which you can communicate through a given cable and the data capacity of the network.

In the 60 Hz and DC world, we can get away with assuming the first order introductory model to how power flows in circuits, without

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Carultch

Senior Member
To finish my last sentence:
In the 60 Hz and DC world, we can get away with assuming the first order introductory model to how power flows in circuits, without the fields and waves models of power propagation really being necessary to consider.

phaset

Member
Another discussion of the transmission line behavior of the real world experiment that Derek proposes, with the excellent Dr. Eric Bogatin and Robert Feranec.

These are guys who design printed circuit boards, and because of the very high clock frequencies, and even higher edge frequencies involved with e.g. microprocessors, the effects mused upon in the video are clear and real at the size of a workbench, due to the much shorter wavelengths involved.

wwhitney

Senior Member
OK, so a few questions from the video, in reference to the diagram below.

The symbol on the left, that just represents "ground" and just means the point we will use as the 0V reference, yes? Let's assume that instead the 0V point is at the left end of the battery, and the battery is 1V with no internal impedance, and the battery and switch are basically a point.

Say the left-right extent of the wires is 1 light second, and define a coordinate x in light seconds that's 0 at the upper left, and increases as you go around the loop clockwise. So x=2 is the same point as x=0, we'll say the wires are a constant distance apart that is near 0. Say that the battery and switch are located at x = 1/4, and say we instantaneously close the switch at time t=0. Say that the transmission line impedance of the pair of wires is 50 ohms as in the video.

I'd like to know the functions V(x,t) and I(x,t) for voltage and current between t=0 and the steady state, for all points on the loop. Starter questions:

A) So V(x,0) = 0 except V(1/4,0) =1, right? I'm thinking of the short segment of wire between the battery and the switch and calling that the point x=1/4. I(x,0) = 0 for any x.

B) How about V(x,1/8)? I think it's 1 for 1/4 <= x <= 3/8, and 0 elsewhere, is that right? The video talks about two way propagation, but I don't see how anything is propagating CCW from the battery when it's all at 0V to start.

C) How about I(x,1/8)? The video talked about a displacement current of 20 ma caused by the change in voltage. At t = 1/8 , where exactly along the loop is the current 20ma, and where is it 0? I think on the basis of the speed of causality, I(x,1/8) has to be 0 except for a subset of (1/8,3/8) and (1 5/8, 1 7/8).

Cheers, Wayne phaset

Member
OK, so a few questions from the video, in reference to the diagram below.

The symbol on the left, that just represents "ground" and just means the point we will use as the 0V reference, yes? Let's assume that instead the 0V point is at the left end of the battery, and the battery is 1V with no internal impedance, and the battery and switch are basically a point.

Say the left-right extent of the wires is 1 light second, and define a coordinate x in light seconds that's 0 at the upper left, and increases as you go around the loop clockwise. So x=2 is the same point as x=0, we'll say the wires are a constant distance apart that is near 0. Say that the battery and switch are located at x = 1/4, and say we instantaneously close the switch at time t=0. Say that the transmission line impedance of the pair of wires is 50 ohms as in the video.

I'd like to know the functions V(x,t) and I(x,t) for voltage and current between t=0 and the steady state, for all points on the loop. Starter questions:

A) So V(x,0) = 0 except V(1/4,0) =1, right? I'm thinking of the short segment of wire between the battery and the switch and calling that the point x=1/4. I(x,0) = 0 for any x.

B) How about V(x,1/8)? I think it's 1 for 1/4 <= x <= 3/8, and 0 elsewhere, is that right? The video talks about two way propagation, but I don't see how anything is propagating CCW from the battery when it's all at 0V to start.

C) How about I(x,1/8)? The video talked about a displacement current of 20 ma caused by the change in voltage. At t = 1/8 , where exactly along the loop is the current 20ma, and where is it 0? I think on the basis of the speed of causality, I(x,1/8) has to be 0 except for a subset of (1/8,3/8) and (1 5/8, 1 7/8).

Cheers, Wayne

View attachment 2558596
Hi Wayne,

Dr. Bogatin's main point for this graphic is that, if the light bulb is at the end of the line, then the displacement current has to propagate all the way to the end before the light bulb lights.

The special case of the light coming on nearly instantly is only when the distance between the switch closing, and the bulb, is 1 meter. In that case the displacement current from the E field only has to traverse the 1 meter of air dielectric.

And, of course the switch closing creates a step(ish) function, and that dV/dt is what launches the displacement current of the field.

wwhitney

Senior Member
Yes, I understand all that, but I'd like to know more about the details.

In particular, is the displacement current only occurring at the voltage waveform as it passes, or is it occurring over the entire region over which the voltage wavefront has passed already?

Cheers, Wayne

phaset

Member
Yes, I understand all that, but I'd like to know more about the details.

In particular, is the displacement current only occurring at the voltage waveform as it passes, or is it occurring over the entire region over which the voltage wavefront has passed already?

Cheers, Wayne
The radiating field (i.e. displacement current) and the opposing field (i.e. return current) propagate along the spaced wire pair. The characteristic impedance along the transmission line (i.e. at each differential slice) is the same, and is purely reactive (L and C).

Behind the switch transient the idealized battery has a constant voltage, so dV/dt is 0, and the ongoing current is conduction current. The e-field is along the wire, the b field around the wire, and the poynting vector into(!) the wire.

The impedance of the transmission line changes at the point where the two wires come together. The displacement current "sees" this new impedance. There will be a reflection, and the field (displacement current) returns back.

Real life wires have resistance, so the reflections are damped with time.

Does that help?

wwhitney

Senior Member
Behind the switch transient the idealized battery has a constant voltage, so dV/dt is 0, and the ongoing current is conduction current.
OK, so say I just look at a point 1/4 light second to the right of the switch (x=1/2 in my previous description). What are the functions V(t) and I(t) there?

V(t) is apparently a step function, 0 until t=1/4, 1 thereafter.

I(t) is apparently 0 until t=1/4. At t=1/4, we get the displacement current from the passing dV/dt. When you say the ongoing current is the conduction current, how much is that at time t=1/4 + epsilon? Is it still 1V of the battery / 50 ohms of the transmission line impedance?

Say the load has a resistance of 10 ohms (let's ignore that an LED may not act like a constant resistance). So in the steady state the current is going to be 100 ma. When will the current at our chosen point becomes a constant 100 ma, and what is the shape of I(t) between t=1/4 second and that time?

Cheers, Wayne

phaset

Member
OK, so say I just look at a point 1/4 light second to the right of the switch (x=1/2 in my previous description). What are the functions V(t) and I(t) there?

V(t) is apparently a step function, 0 until t=1/4, 1 thereafter.

I(t) is apparently 0 until t=1/4. At t=1/4, we get the displacement current from the passing dV/dt. When you say the ongoing current is the conduction current, how much is that at time t=1/4 + epsilon? Is it still 1V of the battery / 50 ohms of the transmission line impedance?

Say the load has a resistance of 10 ohms (let's ignore that an LED may not act like a constant resistance). So in the steady state the current is going to be 100 ma. When will the current at our chosen point becomes a constant 100 ma, and what is the shape of I(t) between t=1/4 second and that time?

Cheers, Wayne

Wayne, sorry for the delay. This is not a full answer, but it's what I have time for right now.

The impedance seen looking into the transmission line stays constant until the propagating field reaches the discontinuity. If the line were an infinite pair of parallel wires, the impedance would always be that of the transmission line.

The transmission line impedance changes with time at the point when the field reaches the connection between the wires.

• Carultch

winnie

Senior Member
OK, so say I just look at a point 1/4 light second to the right of the switch (x=1/2 in my previous description). What are the functions V(t) and I(t) there?

V(t) is apparently a step function, 0 until t=1/4, 1 thereafter.

I(t) is apparently 0 until t=1/4. At t=1/4, we get the displacement current from the passing dV/dt. When you say the ongoing current is the conduction current, how much is that at time t=1/4 + epsilon? Is it still 1V of the battery / 50 ohms of the transmission line impedance?

Say the load has a resistance of 10 ohms (let's ignore that an LED may not act like a constant resistance). So in the steady state the current is going to be 100 ma. When will the current at our chosen point becomes a constant 100 ma, and what is the shape of I(t) between t=1/4 second and that time?

Cheers, Wayne

I've been thinking on this one for a while.

I don't think your description of the circuit matches the diagram in post 46. In your description the battery and switch are located at point 'X=1/4' but in the diagram the battery is at X=0 and there is a resistor and switch located at X=1/4. Going to try to answer based on your description.

First we have the 'transmission line approximation'. In this approximation we consider the speed of light along the length of the line, but ignore it across the line. In this approximation:

A) is correct. The entire circuit is at your defined 0V except for the point between battery+ and open switch.

B) I disagree. The region where things are 'happening' will extend from X=1/8 to X=3/8, and from X=15/8 to 13/8 to the left and right of the just closed switch by 1/8 of a second on both sides of the line. Since you've defined the upper left corner (X=0) as your voltage reference, then I believe that V(1/8...1/4) = -0.5, V(1/4-3/8)=0.5, V(15/8..13/8) = 0, but current is flowing in the conductor from X=15/8 to X=13/8.

C) I agree.
OK, so say I just look at a point 1/4 light second to the right of the switch (x=1/2 in my previous description). What are the functions V(t) and I(t) there?

V(t) is apparently a step function, 0 until t=1/4, 1 thereafter.

Cheers, Wayne

In the approximation where we ignore the width of the line, I essentially agree, but think the step function V(t) goes to 0.5 at t=1/4 and X=1/2.

Now as to the actual function V(t) at this X and around this t? My _guess_ is that it is an exponential with a time constant equal to width/c. In other words if we don't ignore the speed of light _across_ the transmission line, instead of seeing the constant impedance of the line we will instead see a transient that looks something like a capacitor charging with a time constant of 1m/c (for a line 1m wide.)

-Jon

wwhitney

Senior Member
I don't think your description of the circuit matches the diagram in post 46. In your description the battery and switch are located at point 'X=1/4' but in the diagram the battery is at X=0 and there is a resistor and switch located at X=1/4. Going to try to answer based on your description.
OK, this is likely due to my misunderstanding the symbols in the schematic. (As indicated by my first question in post 46). I took the rectangle to be the battery, and the parallel lines of unequal length to mean "ground, i.e. 0V". Sounds like the rectangle was meant to represent the battery's internal impedance, and the parallel unequal lines is the battery. Sorry for the confusion.

So let's stick with my description: there's a rectangle of superconductor that's 1 light second by epsilon, with (0,1) representing one side of the rectangle. There's a 1V cell with no internal impedance and a switch at x = 1/4. And I'd like to define 0V as the negative terminal of the battery.
[If removing the battery internal impedance ends up not being a simplification, we can reintroduce it.]

(post 46) B) I disagree. The region where things are 'happening' will extend from X=1/8 to X=3/8, and from X=15/8 to 13/8 to the left and right of the just closed switch by 1/8 of a second on both sides of the line. Since you've defined the upper left corner (X=0) as your voltage reference, then I believe that V(1/8...1/4) = -0.5, V(1/4-3/8)=0.5, V(15/8..13/8) = 0, but current is flowing in the conductor from X=15/8 to X=13/8.
I had specified a fixed 0V reference to be the negative side of the battery, on the left, at 1/4 - epsilon. [Perhaps it will be better to define some other 0V reference, but for now let's stick with it.]

Regardless, your answer says that the voltage at time t=1/8 on the segment (0,1/4) is not constant. Why? This relates to the 2-way vs 1-way signal propagation question I have.

In the steady state (t >> 1) the voltage will be 1 on (1/4, 1) and 0 elsewhere (voltage drop at the load, voltage rise at the battery). And at t=0, voltage is 0 everywhere (except 1/4 + epsilon). So the picture of the time evolution on the segment (1/4,1) is nice, with the voltage rise traveling down the wire.

But on the complement of (1/4,1), you're saying that even though the voltage before and after the transient is 0V, there's also a voltage wave traveling that way around the loop?

If so, what's a better choice of 0V reference to make the two voltage waves more symmetric? Or is it inherently asymmetric because the switch is on one side of the battery?

Cheers, Wayne

winnie

Senior Member
I don't have time for a detailed reply.

1) I'd misunderstood your selection of reference. I'd used point X=0 in the corner of the rectangle, you used point X=1/4 right at the negative terminal of the battery.

2) Perhaps a more symmetric point for reference is to split the battery into two 1/2V cells, and make the reference the middle.

3) To fully answer I'm going to have to draw pictures, which won't happen today -Jon

wwhitney

Senior Member
Thanks for the response. Maybe I should make this a straight transmission line problem, so let me reformulate it to simplify:

c=1, and the conductor coordinate is -1 to 1, where -1 =1 (a loop); x and -x are a constant distant apart and form a transmission line with a characteristic impedance of 50 ohms. There's a voltage source and switch at x = 0 and a 200 ohm load at x=1=-1.

Let's say the voltage source is 1V, and is internally two sources of 1/2 V in series, with the center our 0V reference. The battery has no impedance and the conductors have no resistance. The battery positive terminal is connected to the positive x side of the conductor.

A) Suppose at time t < 0, neither side of the battery is connected to the transmission line (i.e. we have a double throw switch). But everything is in the fixed geometry specified. Is the voltage scalar field defined on the transmission line?

Now at time t=0, we throw the switch. For t >> 1, the steady state will be that V(x) = +0.5V on (0,1), and V(x) = -0.5V on (-1,0). The steady state current will be +5ma everywhere. For t < 0, the current will be 0 everywhere, and the V(0+) = +1/2V, V(0-) = -1/2V, and V(x) for other x depends on the answer to (A).

B) At time 0 < t < 1, I think we have that V(x) =+0.5V for 0 < x <t, V(x) = -0.5V for -t < x < 0, and the V(x) elsewhere is presumably unchanged from time t < 0, so it depends on the answer to (A). And at time t, I(t) = 20 ma displacement current per the video. But what is I(x) for -t < x < t, is it 20 ma, or 0 ma, or ? For |x| > t, I(x) = 0.

C) Suppose instead of using a double throw switch, we have a single throw switch at x = 0+. How do the answers to A and B change?

D) I'm sure the question of what I(x) and V(x) look like at t greater than but near 1 are of interest, but that's too much for the moment.

Cheers, Wayne

phaset

Member
Wayne, sorry for the delay. This is not a full answer, but it's what I have time for right now.

The impedance seen looking into the transmission line stays constant until the propagating field reaches the discontinuity. If the line were an infinite pair of parallel wires, the impedance would always be that of the transmission line.

The transmission line impedance changes with time at the point when the field reaches the connection between the wires.

Electroboom does a nice treatment of the transmission line case in his latest video:

wwhitney

Senior Member
Electroboom does a nice treatment of the transmission line case in his latest video:
Yes, that's definitely helpful, particularly the graphs of voltage across the lamp vs time. I infer that the lamp is modeled as a resistor, and the current through the lamp will be proportional to the voltage across the lamp.

On the other hand, he comments that both sides of the battery in the original circuit would behave the same, while also noting that with a switch on one side, before it is closed, all of the circuit is at the other side's potential. That seems like an asymmetry to me, so I don't see how both sides of the circuit behave the same.

Cheers, Wayne

winnie

Senior Member
Thanks for the response. Maybe I should make this a straight transmission line problem, so let me reformulate it to simplify:

c=1, and the conductor coordinate is -1 to 1, where -1 =1 (a loop); x and -x are a constant distant apart and form a transmission line with a characteristic impedance of 50 ohms. There's a voltage source and switch at x = 0 and a 200 ohm load at x=1=-1.

Let's say the voltage source is 1V, and is internally two sources of 1/2 V in series, with the center our 0V reference. The battery has no impedance and the conductors have no resistance. The battery positive terminal is connected to the positive x side of the conductor.

Cheers, Wayne

Do I have your description correctly converted into a drawing? : wwhitney

Senior Member
Do I have your description correctly converted into a drawing? :
Nice drawing, thanks. That would be the original geometry from the video (except the switch is now double pole).

My description was the simplification where the source and load were at opposite ends of the loop rather than right next to each other, just to focus on the transmission line behavior to start with.

[Your drawing has x next to 1-x (for x positive), I specified x next to -x.]

Cheers, Wayne