Tear This Guys Theory Apart

[winnie]

Okay, with a revised image:

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?

Yes, though I had to add some resistors to ground, otherwise the transmission line is not connected to our ground reference at all. With the resistors to ground, before the switch is closed V(x) is uniformly 0.

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.

Agreed

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).

Agreed.

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).

Agreed

And at time t, I(t) = 20 ma displacement current per the video.

This is not clear. What is I(t)? The current out of the battery at time t? Or the current at a location?

But what is I(x) for -t < x < t, is it 20 ma, or 0 ma, or ? For |x| > t, I(x) = 0.

For t<1 I(x) for -t < x <t is 20ma except at the transition. In the approximation of a perfect (0 width transmission line) the transition is a step function matching the switch closure. For a real transmission line I am not sure of the shape. When t>1 we should see a change with current dropping to less than 20mA as the 'reflection' from the unmatched termination (200 Ohm load on a 50 Ohm line) starts going back.

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?

I would say that V(x) would start uniformly 0V, but the transition would be no different. In this approximation we are ignoring any sort of coupling to the surrounding earth/ground.

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.

It gets easier if you match the load to the transmission line. Then you don't have to figure out the reflections at the battery

-Jon

 

[wwhitney]

Thanks.

Yes, though I had to add some resistors to ground, otherwise the transmission line is not connected to our ground reference at all. With the resistors to ground, before the switch is closed V(x) is uniformly 0.

That was my question, if you don't have those resistors to the 0V point, then the transmission line potential is undefined everywhere except at x=0 when t <0 (the switch is open). Is that correct? So if V_x(t) = V(x,t), we have V_x(t) is undefined for t < x, and V_x(t) = |x|/x * 0.5 V for t >= x.

This is not clear. What is I(t)? The current out of the battery at time t? Or the current at a location?

That was I(t,t), i.e. at the traveling voltage wave front, i.e. at a time t and a distance t light seconds away.

I would say that V(x) would start uniformly 0V, but the transition would be no different. In this approximation we are ignoring any sort of coupling to the surrounding earth/ground.

Sorry, wouldn't it be uniformly -0.5V, if the single throw switch is at 0+? And so now it seems like the voltage wave is only traveling on the (0,1) side of the transmission line, not the (-1,0) side of the transmission line. And then going back to your first diagram today (post #58), but with a single throw switch, it seems like the voltage signal only travels down one side, not the other. But that's in disagreement with some previous comments, so perhaps I am missing something (hence the question).

Cheers, Wayne

 

[winnie]

That was my question, if you don't have those resistors to the 0V point, then the transmission line potential is undefined everywhere except at x=0 when t <0 (the switch is open). Is that correct? So if V_x(t) = V(x,t), we have V_x(t) is undefined for t < x, and V_x(t) = |x|/x * 0.5 V for t >= x.

Yes, IMHO if you don't have some connection than the voltage is undefined. Of course in the real world you will have leakage and such. But we are talking an idealized system, probably in a frictionless vacuum. https://xkcd.com/669/

That was I(t,t), i.e. at the traveling voltage wave front, i.e. at a time t and a distance t light seconds away.

I would say that right at the traveling wave front its undefined, right in the middle of a step change (in the idealization). Just to clarify, you are asking I(x,t) with x=t. IMHO I(x-abit,t) = 20mA

Sorry, wouldn't it be uniformly -0.5V, if the single throw switch is at 0+? And so now it seems like the voltage wave is only traveling on the (0,1) side of the transmission line, not the (-1,0) side of the transmission line. And then going back to your first diagram today (post #58), but with a single throw switch, it seems like the voltage signal only travels down one side, not the other. But that's in disagreement with some previous comments, so perhaps I am missing something (hence the question).

You are correct, uniformly at -0.5V. When you throw the switch one side starts jumping to +0.5V. Current has to flow on both sides, on the + side to cause the jump to 0.5V, on the - side to keep it at -0.5V

-Jon

 

[LarryFine]

What would be different if the circuit was a circle?

 

[wwhitney]

Current has to flow on both sides, on the + side to cause the jump to 0.5V, on the - side to keep it at -0.5V

(in reference to the post #58 diagram, reproduced below for reference, but with a single throw switch at x=0+)

Can you expand on that a bit more, the "on the - side to keep it at -0.5V"? Is the current flow at each point of time t<1 a complete loop, with Kirchoff's current law satisfied, where the displacement current is occurring at both x = t (and 1-t) and x = -t (and -1 + t)? If so, why is displacement current flowing at x = -t (and -1 + t) when there is no change of potential on the - side?

Cheers, Wayne

 

[winnie]

Going on intuition here, but the results of the experiment should be the same no matter what point we take as reference 0V.

The current flow should be the same if we pick +, -, or neutral of the battery as the 0 reference.

Jon

 

[winnie]

Okay I've been thinking on this topic for a while, and have something of an essay about the original video.

The claim that energy flows in the fields rather than the wires is true but misleading to the extent that it implies that the wires are not important. You _can't_ separate out the wires from the process of delivering energy because the wires change the shape of the field, and in the circuit being described the wires are critically important. If you have a different layout of wires the results (the things that happen when the switch is closed and the lamp lights) are completely different.

The common way of thinking about power delivery (volts * amps) requires both a current flow and a voltage difference between wires. In the DC and low frequency case, this common understanding of electrical power flow is perfectly valid. Current flows in wires, you have a voltage difference between the wires, Kirchoff's laws apply, all is good.

Electrical energy cannot be delivered without fields, even when wires are present. The volts*amps understanding requires a voltage difference between the conductors, and a voltage difference means an electric field. So even if wires are present, if you don't have an electric field you don't have any energy delivery. Similarly, a flow of current means a magnetic field around the conductors, so if you don't have a magnetic field around the conductor you don't have any energy delivery.

Since both an electric field and a magnetic field are present, you can do the Poynting vector calculation, and get exactly the same result as the volts * amps calculation. IMHO it is a meaningless philosophical discussion to ask if the electrical power delivered by the combination of the electrical field and the magnetic field, or by the combination of the electrical potential difference (volts) and current (amps). In the DC case, both calculations give exactly the same result.

The value of the Poynting vector understanding comes into play when we are not working in the DC case. When you have changing electric and magnetic fields you can have energy delivery without any conductors or charges present at all. This is how all electromagnetic radiation works.

The full _AC_ characteristics of a circuit cannot be understood without considering the fields around the wires and how the wires interact with the fields. The switch closure causes the electric fields between the wires to change, and that change can only propagate at the speed of light. Driving that changing electric field is current in the wire, and that means a changing magnetic field which will also propagate at the speed of light. These changing electric and magnetic fields will interact with the conductors and the lamp as soon as the changes hit the respective bits of the circuit.

The original video was also misleading when it said that 'the light turns on at 1/c seconds after the switch closes', because it implies that the the lamp turns on at full power immediately after that short time. (The original video did have the disclaimer that 'the lights on when _any_ current flows through it'.) The reality is that the first inkling at the lamp that something is happening at the switch does in fact happen 1m/c seconds after the switch is closed, but it takes at least 1 second for the lamp to reach full power, as the entire circuit topology gets 'interrogated' at the speed of light.

In general this stuff is not relevant to electricians because at normal power distribution frequencies, normal circuits are much smaller than the size at which AC field speed of light effects would be apparent. But if any of you end up working with very large circuits (long distance transmission lines) or very high frequencies, then these effects will happen. An analogy is that for most building design you can treat the Earth as flat, but for large enough structures you need to consider the reality of a round planet.

-Jon

 

[Carultch]

Alpha Phoenix tested his theory in reality. Obviously not with 2 light-seconds of wire, but with equipment that could measure the essential time scales you'd expect for a 1 km round trip length.

Sure enough, there is some current in the resistor (acting as the bulb in his test), between 3 nanoseconds and 3 microseconds. However, it is nowhere near enough to turn it on, if it really were a light bulb. And if you did get a bulb to noticeably glow with that tiny fraction of the full current, you didn't do a good job designing your circuit, because any realistic bulb would burn out when the full current does get to it, if the full current has a much longer round trip to complete.

So it is true that the field propagates through the space between the wires to get from source to load and will carry energy with it, however, it is still necessary for the field to fill the wires to efficiently guide the energy from source to load. You can transmit electricity with no closed path between the two sections of the circuit, but to get it transmitted as efficiently as possible, you need a complete circuit. Otherwise, the fields will propagate in all 3 dimensions of space, and the inverse square law will spread the energy out so much, that it is barely usable in practice.

 

[jaggedben]

Yeah Derek's original video is click-baity (like all his stuff) and oversimplifies. He could have very much redeemed himself by adding another 30 secs to video at the end explaining how it's a trick question and why a real bulb wouldn't light up so fast, even though he's technically correct about a tiny amount of current flowing at 1/c sec. That said, I gotta hand it to him for getting a video on this subject to 4th or 5th most viewed on Youtube for a day, whatever is was. Beats the hell out of most of the trash that occupies those spots.

 

[wwhitney]

FYI, there's now a sequel to the original video in which the presenter addresses the criticisms and deficiencies of the first video. He does a real world test with 10m of transmission line (made from copper pipe!) and an impedance matched resistive load. The result is that during the time between the direct path field propagation hitting the load and the signal propagating along the transmission line hitting the load, the voltage at the load is ~20% of the steady state voltage. Meaning the power transmission is about 4% of the steady state power transmission. Much more than due to leakage current, but obviously less than steady state.

Cheers, Wayne

 

[tom baker]

I was OK with neutrons and electrons. All this stuff is nice but for now my head has no more room.