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I have a hypothetical scenario that would help me better understand (I hope) electrical fields and propagation of electrical fields better.

So let's have a super simple circuit with one voltage source and one resistor:

schematic

simulate this circuit – Schematic created using CircuitLab

I understand that Ohms law says that I=U/R so using that we would see 50mA flowing through the whole circuit.

Now I have the hypothetical scenario: lets make this circuit veeeeery large.

Large so that the wires would be 10 times the length that speed of light can travel in one second.

If we presume that the wires are made from perfect superconducting material and have no losses in themselves, how fast will electrical field establish itself at location where our resistor R1 is (be at 5V).

Would current flow unobstructed the whole length with the same current strength until it reaches the R1 resistor and would it only at that time "reduce" itself to 50mA when it is impeded by R1 resistance?

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    \$\begingroup\$ Add a simple switch into the circuit at one end. Closing this switch will show the interaction of the electric and magnetic fields as it travels towards the resistor at the speed of light. Google poynting-vector for more information. In very simple terms this vector presents the spatial power product between the electrical field and the magnetic field, in the same way the normal power is the product between the voltage and the current. \$\endgroup\$ – skvery Feb 22 '17 at 17:16
  • \$\begingroup\$ Your circuit would also involve inductance and capacitance distributed along the length of the two conductors so you would need to view this as a (long) transmission wire problem with a termination resistance of 50 ohms. As for speed in the conductors its probably going to be about 0.6 to 0.8c. \$\endgroup\$ – JIm Dearden Feb 22 '17 at 17:16
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You are imagining a 'transmission line'.

Put a switch in series with your voltage source. When you close it, a wave of 5v and some current will progress along the lines at the speed of light (assuming bare wires in a vacuum, slower if there's air or plastic around).

The impedance of the lines is determined by their geometry. The impedance defines how big the current wave is that progresses along with the voltage wave. If the two lines are fairly close together, say 1mm conductors 10mm apart, the impedance will be about 300ohms, something like the old style FM balanced feeder line. If each wire has thin insulation and they're twisted together, then the impedance will be somewhat lower, 50 to 100 ohm range.

What happens when that wave eventually reaches the 100ohm resistor depends on the line impedance. If they're equal, then the current and the voltage waves will be in just the right ratio to be correct for the resistor as well, and the wave will be totally absorbed, leaving a voltage of 5v on the resistor, with the right current through it.

If they don't match, then a smaller wave will be reflected, to take up the portions of the wave that don't match. The wave will continue to bounce between voltage source and load, until eventually the reflection losses attenuate the waves to insignificance, leaving only the DC flowing between source and resistor.

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  • \$\begingroup\$ Quite a good explanation. The voltage wave will travel to the resistor. The current wave will travel back to the source. There will also be a voltage wave travelling back if the impedances are not matched. See this MIT lecture for a good explanation of how these electromegnetic waves can also propagate without wires. \$\endgroup\$ – skvery Feb 22 '17 at 19:28
  • \$\begingroup\$ To see an application that uses the current travelling wave see this short summary for power system fault location. (I do not work for SEL.) \$\endgroup\$ – skvery Feb 22 '17 at 19:36
  • \$\begingroup\$ @User1831847 A voltage and current wave travel source to resistor. If it's mismatched, a voltage and current wave get reflected. They go hand in glove, one cannot exist without the other, together they make the Poynting vector, change of mag field makes electric field, movemtnet of electric field makes mag field. Please make sure you don't answer any questions with your separation of waves view, the topic is complicated enough as it is, please don't spread misinformation. \$\endgroup\$ – Neil_UK Feb 22 '17 at 20:24
  • \$\begingroup\$ I differ, Wikipedia clearly states: "The incident wave travelling down the line is not affected in any way by the open circuit (or resistor) at the end of the line. It cannot have any effect until the step actually reaches that point. The signal cannot have any foreknowledge of what is at the end of the line and is only affected by the local characteristics of the line." \$\endgroup\$ – skvery Feb 23 '17 at 15:28
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    \$\begingroup\$ @user1831847 I've just read the wikipedia page on reflection you linked to. Go down to the section 'arbitrary impedance', which does in algebra what I did with numbers for the 50ohm line 100 ohm load calculations above. \$\endgroup\$ – Neil_UK Feb 23 '17 at 16:52
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When the voltage source is switched on, an electric field is created, coming from the battery. This electric field would travel at the speed of light. As the 'very-long' wires are superconducting, the electrons have no resistance, and when acted on by the electric field, they would accelerate. This happens as the electric field is a force, and that causes the electrons, having some mass, to accelerate (quantum effects notwithstanding). An electric field accelerates electrons in free space. There is no notion of a single current to define the electron motion. Instead the current would rise and would be different at different points in the wire. Why this is so is because current is defined as

I=qnAv

where I is the current, v is the velocity of the electrons as a whole (called drift velocity), q is the electronic charge unit and A is the wire cross sectional area.

This electric field (traveling at the speed of light), and the accelerating electrons (traveling at near the speed of light as they have mass and take time to speed up) would then, after 10 seconds, hit the resistive element. Here the electric field continues to travel at c(speed of light), but now the electrons in the resistor only speed up to a velocity defined by the conductor, because now the electrons face resistance to movement from the resistor atoms.

The accelerating electrons from the superconducting wire, would also hit the resistor a bit later after the field, and they too would slow down, reaching the velocity of electrons already in the resistor. This hitting and slowing down, (and many other transient effects that I am not able to go deeper into), would continue and reduce until the motion of the electrons (as a whole) reach steady state and cause a measurable constant DC current of, as you said, 50 mA.

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    \$\begingroup\$ Electron drift velocity is nowhere near the speed of light. \$\endgroup\$ – pjc50 Feb 22 '17 at 18:30
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    \$\begingroup\$ @pjc50, Obviously, but I was mentioning the motion of electrons as a whole, or as a sort of fluid so to speak \$\endgroup\$ – Transistor Overlord Feb 22 '17 at 18:31
  • \$\begingroup\$ @TransistorOverlord What would be the scenario with electrons if the resistance element was 1ohm (being lower then the characteristic impedance)? In that case the resistor would represent "wider channel" for electron flow, right? \$\endgroup\$ – StjepanV Mar 14 '17 at 17:10
  • \$\begingroup\$ @StjepanV If the wires from the battery to the R1 resistor (1 ohm now) are superconducting, then the same phenomena happens. However this time the steady state current will be 5 A. If the wires have some impedance, then effects such as standing waves would be seen as others have told. \$\endgroup\$ – Transistor Overlord Mar 26 '17 at 17:08
  • \$\begingroup\$ @StjepanV The lower resistance of 1 ohm will represent a "wider channel", but it is still a resistance to electron flow, just lesser this time. \$\endgroup\$ – Transistor Overlord Mar 26 '17 at 17:11
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If you've ever bought some coax you will have noticed that it generally comes in 50 ohm and 75 ohm varieties. This doesn't mean that you can measure 50 ohm with a meter; what it means is that the coax is physically dimensioned so that if 1 volt were applied, the intial current would be 20 mA. In fact, if the coax was infinitely long, with 1 volt applied, 20 mA would continue to flow.

So there is a power of 20 mW flowing down the cable and when it "hits" the load, if the resistance of the load isn't 50 ohm then some of that power is reflected back up the wire; you get a wave-front travelling down and a wave-front travelling back. If you send a pulse down your cable and it meets another cable of different impedance this happens: -

enter image description here

Some energy continues and some is reflected back to the source. This can play havoc with data comms. If you send a pulse down and it sees a short circuit all the signal is returned but inverted: -

enter image description here

Pretty pictures taken from here and note that this website is about physical reflections in a mechanical system and of course, the maths is pretty much the same and to emphasize that here is a pulse travelling down to hit what is in effect an open circuit: -

enter image description here

The bottom line is that if the impedance of the medium matches the load then power is transferred with no reflection. Here's an interesting scenario; a sinewave is sent down a cable and hits a short circuit: -

enter image description here

Notice how the travelling waves from left to right and back from right to left combine to produce a stationary wave (called a standing wave). The ratio of the peaks to the nulls is the standing wave ratio and in this lossless example the SWR is infinite. In practise, cable losses reduce the SWR but a short circuit can produce SWRs of tens or hundreds. A perfect match has an SWR of 1.

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    \$\begingroup\$ your last picture shows nicely how the voltage reaches the resistor and the current wave travels back, with the reflected voltage wave. \$\endgroup\$ – skvery Feb 22 '17 at 19:45
  • \$\begingroup\$ @user1831847 no, that's not quite what is indicated. Voltage travels down and a voltage is reflected back. Because the reflected voltage sees (say) 50 ohm then reflected current is in phase with voltage. Basically power travels back in the reflection but there's no point showing both v and I on the diagram. \$\endgroup\$ – Andy aka Feb 22 '17 at 19:55

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