Help me understand electrical fields better

Question

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:

^{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?

Answer 1

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.

Answer 2

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.

Answer 3

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: -

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: -

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: -

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: -

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.