User talk:Ghitis/Dialoguing with Heron

THIRD DIALOGUING ON ELECTRICITY


 * Jacob, I think there is an error in your description of a generator with no load.


 * Please, Heron, be patient with me. A generator needs a load to convert simple kinesis --kinetic energy—to kinetric?
 * I believe so, yes. With no load, there is no conductive path for the current, and therefore no power transfer.  The generator's output winding will just be a curly length of wire with its two free ends poking into the air.  There is no circuit for the current to flow through.  I don't know what is happening to the electrons in the wire in this case.  Perhaps they are pushed toward one end of the wire, creating an electric field stretched across the two ends of the wire, but the energy required to push them there would be a short burst at the beginning of the rotation, and none thereafter.  --Heron
 * In this case, in contradiction to your view, no current flows. This situation is called an "open circuit" (which is an oxymoron, but a common expression). The generator just spins idly, presenting no resistance to the steam turbine or other engine that drives it.


 * I really do not understand. What resistance?
 * When a generator is connected to a load, the power it delivers to that load comes from a mechanical or engine (or muscle power). Imagine you are turning the generator by hand. First, imagine there is no electrical load: it is easy to turn the generator, and the only mechanical resistance you feel is due to friction inside the generator.  Now connect a load (e.g. a light bulb) to the generator: it is now more difficult to turn the generator, because you are supplying mechanical work to power the load.  (If you have ridden a bicycle with a dynamo-powered light, you will have experienced this effect directly.)
 * A voltmeter connected across the terminals of the generator would read a voltage called the "open-circuit voltage", which is due to what physicists call an electromotive force or e.m.f.


 * Again confusion. I posit that voltage is the expression of the kinetrics, which is an expression of the velocity of the wire mobile electrons or of the “electricons.” E.m.f. is, as far as I see it, a misnomer, it is not a force, and it does not exist.
 * You are right to be confused. I have used inaccurate language here. When you connect a voltmeter to an "open circuit", that circuit immediately ceases to be "open", because a small current flows through the voltmeter to cause some effect in the instrument, such as moving a needle or energising an analogue-to- digital converter.  The meter does not directly measure the emf, because, as you say, emf is not a force.  "Emf" is an archaic term that is still used. --Heron
 * It is well known that charged particles only move at a slow pace compared to the speed of an electrical signal… this slow “drift” speed can be calculated if the current is known, and this speed turns out to be low. It could be thought of as the average speed of the electrons, which are whizzing in all directions, colliding with atoms, but on average drifting with a slow speed in response to an electric field. On the other hand, an electrical signal in a wire travels at a large fraction of the speed of light.  We can prove this by experiment, e.g. by sending signals down very long transmission lines.  If you close a switch at one end, the information that the switch has closed travels to the other end at this high speed.  What actually propagates is the electromagnetic field confined by the wires, not the electrons themselves.


 * I have read about the meager “drift” velocity of conducting electrons. Let us suppose that such is the case, and accept that the “signal” is given by “electricons,” which would be the electric field of each electron carrying kinetric. Then, a voltmeter measures the strenght of these entities, which is actually their velocity. So that we disregard the velocity proper of the metal conducting electrons, and regard these as indispensable, but not  determinants, for the complex mechanism of generating electrical current.
 * I'm beginning to suspect that current is just an illusion, and that electromagnetic energy is the only real thing happening. The electrons just wobble around to create the electromagnetic field, and it is the field that actually does the work.  In semiconductors and similar devices, the behaviour of the charge carriers is much more important, so I can see that thinking about these in terms of currents and voltages is more useful.
 * I have just remembered the answer to the above point about "return current", but I shan't explain it just yet, because it leads down a path that might not interest you. We can return to this subject later, if you like.

A transmission line is a conduit for electromagnetic energy. There are two basic types: waveguides and cables. In a waveguide, an electromagnetic wave simply bounces off the sides, which are conductive, and remains within the guide. Currents may flow in the walls of the waveguide, but in a complex pattern that I don't understand. In a cable, there are two distinct conductors, but they behave in tandem, confining the wave between them.
 * Please, explain. I also have a lot to add on this subject. -- Ghitis 08:18, Sep 6, 2004 (UTC)

Consider a simple transmission line: a pair of parallel wires (called a balanced twin in radio terminology). Imagine connecting a battery to the line. Immediately, an electromagnetic wave is launched into the line, travelling at nearly the speed of light. This wave contains energy, which comes from a current that flows from the battery. This current is the voltage of the battery divided by the characteristic impedance (symbol Z0) of the line. At the instant of connection, the current flows out of one terminal of the battery, across an infinitesimally short segment of the line, feeding energy into the wave as it goes, and back to the other battery terminal. Some time later, when the wave has propagated some distance down the line, the current is still flowing from the battery as more energy is poured into the travelling wave. Eventually, the wave reaches the end of the line and, depending on the load at that point, is absorbed or reflected. The time taken (call it T) for the energy to reach the end of the line is therefore a single journey at almost the speed of light along the length of the cable. It is not the round trip time, which would be 2T. However, the information about the nature of the distant load does not return to the battery until time 2T.

If the load is resistive, then current continues to flow indefinitely as energy is absorbed by the load and converted to heat. In a simple DC circuit, such as a battery connected to a light bulb, the travelling time of the wave is so short that it is usually ignored. People say that the current has travelled down one wire to the light bulb, through it, and back up the other wire - but we know that this is not the whole story. --Heron 21:02, 6 Sep 2004 (UTC)