Hello Frederick David Tombe ,

Please see the attached JPG file,which depicts a single-phase AC power supply (P/S), and a single, full-wave, DC P/S. Why do you say that when I add a full-wave, bridge rectifier to the single-phase AC P/S one of the two infinite wires coming off the secondary of the isolation transformer becomes redundant?

Regards,

Tom Cuff

Dr. Cuff, Thanks very much for sending me this diagram. Before I answer your question, let me first ask a question about your diagram. Consider the first half-cycle of the AC, and let's say that electric energy, irrespective of what its deeper physical nature is, is flowing outwards from the power supply along the positive lead.

Is any energy flowing in the negative lead at the same time, and if so, in what direction?

Hi Frederick David Tombe ,

No need for the honorific, "Dr. Cuff", my highest degree is MS; please, call me Tom. Let us assume that the two infinite wires have the dimensional uniformity so that they can be considered to be a uniform transmission line. There are three types of transmission lines: coax, twin-lead, and open-wire. Further, let us assume the transmission line is lossless, i.e., the conductors have no resistivity, ρ = 0, and the dielectric surrounding these conductors also have no losses, ε = ε0 . Such a transmission line can be modelled as having a characteristic impedance, say, Zchar = 600 Ω for an open-wire transmission line, where Zchar is a real number, Zchar = Rchar. In other words, we need not show the line as being infinitely long, rather, we can just terminate a finite length transmission line with its characteristic impedance to make it act like an infinitely long transmission line because the travelling wave signal from the voltage source will not be reflected by Rchar. Thus, an AC voltage source, or a full-wave DC P/S voltage waveform would both dissipate real power in Rchar: either P = (VDC)2/Rchar, or P(x,t) = (VAC(x,t))2/Rchar, where x is the distance along the transmission line to the termination point (location of Rchar), and t is the time.

One could, of course, calculate the Poynting vector: S = E × H , but this would be more arduous.

Regards,

Tom

Hi Tom, Yes, but what you are doing here is avoiding the issue of which wire the energy travels outwards in, during the first half-cycle. You are modelling the electric current in the wire as a wave between the two wires.

Let's not have a transmission line. Let's have the two wires pointing up and down the page in your diagram, away from each other.

Thomas Cuff So, the question is still outstanding. Does electric energy propagate outwards from each wire during an AC half-cycle? I don't see how it could. It's more likely that the outward energy propagation alternates between the two wires during the full cycle in a single phase system.

Hi Frederick David Tombe ,

I have not forgotten your question, I am simply marshalling my thoughts in order to answer it - if I can. If you take a 1.5 VDC, D-size, so-called dry cell, i.e., sometimes erroneously called a flashlight "battery" and suspend it in a vacuum, then equal and opposite electric charges accummulate on the dry cell's terminals. The conservative electric field due to the accumulated charges exactly oppose the nonconservative electric field due to the cell's EMF (Electro-Motive Force). If the vacuum is perfect, and there are no fingerprints, moisture, etc. on the outside of the dry cell, then there will be no external flow of current. If the dry cell itself is perfect (cell's packaging has infinite electrical resistance, and chemicals inside the dry cell are pure enough not to be broken down by electrolysis due to the EMF), then there are no internal currents flowing. Thus,the Poynting vector, S = E × H everywhere in the volume surrounding the dry cell will be zero because, while E ≠ 0 almost everywhere, H = 0 everywhere.

I will save my analysis of the case where the source of EMF is alternating for a separate answer.

Regards,

Tom

Thomas Cuff Even the example of a battery DC power supply is fine for illustrating the point. But let's take an example where current does actually flow. And let's not have the two wires parallel and close together. In the moments after the power is connected, does the electric energy flow outwards along both wires, or does it only flow outwards along one of the two wires?