@jea48
I seem to recall that in another post you said you learnt your electricity from this forum? Could I suggest reading a couple of recognised textbooks? Also, science advances when researchers publish peer-reviewed papers in scientific periodicals, which other researchers can cite in their work, or can challenge. By all means, read widely but for goodness sake, don’t quote indiscriminately.
So you are saying in a closed circuit the signal flows back and forth from the source to the load. Correct?
Almost. The signal is the voltage, which is analogous to electrical pressure caused by a slight excess or deficit in the balance of protons and electrons. Apply a voltage to one end of a wire, and almost immediately (depending on inductance) the voltage becomes uniform along the wire, including at the far end. If you make the voltage oscillate at one end, it oscillates at the other end. You don’t even need a circuit, though there is an almost imperceptible average movement of electrons when the applied voltage changes.
There is a slight time delay, which Quad make use of in their electrostatic speakers. They want to simulate a point source of sound a foot behind the panel. If you imagine sound radiating from that point, it reaches the centre of the panel first. then radiates outwards in rings. Quad delays the voltage signal to concentric rings using 12 miles of wire wound into inductive coils.
If the source is a CDP and the load is an Integrated Amp, using only one channel, the signal leaves the output on the hot wire of an IC, flowing back and forth, flowing to the input of the Amp, through the input circuit of the preamp, and then returns on the return wire back to the CDP output... Correct?
Vaguely. The source is a CD Player, so it outputs a line-voltage analog signal at audio frequencies, say 20-Hz to 20-kHz. The peak to peak voltage swing is around 1 Volt. Your interconnect (to me, IC also means integrated circuit) looks like a two-wire with RCA connectors rather than a 3-wire balanced connection, so there is a circuit. The amplifier presents an impedance (usually very high) so provides a path for some current to flow. The current flows out on one wire and back on the other. When the voltage reverses, the current reverses. The amplifier and the player could be designed so the current is regarded as the signal but almost universally, voltage is used as the signal instead.
As for the current the "net movement" of charge, it will measure the same on the hot wire as the return wire in the IC back to the analog output section of the CDP. Correct? This whole event takes place in the wires... Correct?
Correct. The current flows in the wires. This is not the whole event though! The current creates a magnetic field outside the wires. Changing magnetic fields create changing electrical fields - the foundation of electromagnetic waves. Note that these waves are not made of charged particles like electrons!
How does the input section in the preamp, for a better word, extract the signal from the IC as it travels through the closed circuit, flowing back and forth, back to the CDP? And what happens to the signal that returns back to the CDP?
Wow! Is that how you are saying it works?
The job of the amplifier is to take the AC input voltage signal (about 1-Volt maximum peak to peak) and increase the voltage enough to drive the speakers at the desired volume. A typical amplifier will have DC power supply rails at several 10s of Volts, plus and minus. The first amplifiers were based on triode valves, where a small voltage applied to one electrode allowed big currents to flow from the DC rails through the speakers. Later the bipolar junction transistor (BJT) and the metal–oxide–semiconductor field-effect transistor (MOSFET) were deployed.
Amplifiers just have to raise the output voltage to some multiple of the input voltage. Would be easy if those pesky speakers stopped allowing big currents to flow! Big currents cause a voltage drop according to Ohms’s Law, making it harder for the amplifier to get the output voltage where it should be.
Energy is what makes the light bulb light, not the current. Current is not consumed by the load. Proof, it returns to the source. Current measures the same on both sides of the load.
Energy is not consumed, it is transferred. It does not return to the source.
Where do I start? Overall, in our universe energy is conserved, so the light bulb converts some electrical energy into heat, some of which is converted into light. The energy comes from the power supply.
Power is usually measured in Watts and is calculated as the current times the voltage drop. Without current there is no electrical power. Electrical energy is power times time. It is usually measured in Watt-hours or kiloWatt-hours. In our light bulb circuit, current is conserved but the bulb consumes power as the voltage drops across its filament which has resistance. Ohm’s Law applies here as well as in speakers. Know any two of current, voltage drop and resistance and you can calculate the remaining quantity and the power.
Do things in the white paper sound familiar? Is Ian M. Sefton wrong too?
Yes they sound familiar but only because you keep repeating them ad nauseum! Yes, he is wrong. "The things the text books don’t tell you" contains things the text books don’t tell you because if they did, the text books would be wrong. I debunked his five major assertions in detail earlier. I marked each response QED, which coincidentally is the name of the maker of my main stranded, silver-plated copper speaker cables. They are spirally twisted around air-core tubes. Finally on-topic!
Ian M Sefton makes me ashamed to be an Australian,.really ...