lambda = c/nu, where c = the speed of light
so, the wavelength of a 60Hz signal is 3 x 10^8 m/s / 60 s^-1 = 5 x 10^6 meters or 5,000 km.
When the marbles roll down the pipe, or back-and-forth, they all move more-or-less at once.
Less actually, not more. The ones at the far end of the pipe move a little bit after the one you push on.
How long after? That depends on the distance, and the speed at which the push goes down the wire/pipe. The speed is the wave-speed, which is the speed of light, which is not the ‘speed of visible light in a vacumn’, but rather the speed-of-electricity-down-the-wire, which is around 50/70% of c.
If I remember correctly, this give you a wavelength of around 2 or 3 thousand miles at power line frequencies.
When the end of the wire is so far away that a push is arriving there at exactly the same time as you are pushing – only one cycle behind - that’s one wavelength difference. If you have a power transmission system that is anywhere near that size, or a fraction of that size, the thing is unstable. You try to put power in, or take power out, and it bounces like a ball. Worse, it bounces like a ball on a moving floor, because people are turning stuff on, and turning stuff off, and whatever, and you don’t hear about it until the wave ges back to you, and that’s already to late, because you just gave a big push.
When I was a lad, this didn’t matter: you couldn’t have transmission grids that big, but you didn’t have transmission grids that big.
This is part of the reason why the country is divided up into three mostly-separate grids: One for the East, and one for the West (and one for Texas, because, well, Texas). You get economies of scale from making a grid as large as possible (fluctuations in supply and demand tend to smooth out better), but because of the phase issues, the only practical way to tie grids together on a continent-wide scale is via expensive DC interlinks.
Ok, back to the electric light that’s turned on, old or otherwise.
Some of those electrons get “used up” at the filament? The “front” ones? Otherwise, (it’s called a circuit for a reason, I know), all the huffing and puffing at the turbine is just to rattle a closed, pre-existing chain of marbles? That can’t be right.
No, no electrons are harmed in the illumination of your filament.
Why don’t you think that can’t be right? All the generator does is to create an EM field that causes the already “loose” electrons to move (in this case, vibrate).
All a piston does is move up and down but yet that can be converted into various forms of work. So why is is hard to imagine that the push and pull of an electric current shouldn’t be able to do the same.
No. The electrons do not get “used up”. They go in one end of the filament and out the other end.
All materials except superconductors will “resist” the flow of electricity through them. Energy is required to make the electrons flow through them, and the energy that ends up being absorbed by the filament gets converted into light and heat.
The wires in your house aren’t superconductors either, so they end up converting some of the energy from the generator into waste heat as well.
Each electron puffs up until it reaches another one, right?
So if you had just one electron, it would fill the jar/room/solar system?
The OP asked about power specifically about generation. Describing AC is appropriate.
Power plants are interesting, because they actually use the ground beneath them to complete the circuit.
Within the power plant, a wire connects from the ground (literally from a spike driven into the ground) to the turbine. As the turbine spins, the magnets connected to the turbine pull and push electrons into and out of the ground. The turbine then pushes and pulls the electrons into the wires leading towards your house (passing through various transformers to raise and lower the voltage).
Alternating current (AC) is used to reduce resistance in the wires. In AC, the electrons only move a discrete distance back and fourth within the wire, at a rate of 60 cycles per second. This reduces heat loss due to pushing too many electrons through the entire length of the wire, as is done in direct current (although DC is used in some extremely high voltage electricity transmission).
At your house, there is actually another spike near your electrical meter, that is driven into the ground to return the electricity consumed by your house to the ground, so that it can “return” to the power plant (in actuality, it just jiggles the electrons in the ground a bit).
I’m pretty sure this is incorrect.
Power plants generate “polyphase” power - the current is balanced in multiple circuits, and ground is only used as protection (current only flows in the ground conductor for load-imbalnce conditions).
Also, AC is used not to reduce the resistance in the wires, but to reduce the resistive losses. The resistance stays the same, but high voltage can be used, which means less current is required, so the power loss (which is proportional to both the resistance and the current) is reduced.
No, they don’t get “used up.” Each electron arrives with a bunch of energy. As it passes through the filament of the light bulb, it gives up energy; that energy is what makes the filament hot enough to glow and give off light. When that electron reaches the other end of the filament, it still exists - it’s not “used up” - but it’s in a low-energy state.
If the light bulb is powered by a DC circuit, then that electron will head back to the power plant (or battery?), where it will pick up more energy and then head back to the high-voltage end of the light bulb’s filament.
If the light bulb is powered by an AC circuit, then at some point the voltage of the circuit reverses; this electron is now back up at a high energy level (courtesy of a bucket brigade of electrons communicating the voltage to it all the way from the power plant), and all the electrons on the other end of the filament are now at a low energy level. This electron marches across the filament again, dumping another into the filament along the way. Repeat ad infinitum.
Square of the current, which is why decreasing current saves so much energy.
This. There’s a grounding spike stuck into the ground at your home, but there are still two wires connecting your home’s electrical system to the powerplant.
With the latest technology, high-voltage DC transmission systems are becoming feasible. They are efficient for the same reason as high-voltage AC is (high voltage facilitates low current, minimizing I[sup]2[/sup]R losses), but also because the use of DC eliminates capacitive losses and skin effect limitations that are inherent in AC systems.
It’s not that AC can be at higher voltages than DC can; it’s just that it’s easier to change the voltage with AC. All you need to change voltages with AC is a transformer, which is basically just a hunk of iron with some wires wrapped around it. There are techniques nowadays for changing the voltage of DC, too, but they’re much more complicated, and are only just now (last decade or two) becoming practical.
There have been power systems that used the earth for their return in the past, but it’s not generally done this way in modern systems. While using the earth as your return does work in general, what you find is that in many areas the ground connection can be a bit poor, especially during dry months. Attempting to run a power system using the earth as your return therefore tends to be a bit problematic.
Modern power systems do not use the earth to complete the circuit, but they do have a ground connection. This is for stability and safety reasons though. It is not part of the power delivery circuit.
If you have high voltage AC, the wires, insulation, standoffs, etc. all have to be designed to handle the peak voltage. However, the actual power you get through the line is equivalent to the RMS voltage and current. This means that for a given wire, insulation, etc. you can pump more power through it with DC. High voltage DC also doesn’t suffer from capacitive and inductive losses the way AC does.
However, DC has disadvantages too. As Chronos mentioned, an AC transformer is much simpler and cheaper (a DC transformer basically uses an inverter to convert to AC, then an AC transformer to change the voltage, and then a rectifier/filter stage to go back to DC). DC switchgear is also more complicated. When a high voltage electrical switch opens, it will draw an arc. AC arcs tend to be self-extinguishing once you get the contacts far enough apart, due to the fact that the AC voltage drops completely to zero twice during the AC cycle. Since DC remains constant, it will sustain the arc, requiring DC switchgear to have arc suppression built into it, which again adds complexity and expense.
What this means is that DC has higher costs for the transformers and switch gear at either end of the line, but lower costs for the wire in between and less loss when operating the system. This makes DC preferable for longer distances, and AC preferable for shorter distances. Over the past few decades, DC transformers and switchgear has gotten less expensive, which is making the break-even distance shorter and shorter. I don’t see that distance ever getting short enough for DC to be used for local distribution systems, but it does mean that DC is getting used more and more for power transmission.
DC also works better for underwater transmission lines since the capacitive losses get huge really quickly under water. DC can also be used to tie systems with different frequencies or phases together.
Thanks for this explanation. I understand the mechanics of power circuits, and voltage and current and resistance and all that, but I always had trouble visualising what it actually is that is getting used up by my electrical appliances. After all, electrons come into my house, and they go out of my house, so why am I paying for them? 
So, a related question - how does the electricity meter in my house know how much energy the electrons are giving up? What exactly is the meter measuring?
There is a good (technical) discussion of Watt meters in this pdf.
Basically, the meter multiples the AC voltage and current together, to get power.
The amount of energy given up by the electrons is proportional to the voltage. The electricity meter only has to count the electrons and multiply by the appropriate factor and hey presto. The actual counting is done by little goblins.
Well, actually the power is also proportional to the current, so the meter uses a little bit of the current you’re drawing to spin a disk, and indirectly measures the power you’re drawing. The more power you’re using, the more current flows and the faster the disk spins.
From the basics:
A Coulomb is a unit of electrical charge; specifically, it’s the amount of charge carried by 6.24X10[sup]18[/sup] electrons.
Electrical current is composed of electrons (i.e. charge) moving down a wire. 1 amp is equal to 1 coulomb of charge (6.24X10[sup]18[/sup] electrons) moving by your measurement point every second.
A volt is a measure of how much energy each electron is carrying; it’s equal to 1 joule of energy per coulomb of charge. So if five coulombs of charge pass through a light bulb operating at 10 volts, the filament receives 50 joules of energy.
Your home’s power meter either measures or assumes a voltage across its terminals (this is usually around 110 volts rms), and then measures the instantaneous current. Voltage (joules/coulomb) multiplied by current (coulombs/second) equals watts (joules/second). The meter integrates over time (power multiplied by time equals energy) to determine your energy usage for the billing period.
So 110 volts times a running average of 10 amps is 1100 watts; for a whole month (720 hours), that adds up to 792,000 watt-hours, or 792 kilowatt-hours. That’ll cost you about $100, give or take.