# About battery and power grid circuits

by Will1987
Tags: battery, circuits, grid, power
 P: 5,632 ac power became a standard largely because it could be transmitted long distances economically. Thomas Edison liked dc better, but lost. By stepping up voltages to very high levels, power (IE) can be transmitted long distances: line losses of i^2(R) where R is the transmission line resistance, can be reduced by keeping current small...that means we want voltage high...say 300,000 volts or more for distance transmissions even though customers may want 120 or 220 volts. That means it musty be stepped up and down, very expensive if dc, much more efficient with transformers and ac. You should get a basic book on electricity as learning a variety of basics all at once is difficult to do in forums like this where responses, explanations are necessarily limited.... and you are more likely to get accurate explanations as well...
P: 1,724
 Quote by Will1987 No Claude, that makes very little sense. It's a bunch of shop talk that doesn't help me much at my current level of engineering background.
You're asking for deeper understanding, but now it feels like you're trying to drink from the firehose? Well, it depends on how much deeper you actually want to go.

 Quote by Will1987 It's my understanding that for charge to flow from point a to point b there must be a difference in electrical potential, and thats basically the "polarity" that makes the whole process occur. So as charge is flowing from point a to point b, opposite charge is also flowing from point b back to point a. Cations in one direction are anions going in the other direction.thats a circuit, no?
This was addressed in one of the earlier posts in this thread by russ_watters. No single electron will go all the way around the circuit, but at any given point, there is a net movement of electrons that flow from higher potential to lower potential. But there is a net flow of electrons at any given point due to the potential difference (and not polarity) from point a to point b. That's a somewhat DC view of things, but we still say that various points in an AC circuit are at higher (or lower) potential than others (which is true if you're looking at the RMS voltage--don't worry about that for now, if ever).

 Quote by russ_watters And for AC.... In the power grid, there are no chemical reactions, so no ions or electron sources. The grid uses only the electrons already present in the wires. The induction by the generator just gives them a push. Regarding their motion, I'm not sure of the actual drift velocity, but the same electrons will basically just oscillate back and forth in a piece of wire if it is long enough.
 Quote by Will1987 PLEASE JUST DESCRIBE IN A SERIAL WAY THE PATH THAT CHARGE TAKES IN THE POWER GRID, FROM ITS POINT OF ORIGIN AT THE POWER STATION, THROUGH THE LOADS, BACK TO THE "SOURCE." So what are the two points of differing potential in the power grid, thats all I want to know. Is the turbine/inductuion coil both the beginning and end of that circuit, just with a bunch of transformers and loads in between?
It's much easier if you think in terms of potential (and current--conventionally, we use the net 'flow' of positive charge rather than negative--blame Ben Franklin for this) rather than the actual charges. And perhaps it's best to just think of the potential rather than current because in AC systems, even the net flow just oscillates back and forth. I think that would help to simplify the picture.

The most simplistic picture has one side (the hot side) of the generator at the high potential. This potential is transmitted along the transmission line to your house (along perfect zero resistance cables) where your appliance (the load) drops all the potential. A return path (the neutral transmission line) back to the other side of the generator completes the circuit, and allows the current to flow.

Got that? Good. That's not a bad picture. In reality, transmission lines have some resistance, and the higher the transmission voltage, the lower the losses in the line, which is still up to 40% (or somewhere around this ballpark) of the power, depending on the distance between the plant and end user. So the power plant (which has many generators) feeds a high amount of power into a transformer station which steps up the potential. Some of the power (and not just potential) is lost along the lines, but the majority makes it to the substation, which transforms this very high potential to high potential (actually, this is usually only an intermediary as there's a transformer at your house or in your block which does the final conversion to whatever mains voltage you use in your country). You use most of the remaining power, and the rest of the power is used up in the return leg (through the various transformers again) back to the power plant.

If that didn't make any sense (but the previous paragraph did), well, take heart in the fact that you've got a high level picture of how things work. That and visit the Wikipedia page on power transmission:
http://en.wikipedia.org/wiki/Electri...r_transmission
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 Quote by Will1987 No Claude, that makes very little sense. It's a bunch of shop talk that doesn't help me much at my current level of engineering background. and since Stewartcs made the oft repeated remark that "the circuit must be complete" then perhaps he can, in simple terms, just describe the "circuit" of a power grid which was my orginial question actually. PLEASE JUST DESCRIBE IN A SERIAL WAY THE PATH THAT CHARGE TAKES IN THE POWER GRID, FROM ITS POINT OF ORIGIN AT THE POWER STATION, THROUGH THE LOADS, BACK TO THE "SOURCE." It's my understanding that for charge to flow from point a to point b there must be a difference in electrical potential, and thats basically the "polarity" that makes the whole process occur. So as charge is flowing from point a to point b, opposite charge is also flowing from point b back to point a. Cations in one direction are anions going in the other direction.thats a circuit, no? So what are the two points of differing potential in the power grid, thats all I want to know. Is the turbine/inductuion coil both the beginning and end of that circuit, just with a bunch of transformers and loads in between?
It's best to just start with a simple AC circuit with a reactive load. You'll find one in your intro to electrical engineering book or even on the web with Google. Read the way they describe how the circuit works. The concept is the same for a large power grid except there will be multiple generators, resistive and reactive loads, transformers, substations, and distribution centers. Those items won't change the way an electrical circuit works though.

In a simple circuit the power source creates a potential difference between its terminals. Power (voltage and current) travels down one leg (the hot leg), to the load where the voltage is "dropped" (i.e. loses electrical potential energy) and returns back to the source on the other leg (the neutral leg). At the load the power is transformed into some other form of energy like mechanical for example as well as some being dissipated as heat due to natural irreversibilities in the system (some energy is also lost due to the resistance in the transmission wire itself known as I^2R loses). Since a potential difference exists between the two terminals (point a and point b) and they are connected in a complete circuit (via the wire or legs previously mentioned) an electrical current flows. This current and the voltage are equal to the power that the source is outputting. Note that the conservation of energy still applies.

All you need to do now is just put in your transformers, think of the source as the generator, and the transmission lines as the "legs" and it's essentially the same.

Does that help?

CS
P: 1,070
 Quote by Will1987 No Claude, that makes very little sense. It's a bunch of shop talk that doesn't help me much at my current level of engineering background. and since Stewartcs made the oft repeated remark that "the circuit must be complete" then perhaps he can, in simple terms, just describe the "circuit" of a power grid which was my orginial question actually. PLEASE JUST DESCRIBE IN A SERIAL WAY THE PATH THAT CHARGE TAKES IN THE POWER GRID, FROM ITS POINT OF ORIGIN AT THE POWER STATION, THROUGH THE LOADS, BACK TO THE "SOURCE." It's my understanding that for charge to flow from point a to point b there must be a difference in electrical potential, and thats basically the "polarity" that makes the whole process occur. So as charge is flowing from point a to point b, opposite charge is also flowing from point b back to point a. Cations in one direction are anions going in the other direction.thats a circuit, no? So what are the two points of differing potential in the power grid, thats all I want to know. Is the turbine/inductuion coil both the beginning and end of that circuit, just with a bunch of transformers and loads in between?
for starters, the power grid is a bad place to start. 3-phase power is not easy to visualize. in fact, engineers use a lot of math tricks that make dealing with it easier.

this is probably not completely well-thought or written, but maybe it addresses part of your questions. i think you really should strive to understand DC well, first, even if that means taking a few things at faith, at first.
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MatlaBdude, Stewartcs, Proton soup, all great answers. Very helpful. So the voltage/current (negligible current?) DOES end up back at the power station! Can you describe the two "terminals" at the power station itself in a little more detail?
I see that when introductory explanations talk about about "current returning to the source" by source they mean the power source which is the two terminals, but its always one source in the sense that its always cycling through, trading places, swapping charge. Confusion was getting created by my thinking that since ground has a net positive charge and electrons negative, that the ground must somhow be the other "terminal."
 In a simple circuit the power source creates a potential difference between its terminals. Power (voltage and current) travels down one leg (the hot leg), to the load where the voltage is "dropped" (i.e. loses electrical potential energy) and returns back to the source on the other leg (the neutral leg). At the load the power is transformed into some other form of energy like mechanical for example as well as some being dissipated as heat due to natural irreversibilities in the system (some energy is also lost due to the resistance in the transmission wire itself known as I^2R loses). Since a potential difference exists between the two terminals (point a and point b) and they are connected in a complete circuit (via the wire or legs previously mentioned) an electrical current flows. This current and the voltage are equal to the power that the source is outputting. Note that the conservation of energy still applies.
I'm interested here in what you're saying about the voltage being "dropped." The hot leg is the volage/current source but what returns on the neutral leg if the voltage has been dropped or translated into mechanical energy and heat? The returning current is also negligible, right? I guess the voltage and the amperage do return, just with less power because its already been used by the load as mechanical work? can you elaborate a little on that?

 The turbine is the active device driving the entire grid. Of course, other turbines are interconnected into that same grid and they also are actively driving said grid. To understand conceptually, it is best to start with just 1 generator. Fuel is burned, and that energy is translated into mechanical power, which then gets translated into electric power. The work done moving charges around the circuit ultimately begins with burning of fuel. Energy conversion is what it is called in the engineering world. Is that easier to understand? BR.
Thanks but now that's too simple Claude, hehehe. Try aiming somehwere between your first response and that one and it'll be about right. by the way what does "BR" mean?

Thanks everybody I'm beginning to UNDERSTAND! Call me a nerd but it feels so good.

Will
P: 2,284
 Quote by Will1987 I'm interested here in what you're saying about the voltage being "dropped." The hot leg is the volage/current source but what returns on the neutral leg if the voltage has been dropped or translated into mechanical energy and heat? The returning current is also negligible, right? I guess the voltage and the amperage do return, just with less power because its already been used by the load as mechanical work? can you elaborate a little on that?
The voltage is "dropped" since it has lost its electrical potential energy. Thus the electric charge has very little potential and returns to the source at about 0 volts (there actually is a very small voltage due to the resistance of the wires: V = IR). Again, referring to a single phase circuit for simplicity, there will be an equal and opposite current flowing in the return wire (or neutral leg) back to the source, and the potential in that leg is just very low (near 0 volts) since the electric charge has already "given up" its energy to whatever device/component is in the circuit (e.g. light bulb, motor, heat).

Make sense?

CS
 P: 5 Yeah I guess that makes general sense. What do the white wires ultimately conenct to ack on the pole? Does the return leg return to the source back at the power station using the ground wire thats on the powerline poles? Will