powerpower supplypower-gridtransmission line
I hope this is not a duplicate, because I am sure other people have the same question…
And could someone explain the reasons why?
N.B. From the 'scienceabc' website linked to below:
Despite being so celebrated, DC exhibited a major problem — transmitting it over longer distances was really difficult. The wires lost power and had to be rejuvenated using additional circuitry.
“Persons were charged with stealing electricity by placing a large coil, somehow, near high voltage AC transmission lines.”
Is this feasible or was it perhaps an April fool’s article that I swallowed hook line and sinker in my youthful naivety.
Entirely feasible.
Farmers were occasionally charged with power theft in this country (New zealand) in the past. I haven;'t heard of a case in a decade or few - maybe they are getting cleverer at it :-).
This is the same principle as used for "IPT" / "Inductive Power Transfer as seen in phone chargers, industrial monorail powering, electric vehicle charging and much more.
I started to say that if the pickup coil was symmetrical with respect to two phases that were perfectly balanced that you'd get zero pickup, and then suddenly realised that I've always done IPT with essentially a single phase, and that with a 3 phase system with 120 degrees phase separation you should get the advantage of the full load current even if the two phases were fully balanced.
You are essentially getting fields produced by the current, not the voltage, and the voltage is essentially irrelevant as long as you observe the normal conventions that apply to any other dealing with xxx kV.
Energy Harvesting from Electromagnetic Energy Radiating from AC Power Lines - FAR more energy can be obtained than they achieve.
Worked example - I suspect some of the conclusions are suspect A Solution to the RWP for Exam 1 - Stealing Power
Low technical content - high relevance
Directly relevant but low technical value Electromagnetic Harvesters: Free Lunch or Theft!
Several related stack exchange questions with variably useful content.
How would an electric company detect wireless theft of electricity?
Stealing energy from radio towers or power lines
Online vehicle transfer - I do not have access to this paper but it is probably at least relevant as it will have examples of dual linear conductors and a pickup coil.
Related:
Capacitive - but impressive:
Power is the product of voltage and current. (P=IV) If you double the voltage, you halve the amount of current required to deliver the same amount of power.
Voltage drop in a conductor is the product of the current running through it and the conductor's resistance. (V=IR)
That's the simple answer to your question: lower current through a fixed resistance equals lower voltage drop, so a higher proportion of the voltage gets through to the point of load.
Say we want to deliver 1 W to a device. We could do it any number of ways, such as 1 V @ 1 A or 1000 V @ 1 mA. Let us say there is 1 mΩ of resistance between that power source and the point of load. What are the consequences of each choice?
1 V @ 1 A: The voltage drop from 1 ampere through 1 milliohm is 1 millivolt. The power lost is thus 1 mV × 1 A = 1 mW.
1000 V @ 1 mA: The voltage drop from 1 milliamp through 1 milliohm is 1 microvolt. The power lost is thus 1 μV × 1 mA = 1 nW.
The first scenario literally wastes a million times more power.
There's no free lunch, though. Higher voltage requires better insulation, greater spacing between conductors, or both.1 It's also harder to make high-voltage semiconductors. This is why the supercapacitor energy density problem hasn't been solved by "just" charging them to a megavolt.2
You may then ask, if the costs from insulation and such go up as a function of the voltage, doesn't that eat up the benefits? The simple answer is "no," which is obvious from the fact that power generation companies do everything they reasonably can to keep their costs down. If they're going to higher and higher voltages, they must have a good reason. We can infer that the cost of insulation as a function of voltage goes up slower than the cost of lost power as the inverse function of voltage.
The 115/230 volt thing is small potatoes. Except in very large buildings, that difference only shows up within shouting distance of the outlet that you're measuring it on.
From the power pole transformer back to the power generation station, international power systems are more similar than different. The power generation station typically boosts the generated power up to tens or hundreds of thousands of volts to take advantage of the lower power loss. They primarily use free air as their insulator, which is of course cheap, as long as you can afford the space it takes, which you can at the top of an electrical distribution tower.
The world's power distribution system does of course use things other than free air as an insulator. Glass and ceramic insulators have been developed to the point of art.
A vast amount of knowledge and expertise goes into dealing with the insulation problem, every day. For example, when a big coil in a motor-generator set arcs over, copper vaporizes and windings can go flying. There are companies that do nothing other than cope with such incidents, skillfully re-winding the coils to bring the M-G set back into operation.
Footnotes:
Increasing the space between two conductors increases the amount of insulation but does not increase the quality of that insulation, which is why I talk about these two aspects separately.
In the energy pseudoscience world, you'll find people pointing out that the energy density in a capacitor is a function of the square of the voltage. That is, charging a capacitor to twice the voltage requires storing four times the energy. So, the logic goes, the energy storage problem is easy, innit? Just charge a capacitor to a billion volts, and you'll have all the power on tap you could want.3
That's one problem solved; let's move on to world peace.
Too bad it's only easy when you ignore the costs and difficulties involved in creating and using such capacitors.
1 farad at 1×109 volts is approximately sufficient to power Norway for a year. Just scale it up a bit and you can vaporize the moon, no problem.
Best Answer
When the OP writes \$\boxed{\text{more efficient}}\$, I take it to mean efficient in terms of minimizing power loss
So, basically at any distance DC is more power efficient than AC.
DC is a more efficient transporter of power than AC due to skin-effect increasing AC power line losses and reactance causing AC regulation problems at the load. However, if you are looking for the reason why we normally transport power using AC then, there are a whole number of other things (unrelated to power efficiency) to consider.
AC is more efficiently \$\boxed{\text{generated}}\$ than DC and, AC transportation can make use of transformers to change voltages to suit the transport section.