I've read about it, checked a few youtube videos about the subject but I still can't understand the process.

What is it that makes high voltage + AC current a better process to transport energy? I have seen formulas, but I need to understand how it works conceptually first.

If we use High Voltage (to cope with the resistance over a long distance) together with DC current a lot of energy is lost due to the continues resistance and the same happens with AC, so how is that one is better that the other one?

The only difference I can see, is the relief on the cables because of the polarity switch, is that why AC is better to transport electricity over long distances? Please don't use formulas to explain it, I need to understand how it all works first.

And if the voltage is higher how can the current be lower? I mean, there's a stronger force to push the electrons, so shouldn't it be the opposite?

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    \$\begingroup\$ Nobody ever said that losses do not happen in AC. I think you have taken a wrong conclusion. And "don't use formulas" are probably the reason you have difficulties understanding. The reason is all about the result of formulas. It can only be explained in formulas. \$\endgroup\$ – Oldfart Apr 14 '20 at 15:39
  • \$\begingroup\$ but there weren't formulas in the beginning. The formulas only appear after we have observed and understood the problem but I'm not there yet. I need to understand some of the points I mentioned. Of course there is power loss is AC transmission, I know that, it's the other points I am not sure about. Thanks \$\endgroup\$ – MeTitus Apr 14 '20 at 15:42
  • \$\begingroup\$ @Oldfart I've updated the post a bit to make it more clear. \$\endgroup\$ – MeTitus Apr 14 '20 at 15:45
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    \$\begingroup\$ It should be noted that there are some long-distance DC transmission lines in the US (or at least there used to be, 20 years ago). DC is less "lossy" over a long distance, but it is complicated (and somewhat wasteful of energy) to convert from AC to DC and back. And it's impractical to use DC for standard power transmission within a city. \$\endgroup\$ – Hot Licks Apr 15 '20 at 0:23
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    \$\begingroup\$ You're conflating that we need to use high voltage to more easily transmit more power, but we use AC because it is easier to drop that voltage back down for households. \$\endgroup\$ – rrauenza Apr 15 '20 at 6:18

The benefit of using high voltage is that we can deliver the same amount of power with lower current through the transmission lines. Lower current reduces the losses due the resistance of the lines. This is true whether we use AC or DC at high voltage (and, in fact, high voltage DC transmission is becoming more common)

The benefit of using AC is that, given the technology of the late 19th and early 20th century that was present when our transmission network was developed, it is much easier to convert high voltage AC to medium or low voltage AC for delivery to the end customer. We can do this using transformers. No comparably affordable and reliable technology was available for converting between DC voltages when the power network was designed and deployed.

And if the voltage is higher how can the current be lower?

Notice I said above "to deliver the same amount of power".

A 100 W lightbulb in a 240 V country uses the same amount of power and produces the same amount of light as a 100 W lightbulb in a 120 V country. But in the 240 V country, the lightbulb is designed with higher resistance so that it draws less current than the 100 W lightbulb for use in the 120 V country.

Similarly, if we have 20 residential customers drawing 20 kW of power in aggregate, and we feed them with a 20 kV line (using a transformer to step that down to 240 or 120 V before delivering it to their homes), that line will carry less current than if we feed those customers with a 10 kV line.

The only difference I can see, is the relief on the cables because of the polarity switch, is that why AC is better to transport electricity over long distances?

This is a bit off the main focus of your question, but actually AC isn't better than DC as far as the wire losses are concerned.

First, because the AC signal spends some of its time near 0 V, the peak voltage of the AC waveform must actually be higher to deliver the same power as a given DC voltage. For example, when we say we have a "120 V AC" power source, we mean the AC voltage has a root mean square (rms) voltage of 120 V, as this is able to deliver the same power to a resistive load as a 120 V DC source. But the peak voltage of this AC source is about 170 V. This means the wire must be insulated to prevent arcing at 170 V rather than just 120 V.

Second, because of the skin effect. This means that AC currents tend to mostly flow on the outer surface of a wire, while DC currents can flow through the full cross-section of the wire. The effect is small at the fairly low frequencies we use for power transmission, but it still means that the transmission wires have effectively higher resistance when carrying AC than when carrying DC.

So again, the main reason for choosing AC power transmission is to be able to use transformers to convert between voltages rather than because AC is inherently better.

  • \$\begingroup\$ If we increase the voltage we have more potential energy to push the electrons over the network and if the resistance of the circuit is the same, shouldn't higher voltage means higher current, after all we can push more electrons right? I know it doesn't but I just can't understand it. Thanks a lot for your reply. \$\endgroup\$ – MeTitus Apr 14 '20 at 15:50
  • \$\begingroup\$ @MeTitus, edited to address that question you added. \$\endgroup\$ – The Photon Apr 14 '20 at 15:51
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    \$\begingroup\$ @MeTitus, Potential difference is power per charge. A 1 C charge at 100 V has 10x the energy of 1 C at 10 V. So it's inherent in how we define potential difference (voltage) itself that the amount of energy stored by a given charge depends on its potential. \$\endgroup\$ – The Photon Apr 14 '20 at 16:11
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    \$\begingroup\$ @MeTitus, We never push the current from A to B. We make a voltage available to a device (or a network), and the device (or the full network) pulls the amount of power it needs. If the available voltage is higher, the device needs to pull less current to achieve the same power. \$\endgroup\$ – Hoki Apr 15 '20 at 8:19
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    \$\begingroup\$ @hoki I'm not sure that's entirely accurate, especially in the context of AC and transmission lines, but as the subject gets deeper the question of what is meant by "push" gets complicated. \$\endgroup\$ – pjc50 Apr 15 '20 at 11:06

High voltage in general allows more energy to be sent down a given wire size, as lower current can be used. The lower the current, the less the wire (resistive) losses. That is:

  • Power delivered is current * voltage, or W = E*I
  • Power lost to heat is resistance time the square of current, or W (loss) = I^2 / R

It's that I^2 term that gets you. Minimizing current is a big win in terms of reducing losses. (I'm not considering reactance here yet. That's another discussion.)

As to why AC is used, it's easier to generate and work than DC at almost every stage, especially when you consider that most of the core technology for power generation and distribution was developed in the late 19th century:

  • The source generator, essentially a rotating magnet, makes AC to begin with.
  • This AC is stepped up via transformers to high AC voltage and sent down the wire. Transformers are simple and reliable: no moving parts, no electronics.
  • The high-voltage transmission line does care about AC vs. DC (DC is better). More about this below.
  • The HV network also cares about phase-alignment when power is moved from grid to grid (DC is better). Again, more below.
  • Near the consumer side, again transformers step down the AC to a friendlier voltage for local use. Again - simple, no moving parts, no electronics.
  • At the consumer, 3-phase AC is ideal for most big motors. Single-phase is easy to step down to a safe voltage, like 240/120V for appliances and lighting.

Now, let's talk about DC. High-voltage DC (HVDC) is a technology that was originally developed in Sweden (by ASEA, now ABB) to solve a problem with undersea cables: dielectric and shield loss. More here: https://mycableengineering.com/knowledge-base/dielectric-loss-in-cables

The Swedes long knew that the constantly-changing electric field in an underwater AC cable resulted in large coupling losses to the surrounding armor material. This coupling becomes heat, that is, loss. So for taking power across the fjord from one island to another, it proved worthwhile to convert to DC prior to sending down the cable, then convert back to AC for use. More here from ABB.

And a bit about the Nazi-hating Swede who brought it about: Uno Lamm.

There's another benefit to using HVDC, be it overhead lines or buried: no skin effect. AC current in a cable produces localized eddy currents in the middle of the cable which oppose current, resulting in the main current being concentrated in the cable outer perimeter. This concentration of current increases the cable resistance, so more energy is lost as heat. More here: https://www.electrical4u.com/skin-effect-in-transmission-lines/

DC current doesn't form eddies to oppose current, and so has almost no skin effect. This means all the cable is being used, allowing more current to be sent down the same size wire at lower losses.

Finally, there's the intertie problem. When moving AC power between grids, their phases and voltages need to be closely matched. This is very difficult for large-scale systems. More about this here: https://www.testandmeasurementtips.com/how-ac-power-sources-get-synchronized-faq/

DC mitigates this issue for interties - no phases to match, and it's easier to adjust the step-up voltage and add it to the network as a new source comes on line. It's used in many large power corridors as an intertie, including this one: the Pacific DC Intertie which takes power from the Bonneville Dam on the Columbia River and ships it to southern California.

Since then the use of HVDC undersea cable has been deployed not only for submarine power cables, but also for tying offshore wind power to onshore stations. (This uses a variant called HDVC Light, more from the Swedes here: https://new.abb.com/systems/hvdc/hvdc-light)

As inverter technology matures and costs come down, the trend for long lines is to migrate to HVDC, while retaining AC for local loops owing to its continued advantage for motors and other large machines.

Even then, the same electronic technology that makes HVDC practical can and does get applied to the consumption side, so we will continue to see more DC in the local side too. This is already happening with data centers, which are beginning to use 48V DC for server rack power. Even induction motors, the machines that really like AC, can move to Inverter / VFD drives for greater efficiency and flexibility, at some expense.

  • \$\begingroup\$ Thanks a lot for you detailed answer, it will take me some time to digest it but it's helping me understand the whole process a lot better. A just have a quick question regarding high voltage: when we scale voltage up we also increase the current, so to make it efficient to transport electricity over long distances, we need to decrease the current, but how can we limit it when we have just increase the voltage? \$\endgroup\$ – MeTitus Apr 14 '20 at 17:16
  • \$\begingroup\$ No, that’s the point of increasing voltage: to increase the power delivered while not increasing current, and thus, losses. That is, W = E times I. At the same time, the heating losses of an overhead cable is proportional to the square of the current, that is, loss (power) is W = I^2*R. \$\endgroup\$ – hacktastical Apr 14 '20 at 17:51
  • \$\begingroup\$ Those equations make sense but I feel like I'm still missing something. If we are sending less particles through the wire how can we get the same power? The equations say so but how? For the same conductor does a higher voltage implies a higher speed at which electrons travel, thus generation more energy (joules) when the find resistance? I'm sorry if this does make any sense. \$\endgroup\$ – MeTitus Apr 14 '20 at 18:12
  • \$\begingroup\$ Ah, ok. So you're asking about the physics of energy flow. Using the water analogy, think of current as the volume of flow, and voltage as the pressure. We can increase the pressure and deliver more power to the far end, like powering a hydraulic drill with very high pressure fluid. \$\endgroup\$ – hacktastical Apr 14 '20 at 18:24
  • \$\begingroup\$ Exactly, but using your analogy, the higher the pressure the higher the spread of the H2O atoms, does the same apply to the electrons, when applying a higher voltage? Thanks a lot. \$\endgroup\$ – MeTitus Apr 14 '20 at 18:42

transformers, giant transformers, are cheaper than huge stacks of rectifiers and choppers to convert the HV DC into lower voltage DC.

And the transformers are more robust.

  • \$\begingroup\$ Thanks a lot for your answer, but I need to know things from a different perceptive, \$\endgroup\$ – MeTitus Apr 14 '20 at 16:07

It's really hard to deal with it in pure theory.

Let's try a practical example.

Let's suppose I've built an off-grid house with a perfectly functional home-power system. Batteries, most lighting and auxiliary loads are DC, inverter runs a few things as needed. Battery voltage is 12 volts.

500 metres away, I have a 480 VA (basically 480 watt) windmill, whose furling makes it run at basically one speed. It isn't hard to wind it so you get 60-ish Hz out of it, single-phase. I'm stuck with the location because that's where the nape of the hill is. How do I wind the generator? What voltage?

My system voltage is 12 volts. So let's just wind the generator for 12V, giving 40 amps. Now, I need to get my 12V @ 40A from the windmill to the house 500 metres. What wire will I use????

The minimum Code allowed wire for 40A is 8 AWG (8.37mm2). That stuff is 2.061 milliohms per metre, so my 1000m round trip is 2.06 ohms. E=IR voltage drop is 82.4 - well that doesn't work!

Let's go up a wire size to 6 AWG (13.3 mm2). 1.3mohm/m, or 1.3 ohms at 1000m, or 52 volt - No, that won't work either.

Let's go to the biggest - 0000 or 4/0 AWG (107 mm2). That is 0.161 mohm/m, or 0.161 ohms for our 1000m run. 6.4 volt or 53% voltage drop, *wow, we are losing half of it, and we are paying $10/metre for the wire (actually for 300kcmil aluminum at $1.59/ft; we wouldn't use copper at this size).

Let's go for the biggest wire made. 2500 kcmil AAC "Lupine" the size of your wrist, at $18/metre each way. 0.023 ohms/km. So 0.91 volt or 7.6% voltage drop (finally! A reasonable number!) but that's still considered not a good number in practice.

well, this ain't good.

But look. This XHHW wire we're using actually says "600V" on it. How about if we bump the voltage and step it down at the house? 600V transformers are weirdobtanium, so let's try 480V since it's common. 480 VA, at 480V, happens at 1 amp. Now let's go back and hit that voltage drop calculator.

14 AWG copper wire (2.08 mm2) @$0.23/metre is the smallest THWN/XHHW wire available. Let's try it. Resistance is 8.282 milliohms per metre or 8.282 ohms for us. Horrible! Oh snap, we're in trouble. This is not working out, but let's press on just for academic sake to see how bad it is. Let's see, 8.282 ohms x 1A = 8.282 volt or 1.7% voltage drop. Wait.... that's perfectly acceptable, why did that work???

But that seems weird, like something went wrong there. Let's try it one more time with the smallest available aluminum wire, 6 AWG.

6 AWG aluminum (17.16 mm2) @$0.55/metre. Resistance is 2.16 milliohms/metre or 2.16 ohms (uh oh!) for our run. Giving 2.16 volt or 0.45% voltage drop. That really, really works.

All we did was change voltage from 12 to 480.

Ohm's Law, meet Watt's Law.

Here's the thing. Ohm's Law is

 E = I R 

 Voltage (drop) = Current x Resistance 

So voltage drop is proportional to current. Above, we're flowing 1 amp instead of 40 amps, so voltage drop is obviously 1/40 as much. But there's more.

Watt's Law says

  P = E I 

  Power = Voltage x Current

Remember, in our application, power was constant: 480 W / VA. When we raised the voltage, this caused a proportionate reduction in current for the same power. We dropped current by a factor of 40.

So back to Ohm's Law, voltage drop (in absolute volts) fell by a factor of 40. However another thing was happening. Voltage increased by a factor of 40. That means that the bite that voltage drop was taking also shrunk by a factor of 40. Relative voltage drop compared to system voltage, dropped by 40 squared.

Ka-zinga! You can see the power of this increase in voltage (for a given static power requirement).

Try it again in a very practical, typical application.

1500 feet away, you want to power a set of driveway post lights. They draw 240 watts. You can power them either with 120V or 240V (the lights will happily handle either). 3.5% voltage drop is acceptable.

Sidle up to your friendly neighborhood voltage drop calculator and see what makes sense. You'll also be pricing the options here selecting /2 UF-B w/safety ground.

  • Run 120V. Current draw is 2 amps.
  • Run 240V. Current draw is 1 amp.

Which would you rather pay for?


Regarding resistance, DC is better than AC (see "skin effect").

DC also requires less insulation for the same effective voltage (AC peaks are 41% higher than the effective voltage)

It is also better in a sense that the energy transfer is continuous. In the single-phase AC the energy comes in pulses 100 or 120 times per second. (3-phase AC network, when balanced, also has continuous energy transfer.)

DC is also better because it doesn't need synchronization. Adding a generator to an AC network is a major pain even today, it was far worse 30-40 years ago when computers weren't everywhere.

In a DC network, you don't have "reactive power" and all the hassle associated with it.

Most electronic devices require DC in order to operate. Getting a good DC from mains AC is a great engineering effort.

In a small, insulated power network (e.g. a motor car) you are OK using DC. You can use power sources and power consumers that use a single DC voltage directly.

T. Edison tried to use the same approach city-wide. He almost succeeded. A century later, we still don't have all the technology he needed in order to succeed.

What happens in a large scale:

You need 100-500kV long distance network in order to keep the ohmic loses AND wire diameters manageable (copper is also expensive, important when ordering 100's of kilometers wire).

Neither the generators nor the consuming devices are practical at 100's kV voltages. So you need a method to change the voltage. Here comes the Transformer. A century old, mature, efficient technology, simple principle of operation, easy production.

A caveat: the Transformer works only with AC. With DC, you need an inverter of some sort to make AC, then transformer, then rectifier of some sort in order to get back the DC.

The high-power, high-efficiency inverter technology exploded in the 21st sentury and is still not on par with the simple Transformer. It also struggles at scale. That's why you can count high-voltage infrastructure DC networks world-wide on your fingers.

  • \$\begingroup\$ Thank a lot for your input. \$\endgroup\$ – MeTitus Apr 19 '20 at 18:28

And if the voltage is higher how can the current be lower?

Have you ever used a lever, gearbox, hydraulic jack, or pulleys? These are devices that provide mechanical advantage. To do the same amount of work, like lift a sack of potatoes one meter, you can choose between:

  • providing a lot of force, for not a lot of distance, or
  • providing not a lot of force, for a lot of distance

The transformers in the distribution system is the same idea, except instead of force and distance, it's voltage and current.

Specifically, for mechanical systems power \$P\$ is the product of force \$F\$ and velocity \$v\$:

$$ P = F v $$

and the electrical equivalent is the product of current \$I\$ and voltage \$E\$:

$$ P = I E $$

If one factor decreases and the other increases by the same factor, power (the rate of energy transmission) remains the same.

What is it that makes high voltage + AC current a better process to transport energy?

Let's take this in two parts, high voltage, and AC. First, why high voltage?

Imagine you have an engine powering a machine through a shaft. But attached to this shaft is a fan. The fan creates resistance: the faster the fan spins, the more resistance. Overcoming this resistance is wasted power.

Adding a gearbox so the shaft spins at a lower speed but with a greater torque transmits the same power, yet the shaft spins more slowly so the fan generates less wasted power. If this gain is greater than the gearbox losses, the system becomes more efficient.

Specifically, the power \$P\$ wasted in the transmission lines is:

$$ P = I^2 R $$

The resistance \$R\$ can be reduced by using fatter conductors, but metal wires and the towers to support them are expensive. But reducing the current \$I\$ is cheap: it requires only a transformer at each end to increase the voltage and then step it back down, and maybe some upgraded insulators to withstand the higher voltage. Insulators are cheaper.

Why AC? Because during the war of the currents, the only economical way to step-up or down voltage was with a transformer, and transformers only work with AC. There have since been inventions that make DC transmission at high voltage feasible, and so now some high-power transmission lines use DC

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    \$\begingroup\$ HVDC inter-tie lines use converters called "Valve Halls," named because of the big warehouses full of giant thyratron tubes used in early versions. Search Goog Images for EXTREMELY cool photos: google.com/… (The arrays of thyristors must suppress corona, and survive occasional hundred-KV arcing caused by lightning strikes to power lines.) \$\endgroup\$ – wbeaty Apr 16 '20 at 19:45

Ignore that the electricity is AC, that high voltage is more efficient is true for AC or DC. AC was chosen for different reasons.

To use an analogy of water in a flowing under pressure down a hose to drive a turbine:

Current is analogous to the flow rate of water such as gallons per minute

Voltage is analogous to the pressure of the water in the hose

The aim of the water flow and pressure is to deliver power to a turbine at the end

the power delivered to the turbine is flow rate x pressure which is analogous to electricity where volts x current = power and if either is low you have less power -

To increase the power you can increase the flow rate or increase the pressure. Increasing the flow rate leads to pressure loss at the end of the hose unless you increase the size of the hose. So you increase the pressure (ignoring the problem of bursting the hose) instead and design the turbine accordingly to keep the flow rate the same.

Electricity is the same: pushing current through a wire causes voltage drop along the wire due to resistance so some of the power is lost in heating the wire, that is minimized by using high voltage and low current.

High flow rate(current) kills you doubly because not only is the pressure(voltage) loss higher due to the flow rate(current) the lower working pressure(voltage) means it's a greater percentage too.

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    \$\begingroup\$ AC was chosen because they could use simple transformers to convert high voltage/low current AC to low voltage/high current AC at the end, Transformers only work for AC. \$\endgroup\$ – Chris Glendinning-Smith Apr 16 '20 at 15:27
  • \$\begingroup\$ Thanks for your input, I liked the part where you touch the voltage drop, still very new to me. \$\endgroup\$ – MeTitus Apr 19 '20 at 18:27

Corona loss if also often a factor in AC transmission, but it's not often mentioned. General Electric was also a pioneer in HVDC conversion and transmission. They built the Sylmar terminal in Sylmar C.A. , which was equipped with solid-state thyristor rectifiers. ABB built the original terminal at Celilo Oregon with Mercury Arc Rectifiers. That station has been upgraded with solid-state valves. One nice feature of an intertie is that the phase of the voltage waveform can be changed very rapidly, so that the DC line can be used to damp out instabilities in the connected AC lines. This capability has gotten more and more powerful over the decades as the reaction time of the valve control electronics has decreased. Also, many AC lines are limited not by resistance, but by inductive reactance. That's not the case with DC. They were talking about putting a large DC line under one of the great lakes. Since practical transmission of AC under water is limited to about 100 miles, due to CAPACITIVE reactance, it couldn't have been done with AC. When I worked in the field, the break-even point for DC versus AC was about 400 miles. Maybe somebody knows what it is today? Anyway, GE was gung-ho on HVDC until the oil crisis (1974?) caused people to start conserving electricity so that new lines were not that much needed. HVDC is all the rage in China and other places, though. Look up "battle of the currents" to find some interesting history of AC versus DC in the US.

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    \$\begingroup\$ Welcome to the site :-) You have tried to edit your answer using a different account, but with the same username (so I guess you have two accounts - that's not a good idea for other reasons, but anyway...). Instead, please don't use the duplicate account which you somehow created. Instead, login using the same credentials which you used when you wrote the answer, and then you can edit the answer directly. (I also recommend that you read the tour and help center to learn more about Stack Exchange sites and how they differ from typical internet forums, if you haven't already done that.) Thanks. \$\endgroup\$ – SamGibson Apr 15 '20 at 22:03
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    \$\begingroup\$ " thyristors. That station has been upgraded with solid-state valves" Thyristors ARE solid-state. \$\endgroup\$ – winny Apr 16 '20 at 7:58

I'm going to to write this in the simplest form possible so anyone who isn't that much into EE can understand. I will just use the basic formulas to make it as simple and as intuitive as possible.

The basic formula for power is: P = U * I. Now if you change U as R * I you will get: P = R * I * I or P = R * I^2. You see that the power is related to the amount of current which is sent through the conductor. If we reduce the current we reduce the power and thus reduce the power loss.

Now if we want to lose as less power as possible we need to reduce the current. If we want to deliver the same amount of power and decrease the current and power-loss we can increase the voltage. P(const) = U⬆️ * I⬇️.

Now we understood the high-voltage part, why use AC over DC?

Generally AC is more used because it was cheaper to setup and way simpler. Building an AC-AC and AC-DC transformer to stepdown and setup-up the voltage is quite easy. For AC-AC you need insulated wiring (you can use a metal core but it's not a must for experimental purposes) and for AC-DC you just need AC-AC transformer with a capacitor and a FULL-BRIDGE RECTIFIER at the end. A 10 year-old could make it.

But there is a downside, AC has the so-called skin-effect. It basically means more electrons go to the surface of the conductor than in the center off iy, not distributing well.

That's where HVDC jumps in! HVDC means, believe me or not, High-voltage DC. It uses kilo-volts to transmit power. It wasn't used at the beginning because you would need electronics to create it and control it on both ends (the most sophisticated electric element at the time was a heating light-bulb).

Even with the power loss of electrons smashing into molecules, it's less than AC's skin-effect over large distances.

HVDC is slowly starting to be used but AC is just cheaper to maintain than converting to HVDC.

  • \$\begingroup\$ I agree with everything until the last statement, ac is actually used not because the dissipation is lower that using ac, actually it's worse, with ac you have skin effect. however in the time in which the ac networks were deployed for the first time there were no electronics, so you couldn't change voltages easily unless you used transformers, so it was better to generate in ac and then step up, plus AC generators are simpler than DC ones. Today DC is used for long distance high voltage transmissions where the losses over distance make it cost effective, take a look at HVDC networks \$\endgroup\$ – diegogmx Apr 15 '20 at 12:34
  • \$\begingroup\$ Thanks @diegogmx I'll rewrite that last part, I totally forgot about that! Honestly, I can't believe what I learned in high school ahahah! My school was EE-based (I graduated in 2019), and there we, I swear to God, were tutored that the skin effect is wanted. I totally forgot about high-voltage DC powerlines and that they're not that used because it's a drastic change for the power-lines. \$\endgroup\$ – Zastrix Arundell Apr 15 '20 at 12:46
  • \$\begingroup\$ omg why would you ever want skin effect in transmission lines LOL \$\endgroup\$ – diegogmx Apr 15 '20 at 12:50
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    \$\begingroup\$ i think what they actually said there is that since there is skin effect, most of the current will flow in the outer diameters of the conductor, so you could hollow it out saving material without losing performance, that is different than to say that that skin effect improves anything \$\endgroup\$ – diegogmx Apr 15 '20 at 13:05
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    \$\begingroup\$ "skin-effect. It basically means that electrons go to surface of the conductor and they can jump in the air and you just have less output." - This is just wrong \$\endgroup\$ – Rev1.0 Apr 15 '20 at 13:11

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