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Super nooby question involving Ohm's law, but this has been on mind this morning.

Say I have a 60W device, and I want to power it. Usually this calls for a 120V source or something. However, why not use a 5V source and draw 12A with really low resistance? Is it for safety purposes mainly? Or is there an issue with getting the resistance low enough to achieve the 12 amps?

I tried googling this but not much came up. Probably really obvious but just wondering..

EDIT for duplicate mark: The duplicate suggestion is similar; however, it discusses series vs. parallel cells and adds interesting information, but isn't exactly what I was asking about. The answers provided on this post were much more useful to me.

EDIT 2: I added my original edit back now that the duplication mark has gone through.

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    \$\begingroup\$ Ohms law shows the lower the voltage for a given power the current increases. The power loss in feeding a given power is the current squared, so feed losses are greater at lower voltage. \$\endgroup\$ – Optionparty Jul 31 '17 at 14:20
  • \$\begingroup\$ Kelvin's law is worth a look. \$\endgroup\$ – Andy aka Jul 31 '17 at 14:33
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    \$\begingroup\$ Oh, and on topic - one example of a high power device which uses high current/low voltage is spot welding machines. They work by using the resistance of the metal to be welded to produce heat at the weld spot. \$\endgroup\$ – pjc50 Jul 31 '17 at 15:30
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    \$\begingroup\$ Did you ever look at starter motors in cars? They are powerful (>1kW) electric motors, powered at 12V (so around 100A). Compare their cables size with the cable of your hair drier (again around 1kW)... \$\endgroup\$ – frarugi87 Aug 1 '17 at 8:05
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    \$\begingroup\$ EDIT for duplicate mark: The duplicate suggestion is similar; however, it discusses series vs. parallel cells and adds interesting information, but isn't exactly what I was asking about. The answers provided on this post were much more useful to me. \$\endgroup\$ – Capn Jack Aug 1 '17 at 15:09
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You are right in that power is the product of voltage and current. This would indicate any voltage x current combination would be fine, as long as it comes out to the desired power.

However, back in the real world we have various realities that get in the way. The biggest problem is that at low voltage, the current needs to be high, and that high current is expensive, large, and/or inefficient to deal with. There is also a limit on voltage above which it gets inconvenient, meaning expensive or large. There is therefore a moderate range in the middle that works best with the inconvenient physics we are dealt.

Using your 60 W device as example, start by considering 120 V and 500 mA. Neither is pushing any limits that result in unusual difficulties or expense. Insulating to 200 V (always leave some margin, particularly for insulation rating) pretty much happens unless you try not to. 500 mA doesn't require unusually thick or expensive wire.

5 V and 12 A is certainly doable, but already you can't just use normal "hookup" wire. Wire to handle 12 A is going to be thicker and cost considerably more than wire that can handle 500 mA. That means more copper, which costs real money, makes the wire less flexible, and makes it thicker.

At the other end, you haven't gained much by dropping from 120 V to 5 V. One advantage is safety rating. Generally at 48 V and below, things get simpler regulator-wise. By the time you get down to 30 V, there isn't much saving in transistors and the like if they only need to handle 10 V.

Taking this further, 1 V at 60 A would be quite inconvenient. By starting at such a low voltage, smaller voltage drops in the cable become more significant inefficiencies, right when it becomes more difficult to avoid them. Consider a cable with only 100 mΩ total out and back resistance. Even with the full 1 V across it, it would only draw 10 A, and that leaves no voltage for the device.

Let's say you want at least 900 mV at the device, and therefore need to deliver 67 A to compensate for the power loss in the cable. The cable would need to have a out and back total resistance of (100 mV)/(67 A) = 1.5 mΩ. Even at a total of 1 m of cable, that would require quite a thick conductor. And, it would still dissipate 6.7 W.

This difficulty in dealing with high current is the reason that utility-scale power transmission lines are high voltage. These cables can be 100s of miles long, so series resistance adds up. Utilities make the voltage as high as they can to make the 100s of miles of cable cheaper, and for it to waste less power. The high voltage does cost some, which is mostly the requirement to keep a larger clearance around the cable to any other conductor. Still, these costs aren't as high as using more copper or steel in the cable.

Another problem with AC is that the skin effect means you get diminishing returns in resistance for larger diameters. This is why for really long distances, it becomes cheaper to transmit DC, then pay the expense to convert that to AC at the receiving end.

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  • \$\begingroup\$ That's a really good point on the mention of voltage drop being much more noticeable at a low voltage. Thanks for such a great answer. I love it when I get an answer to my question and then some! : ) \$\endgroup\$ – Capn Jack Jul 31 '17 at 14:44
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    \$\begingroup\$ I might add that we usually ignore resistance when dealing with high voltage transmission lines, because the inductance is so much larger compared to the resistance. The active power flowing through a transmission line is (V^2/X)*sin(theta), where V is the voltage, X is the inductive reactance, and theta is the phase angle between the ends. So even in this case a high voltage is highly beneficial. In fact, this is the reason why transmission lines use high voltages -- the limiting factor is often static angular stability. \$\endgroup\$ – ntoskrnl Aug 1 '17 at 6:58
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    \$\begingroup\$ @ntos: Good point about the inductance dominating. The resistance is still important in terms of power loss, and dissipation in the power lines. Sagging power lines due to high ambient temperature plus heating due to high load have caused power outages by shorting against trees and the like. Resistance can be ignored for some purposes, but not others. \$\endgroup\$ – Olin Lathrop Aug 1 '17 at 14:47
  • \$\begingroup\$ A kind-of-related question: why electric locomotives uses a relatively low motor voltage (KV or sub-KV level) compared with transmission line voltage (tens of KV level)? \$\endgroup\$ – user3528438 Aug 1 '17 at 15:58
  • \$\begingroup\$ @user3528438 TGV (and probably other catenary-fed) trains can use 25 kV, but metro "third-rail" trains (the Chicago 'L' uses 600 V DC) need to worry more about things like arcing, safety, and parasitic resistance when it rains. I'm willing to bet third rails are cheaper to maintain and operate than catenaries, and work just fine when your top speed is 55-70 MPH. \$\endgroup\$ – Nick T Aug 1 '17 at 17:39
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Combine $$ P = V \cdot I $$ with Ohm's law $$ V = R \cdot I $$ to obtain:

$$ P = I^2 \cdot R $$

where \$P\$ is the power dissipated on the supply wires, \$I\$ is the current flowing through the wires and \$R\$ is the wires' resistance.

For every doubling of the current, the power lost on the wires quadruples. To compensate for that, one would have to make the resistance four times smaller i.e. increase the cross section of the wire by a factor of four (double the wire's diameter) meaning four times more copper.

For the very same reason the power grid uses up to several hundred kilovolts to transport electricity (transport at household level voltages would require of the order of a million times more copper to keep losses the same).

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    \$\begingroup\$ +1 this is a really good explanation of what was previously posted about power loss in conducting components. \$\endgroup\$ – Capn Jack Jul 31 '17 at 15:53
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High currents are undesirable for a couple of reasons. Firstly larger currents require larger conductors and larger contacts in switchgear. Secondly high currents are a fire risk, in a high current system a small ammount of extra resistance from a bad connection can easilly get very hot.

High voltages are also undesirable, they require thicker insulators, require larger contact gaps in switchgear and larger spacing between terminals and pose more of an electric shock hazard.

Of course for a given power reducing voltage will increase current and vice-versa.

So we need to find a happy medium, the happiest medium will depend on the power level involved and to some extent on the details of the load. In practice we also have to compromise for compatibility, people want to have one set of wiring in their house into which they can plug everything.

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Achieving the really low resistance reliably is a major issue. Until room temperature super conductors exists it will remain a big issue.

Many PC power supplies will feed high power over low voltages. They have a sense wire on the power rail that is bonded to the end of the cable. This feeds back to the regulator circuit to boost the voltage to compensate for the voltage drop from the high current draw and the internal resistance of the wire. However modern motherboard will draw most of their power from the highest voltage rail to avoid the losses and regulate it down internally.

High amp loads also need beefy conductors that won't heat up and melt under that high current. If the conductor is damaged in any way that spot will have higher resistance and heat up more.

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  • \$\begingroup\$ This is a lot of what I suspected, thanks! Interesting mention about the PC power supplies too. Really cool. \$\endgroup\$ – Capn Jack Jul 31 '17 at 14:29
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As others have noted, the higher the voltage the lower the power loss over the cables connecting the power to the device.

Consider mains power that gets boosted up to hundreds of kilovolts for long-distance transmission over the electrical grid. These are carried on the largest electrical transmission towers that need a huge amount of space to keep the wires away from each other and anything they may arc to. They are very dangerous voltages and completely inconvenient when you need to use the power in a normal setting - it does allow the power to be efficiently transported over very large distances, however.

When it gets to a local substation it will be reduced in voltage to something on the order of tens of kilovolts and carried on smaller towers and poles (or underground) to large facility customers and neighborhood distribution transformers. These then lower the voltage again to your household mains level (100-240V). At this level the voltages are high enough to allow efficient transport of the power around your house (on reasonably sized wires) but low enough that they don't have many of the issues of high transmission voltages (RF interference, arc hazard, etc).

Consider now something like a computer - the mains voltage makes its way at low loss through the wires in your house until in reaches the power supply. At this point it is further reduced to 5V and 12V (DC). Here the power only needs to make its way a very short distance to the motherboard and components, and having very thin wires at mains voltage levels inside such a case is not really convenient. None of the internal devices in a computer can operate on such high voltages directly anyway, so the PSU is there to conver the power to a form that is useful for the end device.

On the motherboard itself, the voltage is again reduced to feed the RAM, chipset, and CPU - the latter a delicate piece of hardware that would be destroyed by voltages much higher than about 1.3V. Here the power only needs to move a few centimeters or less, and a typical CPU can draw something between 60-80 amps of current at that very low voltage. So here you have, say, a 90W CPU drawing 70A at 1.3V from a voltage regulator drawing 7.5A at 12V from the PSU which is drawing 0.75A at 120V from the plug in the wall which is drawing 23mA at 4kV from the neighbourhood transformer which, up the line, is pulling 230 microamps from the long distance lines on the grid.

At the end of the day, it's about matching the power supply to the load in an efficient way. This usually means transforming the electrical power numerous times, at each point to a voltage that suits the application.

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Simply put, a low voltage requires high current. High current puts a lot of thermal stress on all components on circuitry. And you need to have thicker wiring as a bonus. High voltages do not stress most of the components as long as you don't short anything..

You can definitely power a 60W device from 12A@5V PSU but 12A is already a rather high current for connectors, ferrites, inductors ..

From safety point of view, 24VDC is often used, especially on a medical setting. Higher voltages may be used depending on jurisdiction but the popular option is to just insulate the device so you can't stick your finger on live circuitry.

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As an anecdotal addendum to the other answers, there's an old rule of thumb that the appropriate power transmission distance for some voltage V is around V feet. If you think about how far you'd want to run, say, 12V to a light fixture drawing a significant current (e.g. the halogen lamps that became very fashionable in the 90s and are now, glory be, being displaced by LEDs), 12 foot isn't a bad guide. Likewise for 230V, 230 feet from transformer to domestic light bulb works pretty well.

Never a hard and fast rule, just an approximation of course.

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