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Since lower voltage means proportionally higher current for a given power, and higher current means more impedance, presumably having a lower voltage at the outlet means transmission power losses are greater than if a used a higher voltage.

How much more energy is lost per year due to the USA using 110V than if it used Europe's 220V? What is the dollar cost?

Assume the USA consumes 5 billion MWh per year, and 1 MWh costs $US0.20 to produce so total current (no pun intended) cost is $US1B p.a.

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  • \$\begingroup\$ "the USA using 110V" Appliances use 110V. Houses are supplied 220V regardless. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Dec 23, 2016 at 9:38
  • \$\begingroup\$ @IgnacioVazquez-Abrams OK. So loss is over wiring within the house. \$\endgroup\$
    – Bohemian
    Dec 23, 2016 at 9:40
  • \$\begingroup\$ The USA has a lower voltage in the home circuits, but the power is transmitted at higher voltage across the large distances. Look up the US nation grid on google and you'll see the power is transmitted at 110kV and upwards. Within a house, the power you're looking at is extremely low. However, in the rest of the world, electric cookers and hobs are a lot more common than in the US (apparently), due the higher power available at the higher voltage. \$\endgroup\$
    – Puffafish
    Dec 23, 2016 at 9:43
  • \$\begingroup\$ @IgnacioVazquez-Abrams: Usually "using 110V" is referred to the voltage of one phase to GND. If you want to compare the often two phases of center tapped 220V that are often supplied to houses to the common three phase installations in lots of parts of europe you could say those housholds are supplied with 400V, which doesn't help for the thing in question which is one nominal mains voltage appliance and the current consumption for a given power. \$\endgroup\$
    – PlasmaHH
    Dec 23, 2016 at 9:45
  • \$\begingroup\$ Assuming the US house load is split evenly across both 110V phases, there will be very little difference. \$\endgroup\$
    – user16324
    Dec 23, 2016 at 10:19

2 Answers 2

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Only the final leg of wiring to the house is 110V and compared to the transmission losses upstream the cost is minimal. The decision to use the lower voltage of 110V was driven by safety. I am well aware that 110V can be lethal but a compromise had to be made somewhere. Given the low cost of electricity in the USA I would say that the decision is a good one.

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    \$\begingroup\$ How does this answer the (current?) question? \$\endgroup\$
    – Wouter van Ooijen
    Dec 23, 2016 at 12:44
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In 110V-land, high-current devices, typically resistance heating like cookers, ovens, water heaters, clothes dryers, etc. are wired to operate at 220V volts just like in the rest of the world.

The waste of power from hundreds of millions of "wall-wart" chargers using power with no load connected is an issue that is now being addressed by government passing laws requiring circuits to be designed for minimum power drain when not supplying any power. This appears to be a much more significant issue than whatever resistance losses in the "final-leg" 110V distribution wiring which is typically only a few meters long.

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  • \$\begingroup\$ The last paragraph is a (relevant) comment, not an answer. However I didn't know that permanently installed high current devices operated on 220V. Also, the rest of the world doesn't use 220V; there are roughly equal numbers of countries using 220, 230 and 240. \$\endgroup\$
    – Bohemian
    Dec 23, 2016 at 16:49
  • \$\begingroup\$ The last paragraph is relevant because it demonstrates that losses from using lower voltage/higher current is not even as significant a problem as the rather minuscule loses from idling wall-warts. The terms "110" and "220" are generic and not specifying the exact line voltages, but the general utility standards in a particular area. Indeed the differences between the standards in different countries vs. the European Union standard are simply "defined-away" by changing the terminology of the specifications vs. sending workers out to re-tap millions of transformers. \$\endgroup\$
    – Richard Crowley
    Dec 23, 2016 at 16:54

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