I was in a seminar and heard the presenter say something I am not able to understand. The statement was that if you use a higher gauge wire it would help reduce your electricity bill. Since I was not an official guest of this seminar I was not really in a place to ask how this is possible.

I doubt that this is entirely true as I would have heard it by now. What I am trying to figure out is if there might be some truth to this. I want to know if this is true and to what extent. I believe that the reason on why you would want to use a higher gauge wire is for safety rather than lowering your electric bill.

Currently here is my argument to why this might not be possible. So what happens when using thinner gauge wires is the wire resistance increases and when the resistance increases this creates a small current limiting effect like how you use a current-limiting resistor to an LED. So you could say that using thinner wires will save your electric consumption. I mean at higher load the wire gets hot and would not be safe to handle.

Any thoughts?

  • 1
    \$\begingroup\$ "Higher wire gauge" is an ambiguous term which should not be used since lower numbers means larger diameter wire. "So you could say that using thinner wires will save your electric consumption. I mean at higher load the wire gets hot and would not be safe to handle." That is a good point and would depend on whether your loads are constant current or constant power or not \$\endgroup\$
    – DKNguyen
    Commented Feb 2, 2022 at 2:19
  • \$\begingroup\$ When you did a web search on "wire gauge and its impact on energy efficiency in homes", what results did you get? \$\endgroup\$
    – TimWescott
    Commented Feb 2, 2022 at 2:29
  • \$\begingroup\$ Unless there’s significant loss in your wiring, there’s not much to gain. The wiring codes will specify a maximum voltage drop and it is around 3% if my memory serves. I’d say there are savings to be had, but you’d be scraping the bottom of the barrel. I dare say there’d be better low hanging fruit than wire size. \$\endgroup\$
    – Kartman
    Commented Feb 2, 2022 at 6:07
  • \$\begingroup\$ @DKNguyen, the first part of your comment completely depends on if they are talking about (the very non-intuitive) AWG or metric (square mm). \$\endgroup\$
    – StarCat
    Commented Feb 2, 2022 at 7:41
  • \$\begingroup\$ As an European I've never heard of the term "wire gauge" referring to anything else but the AWG system. So "higher wire gauge" seems pretty senseless to me, thinner copper wires can only make things worse and never better. \$\endgroup\$
    – Lundin
    Commented Feb 2, 2022 at 9:51

3 Answers 3


In some systems such as North American AWG, higher gauge wires are smaller.

There are minimum wire sizes which you cannot go below. Generally speaking, all wiring installed in buildings is already at this minimum and cannot be thinned further. For instance a 20/23 amp dryer in North America (20 amps 240V resistive load, 3 amp 120V motor load) requires 10 AWG wire, because of NEC 240.4(D) Going to thinner 12 AWG is not an option. Going to fatter 6 AWG is the only thing that can be discussed.

As for any potential savings, you can crunch the numbers for yourself, using any of the voltage drop calculators located across the web and intended for commercial mains electrical wiring.

Let's talk about "reducing voltage saves energy".

Your misconception is that a load supplied less voltage will consume less current. That's not true for all loads. It works for resistive electric heating loads, but people who care about saving energy are using fewer and fewer of those.

Even so, the issue is that you are not simply "saving" the energy like you would if you were bucking the voltage down with a transformer. You ARE spending the energy - but you are spending it in an uncontrolled place - the wires themselves in the form of voltage drop, which heats the wires.

It doesn't work. The appliances will still do their job.

Let's say you take your power to a water heater. It is a 4800W resistive element, so 20 amps and 24 ohms. Clever you -- you add 1.2 ohms to the resistance of the wires, by running hundreds of meters of wire back and forth (so you meet the minimum size requirements and distribute the heat). So the 20A results in a 24V drop (-10%) giving 216V.

On paper, this sounds great. By reducing the voltage 10% to 216V, you reduce current by 10% to 18A. The heater is now drawing 3888W (formerly 4800W) or -19% from its former draw! *But hold on. The dropping resistance wasn't free. 24V x 18A of flow means 432W of "voltage drop losses" in the wire. Still not so bad, since for our 432W investment we save 912W. Time to practice the Nobel speech!

Except for one thing. It used to take 81 minutes for the water heater to heat its water from cold to target temperature. For some reason, it's now taking 100 minutes. Could this be related?

You bet it is. The core job (warm cold water) is still sitting there waiting to be done. By clipping the wings of the water heater, we only make it work longer. It still finishes, and the water heater uses the same amount of energy:

  • Before: 4800W x 81 min = 388,800 watt-minutes (6480 Wh)
  • After: 3888W x 100 min = 388,800 watt-minutes (6480 Wh)

Well, that was a waste of time.

Externalties, though

Oh, snap! We forgot about the 432W we're losing in the voltage drop!

  • Before: (negligible) x 81 min = 0 Wh
  • After: 432W x 100 min = 43200 W-min = 720 Wh

Even worse, the wasted 720Wh is showing up in uncontrolled locations, like the walls of the house. Which means now an air conditioner must remove that extra heat. Assuming a 4:1 efficiency, that's another 180Wh. Or even if we want heat, we're getting electric resistive heating when we have far more efficient ground-source heat pump (3:1 better) so we're spending 720Wh to make heat we could have gotten for 240Wh with the heat pump. (so 480Wh spent the hard way).

So we went from 6480Wh before meddling, to ~7600Wh after meddling. We are now less efficient by 15%!

So no, increasing wire resistance to improve efficiency of appliances will tend to backfire. Generally, wire losses are considered to be a total loss.

Ah, so enlarge wires then?

Maybe. However in most installations, the wires aren't really long enough to account for significant voltage drop. This is a worthy area of study, but you have to "crunch the numbers" as to whether the savings in electricity is worth the extra cost of wires. It's "marginal at best" in most cases.

But here, we can use a quintessentially American concept, found in highly electric jetliners like the Airbus A350: Aluminum wire. America has had its "trial by fire" with aluminum and has dialed in how to use it properly. The laughably low cost of the stuff allows you to "throw cubic metal at the problem".

For instance, when people talk about running a 30A subpanel out to a garage for electric vehicle charging, we'll immediately advise "use 2 AWG aluminum". It's vast overkill, with 90A ampacity, but it's a "sweet spot" in wire pricing and costs no more than 30A wire. (it barely costs more than 65A aluminum wire). However its resistance and thus voltage drop is vastly superior, and results in real energy and cost savings over the life of an EV charge installation.

When loads are large (like EVs) and wire is long enough for voltage drop to arise as an issue, aluminum makes the difference between "economical" and "not economical" to enlarge wires. The rules of its handling must be faithfully followed, but they're not hard if suppliers cooperate by providing aluminum rated terminals.

  • \$\begingroup\$ I like the example with the heater in the home appliance. Also, most smaller power supplies today use a control loop, drawing whatever input power to maintain the required output power. If your power supply needs 100 W at a high input voltage, it will still need 100 W at a lower input voltage, caused by thinner wire. The wire losses will thus be added to the equation. Oh, and I liked the part about the air conditioner having to cool the wire. That's if you're lucky. If you're less lucky, it will be the fire department taking care of the heat, not the air conditioner. \$\endgroup\$
    – zebonaut
    Commented Feb 2, 2022 at 7:30
  • \$\begingroup\$ Negligible difference. There are cheaper and easier ways to save on electricity. \$\endgroup\$
    – winny
    Commented Feb 2, 2022 at 7:39
  • \$\begingroup\$ @zebonaut yes, I presume one followed my first paragraph's advice and used wires of a legal size, just a very great length of them zig-zagged and not coiled. (packed tightly, they melt). \$\endgroup\$ Commented Feb 2, 2022 at 9:31
  • \$\begingroup\$ The fire department will know this is a bad idea. The gauge will be legal, but the resistance of the long wire may become so high that the circuit breaker won't trip if there is a short circuit... \$\endgroup\$
    – zebonaut
    Commented Feb 3, 2022 at 10:07
  • 1
    \$\begingroup\$ @Graham good catch, thanks. Fixed. \$\endgroup\$ Commented Feb 6, 2022 at 20:37

The electric bill is the cost of the Power over time "kWHr" that you consume. For a resistive load the power is V*V/R. So, if you increase the wire gauge, R will increase and reduce your power consumption, hence your electric bill.

However, the power that you consume "Pin" is in part been dissipated in heat "Ploss" in the wire and the remaining to the load "Pout". The efficiency Pout-Ploss/Pin goes down. Furthermore, you will no longer comply with building code, a could cause a fire. The wire gage is function of the breaker ampacity

If your load is constant power (such as some electric motor), then your consumption, Pin = Pout+Ploss. In this case you would go with a lower gage wire to reduce the losses.


Your energy supplier is measuring your consumption after the power meter. It doesn't matter whether if you spend the energy you pay for in the cable or in the loads.

better cables will cause, with resistive loads, that the energy which is not consumed in the cable now is consumed by the load - but you will still pay the same. e.g. if your load is a heater you might improve the energy utilization to some fractions of a percent... if you will notice a difference is unlikely.

with active loads (a switching power supply) a better cable will reduce the voltage drop of the cable and thus, the active load will reduce its current consumption - which will reduce the voltage drop further. Here you can expect an overall improved electrical efficiency. But the more power the active load has (relative to the power looses of the cable), the smaler the effect is. And with small active loads, you might safe some fractions of a cent...not worth the effort


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