Generally, one would like to minimize power losses via Joule heat. Intuitively, using low resistance cables (for example by using materials with low resistivity like copper) should do a better job than highly resistive cables.

However this is not what I get mathematically. Here's my reasoning:

Let's assume that Ohm's law holds, so \$V = RI\$, and let's assume that our power source is a voltage source, so \$V\$ is constant and fixed. Let's consider 2 cases, one with a resistor R and the other with a resistor R/2, where "resistor" means the total resistance of the electrical circuit, so cables + load. In the 1st case, the power dissipated is \$I^2R=V^2/R\$. In the 2nd case, it is \$I^2(R/2)=2V^2/R\$, i.e. twice as much power is dissipated if we halve the total resistance of the circuit. It seemed counter intuitive at first to me, but when one halves the total resistance of the circuit, we double the current, and since the power losses go like \$I^2R\$, the increase in current more than offsets the decrease in resistance, so overall the power dissipated increases. So, I am lead to wonder why would one want to use lower resistive cables? Wouldn't it mean a higher current, thus a higher dissipated power in the circuit?

The only reason I have found so far is that the voltage available to the load is greater when the cables have a low resistance. I am wondering if that's the reason why we use low resistance cables, because it seems like it still dissipates more power than if we had picked a material like iron instead of copper. So it seems that the premise of the question may be wrong, i.e. in reality we might not seek to minimize power losses.

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    \$\begingroup\$ You've not separated out "load" and "cable" in your calculations, so your results are numerically correct but highly misleading, and you've not accounted for a third factor at all: the internal resistance of the power source. If you put a resistor in series with a light bulb, you get less power consumed and less light. Removing the resistor increases the amount of useful power delivered. \$\endgroup\$
    – pjc50
    Aug 16 '19 at 8:45
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    \$\begingroup\$ Having a high resistance cable is like driving your car with the clutch permanently slipping. Higher resistance means more slippage. Think about it. \$\endgroup\$
    – Andy aka
    Aug 16 '19 at 9:21
  • \$\begingroup\$ I've addressed that issue @pjc50, with an edit that basically contains an answer. \$\endgroup\$ Aug 16 '19 at 9:34

You are assuming cables and load to be one "unit". But this does not make sense in this case.

Typically, your load requires a specific power and for that it needs some specific current (calcualted by Ohms Law). This current now creates additional losses in the cables and these losses are lower when the resistance is smaller.

Take, for example, a 500 W computer power supply. When the computer requires 500 W, the PSU will deliver this power - independent from the input voltage, at least over some specified input voltage range. But a resistance in the cable creates a voltage drop, so the power supply needs a higher current to deliver the 500 W. So you get high losses in the cable.

  • \$\begingroup\$ I've edited my question by including an answer. I have followed your tip to split the total resistance into the wire and the load one. What I found is that the power dissipated in the wiring decreases when the resistance of the wires go to 0 or when that resistance increases compared to the resistance of the load. This goes against your claim that a high resistance in the cable necessarily implies a high loss in the cable. Did I go wrong somewhere? (I moved my edit to an answer) \$\endgroup\$ Aug 16 '19 at 9:33
  • \$\begingroup\$ Your explanation doesn't hold water. Reading between the lines, you mean mains input cables going into the regulating PSU. But it describes all cables. Some outputs are post-regulated by the motherboard, some not and the latter get voltage drops. Downvoting til rewritten and corrected, I'm afraid (which isn't just adding 'input' in a few places). \$\endgroup\$
    – TonyM
    Aug 16 '19 at 11:54

The higher the internal resistance of the cable, the higher the voltage drop will be across the cable. We can calculate this with Ohm's law and a voltage divider.

So we have a Vin than a cable with resistance Rcable and than we have the load resistance Rload

This gives us Vin = I * (Rcable + Rload) = Vcable + Vload This is fixed, can't change it.

And Vload = Vin * Rload / (Rload + Rcable)

So, if Rcable rises, Vload will be decreasing. And we lose power delivered to the load as there will be more voltage across the cable.


You are basically correct.

For any device that can be approximated as a resistor, using higher resistance wires will decrease power consumption.

This is usually not the primary goal of electronic devices. They need to use power to perform their function, and they are internally designed to use the power they need based on assumptions about the power supply. When you make the wires more resistive, it violates those assumptions and can cause the device to behave abnormally.

Saying you can decrease the power loss in a resistive system by increasing the resistance of the wires leading to it is roughly the same as saying you can reduce the power consumption by unplugging it. It's true, but not really a useful result for most applications.

The simple example is heaters. If you put a resistor in series with a heater, it will consume less power. It will also put out less heat, and a higher percentage of the heat that is produced ends up inside your power supply instead of whatever you are trying to heat. In the case of an electric oven, you can probably see how this might be a problem.

If you have a load that will adjust until it gets the power it wants, such as the computer power supply from another answer, then putting higher resistance wires will not decrease power consumed by the load, and will add additional losses due to power dissipated by the wires.

So to bring it full circle, if you redo your math with the additional assumption that the power consumed by the load is a constant, and only adjust the resistance of the wires leading to that load, you should end up with a result that makes more intuitive sense in the context of real appliances / electronic devices.

  • 5
    \$\begingroup\$ "Saying you can decrease the power loss in a resistive system by increasing the resistance of the wires leading to it is roughly the same as saying you can reduce the power consumption by unplugging it." I think this is one of the rare cases where taking something to an extreme actually makes the argument clearer. It should be mind-numbingly obvious that an unplugged appliance consumes the least power, but even so it proves a very useful way to poke and prod at the real question! \$\endgroup\$
    – Cort Ammon
    Aug 16 '19 at 18:36
  • \$\begingroup\$ I've done the math but was a bit "stuck" in that I found out that it was not always possible to keep the power consumption constant, \$R_\text{load}\$ cannot adjust enough if \$ R_\text{wire}\$ gets big. I will redo the math... \$\endgroup\$ Aug 16 '19 at 20:06
  • 3
    \$\begingroup\$ That at least is probably not a mistake, but rather a useful bit of knowledge gleaned through math :) If the resistance of the wire is too large, the load cannot adjust enough and stops working. Kind of like how older LED and Fluorescent lightbulbs didn't work with dimmer switches. Not enough power to the load = load doesn't work. This is the core of the answer to your question, actually. If the load has a minimum power requirement, then there is a corresponding maximum wire resistance before the load can't compensate. \$\endgroup\$
    – Kyle
    Aug 16 '19 at 20:14


No, not really. Example: space heater. You want the joule losses. It's not about total "loss", it's about where it happens. You want the power at the heater in your room, but not at the cables on their way to power plant.

Power in the load is "gain", power in cables is loss.


Yeah, that's exactly what we want. Higher dissipated power = more heat in your room at cold winter night.


The premise in your question is right, what wrong is treating all power as "loss".

Electricity is about power delivery. About getting as much as you need where you need it. Your question is about reducing total power usage. And we are doing it all the time. But rather than switching from copper cables to iron ones in order to increase resistance of the whole circuit, you use the power switch at your heater to increase the resistance of the heater alone. So you still get (almost) twice the resistance and less power dissipated. You pay smaller power bill, but your room is colder now.

The point of copper cables is: nobody is paying for useless heating of the power lines outside.

/edit: We do use the method you've mentioned. In a lab, we often use rheostats (a variable resistance device) connected in series in order to regulate the power in the load. Which exactly implements your idea of increasing resistance of the cables, rheostat simply serves as one of the cables. This is the simplest way of regulating power of a device that doesn't have power regulation feature, eg a lightbulb. Real world dimmers don't work this way, because they would heat up. And we don't want the dimmer heating, we want only lighbulb's filament heating.


Let's try a practical example

Let's suppose we want to power a string of pole lights at the end of a very long drive, say 2000 feet. We need 60,000 lumens to get the job done. That will require 600 watts of energy.

Supply is constant-voltage 120W, North American supply.

Got the parameters so far? 600 watts. 2000 feet (600m) with 10% deviation acceptable, so 540-660W.

I am crazy, so I use #14 wire (2.0mm2). This has .0025 ohms per foot (.008 ohms/m). So crunch the numbers for our 4000' round trip, I get 10 ohms. Uh-oh.

P=EI. 600=120*I. I=5A current.

E=IR. E=5*10. E=50=voltage drop in the wires.

usable voltage @5A is 70V. Giving 350 watts. This does not meet the spec.

If we pull 6A instead we get 60V drop and 360 usable watts. At 7A we have 70V drop and back to 350 usable watts; clearly 6A is the peak. #14 wire cannot deliver 600W.

How about larger 12 AWG (3.3 mm2, 0.0016 ohms per foot)? It can't carry 600W either. I crunched the numbers and at 8 amps, it has 51.2 volts drop, giving 68.8 volts usable, giving 550.4 watts usable. That's in spec, call it good. However, this uses 960 watts to deliver 550.4 watts. That wasted 410 watts will cost 5 cents an hour, and this thing is on half the time, so $215/yr.

How about still-larger 10 AWG (3.86 mm2, 0.000995 ohms/ft)? Here, at 6.33 amps we drop 25 volts and can deliver the full target 600W. We waste 159 watts at a cost of about 2 cents an hour or $83/year.

How about we quit fooling around and go with fat 1 AWG (42 mm2) aluminum (nobody uses copper at these sizes). This is 0.000207 ohms/foot. Here, at 5.2 amps, we drop 4.3 volts and deliver target 600W. We waste 22 watts at a cost of 0.3 cents an hour or $12/year.

So you can see where when you actually have work to do, thinning out on wire doesn't save anything

More realistic in this case is punch it down onto a 240V breaker, easy in the North American system and the default in Europe.

Now with #14 wire, 600W happens at 2.84 amps, 28.4 volts drop (on 240V giving 211.6), wasting 80 watts/$42.

With #12 wire, 600W happens at 2.7 amps (17.2V drop, 46W or $25/yr wasted).


Thanks to the comments I got, I decide to answer my own question.

I am told that I should consider the resistance of the wires and of the load separately. When I do that, I get that the power dissipated in the wire(s) is worth \$ P_\text{Joule, wire} = \frac{V^2R_\text{wire}}{(R_\text{load} + R_\text{wire})^2} \$ which can be thought as a function of \$ R_\text{wire} \$ and displays a maximum when \$ R_\text{wire} = R_\text{load} \$ which is something we would like to avoid. So we have two ways to minimize power losses in the wiring. Either increase the resistance, or decrease it (compared to the resistance of the load). If we increase the resistance, it turns out that there will be a voltage drop very big so that the load will not have enough voltage or power to operate as it is designed to (inline with Swedgin's answer). While if we decrease it, the load will have a voltage closer to that of the power supply, and more power. Note that it also means that it will dissipate more power than if the resistance was very big, but it doesn't matter, we aren't looking to minimize power losses in the load when we design wiring cables. We are looking to minimize power losses in those cables while still providing enough energy to the load for it to operate normally.

  • 1
    \$\begingroup\$ With your assumption the power in the load does not stay constant. If you increase the wire resistance to infinity, sure you have no power losses. But you also do not have any power in your load. This basically works as an open switch. But you want he power in the load so stay the same. A switching power supply for example will increase its current consumption with higher cable resistance to compensate for the voltage drop. This again increases losses in the cable \$\endgroup\$
    – jusaca
    Aug 16 '19 at 10:01
  • \$\begingroup\$ @jusaca Alright, if I understand you well, you are saying that generally the load has a variable resistance and it will adjust so that the total power available for the load stays constant? My analysis would be valid only if the load has a fixed resistance. Is that correct? \$\endgroup\$ Aug 16 '19 at 10:21
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    \$\begingroup\$ @thermomagneticcondensedboson A load that uses a switching power supply will adjust to keep power constant (assuming the load requires a constant amount of power, anyway). A load that is just a resistor (such as a heater) will have constant resistance (with fluctuations based on temperature). A load with a linear regulator will adjust to keep current constant (again, assuming it requires a constant amount of power). \$\endgroup\$
    – Hearth
    Aug 16 '19 at 14:44

You're overthinking this. Just unplug the device. That is the exact equivalent of increasing the cable resistance to infinity. Yes, it minimizes power losses. In fact, it minimizes overall power consumption: the device no longer consumes any power at all. It also no longer does anything, but… planet saved.


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