# Resistance of wire

I am currently studying Practical Electronics for Inventors, Fourth Edition, by Scherz and Monk. Chapter 2.5.1 How the Shape of a Conductor Affects Resistance says the following:

The resistance of a conducting wire of a given material varies with its shape. Doubling the length of a wire doubles the resistance, allowing half the current to flow, assuming similar applied voltages. Conversely, doubling the cross-sectional area $A$ has the opposite effect -- the resistance is cut in half, and twice as much current will flow, again assuming similar applied voltages.

Increasing resistance with length can be explained by the fact that down the wire, there are more lattice ions and imperfections present for which an applied field (electric field instigated by added electrons pumped in by the source) must shove against. This field is less effective at moving electrons because as you go down the line, there are more electrons pushing back -- there are more collisions occurring on average.

Decreasing resistance with cross-sectional area can be explained by the fact that a larger-volume conductor (greater cross-sectional area) can support a larger current flow. If you have a thin wire passing $$\ 0.100 A \$$ and a thick wire passing $$\ 0.100 A \$$, the thinner wire must concentrate the $$\ 0.100 A \$$ through a small volume, while the thick wire can distribute this current over a greater volume. Electrons confined to a smaller volume tend to undergo a greater number of collisions with other electrons, lattice ions, and imperfections than a wire with a larger volume.

I found this interesting, because I have never heard of people considering the resistance of the wire in their calculations (as they would a resistor, or some other component) when doing electronics projects; although, I have often heard people discussing the appropriate wire gauge to use for a project, but this doesn't seem to be a matter of resistance calculation, and is more-so a matter of physical (not physics) considerations.

How common is it to consider the resistance of a wire? Is the resistance of a wire an important consideration when doing electronics work? Does this resistance need to be factored into calculations, as would other components (such as resistors)?

I would greatly appreciate it if people would please take the time to clarify this.

• It is very often considered, especially in low voltage, high power applications. Resistance is seldom a problem in mains voltage wiring, because the current carrying ability of the wire normally is more limiting. Resistance could be a consideration in very long runs of mains wire (let's say 30 meters/100 feet or more). – mkeith Jan 26 at 0:27
• "I have often heard people discussing the appropriate wire gauge to use for a project, but this doesn't seem to be a matter of resistance calculation, and is more-so a matter of physical (not physics) considerations." - What do you mean a matter of physical considerations? – marcelm Jan 26 at 12:59
• @ThePointer At the distances of a couple of breadboard the resistance of a wire can be approximated as zero so most people just do that – slebetman Jan 27 at 2:19
• Wire gauge is resistance, just measured by diameter of copper instead of ohms, in the same way as measuring air pressure in inches of mercury. – Graham Jan 27 at 8:39
• I have often heard people discussing the appropriate wire gauge to use for a project, but this doesn't seem to be a matter of resistance calculation - Yes, it is. The appropriate gauge to use is one that limits voltage drop to a level that effectively allows you to discount the resistance of the wires in the rest of your calculations. So, really, it's all about resistance. – J... Jan 27 at 15:39

Sometimes, a wire is negligible in terms of its resistance. Other times, impacts of the resistance of a wire can become significant. I'll first show the resistance of a wire, and how you can ignore it in most cases, and then show examples when its impact is significant, and finally a few applications.

## The Resistance of a Wire

Ideally, the formula of the resistance of a conductor is...

$$R = \rho \frac{L}{A}$$

Given the cross-section area (A), length (L), and resistivity ($$\\rho\$$) of the material. For copper, $$\ \rho = 1.68 \times 10^{−8} \Omega \cdot \text{m} \$$ at 20 °C.

For cylindrical conductors (like a wire),

$$R = \rho \frac{L}{\pi r^2}$$

Example: What is the resistance of 5 cm of AWG-30 (0.255 mm in diameter) copper wire?

Answer: First, the radius of a AWG-30 wire is $$\ 1.275 \times 10^{-4} \text{m} \$$, find the resistivity of copper from a textbook, which is $$\ 1.68 \times 10^{−8} \Omega \cdot \text{m} \$$ at 20 °C. The formula yields $$\ R \approx 0.0164 \Omega \$$.

Example: What is the resistance of 5 cm of AWG-24 (0.511 mm in diameter) copper wire?

Answer: $$\ R \approx 0.004 \Omega \$$.

• Remark 1: As we see, resistance of a wire is lower when the wire gauge is thicker. Specifically, when the diameter of a cylindrical wire doubles, its resistance decreases to one-fourth of the original wire. Thus, wire gauge is not only an indication of its shape. It is indeed a metric of its electrical property when its material (almost always copper) and length are given.

• Remark 2: A quantitative calculation of wire resistance is not always performed. Sometimes rules of thumb are used. Often the consideration is only "whether the wire is thick enough", not "how much resistance/voltage drop/temperature rise does this wire have". On the other hand, to analyze a wire quantitatively, knowing its gauge is the first step. Not to mention that wires are sold by gauge, so people talk about "wire gauge" (or "trace width" in circuit board design) more often than wire resistance.

On a Printed Circuit Board, you can calculate the resistance of traces in a similar way from the thickness of copper and the length of a trace. The only difference: Wires are cylindrical, while traces are rectangular.

Example: What is the resistance of a 10-mil, 10-cm trace on a 1-oz circuit board?

Answer: 1 mil is a thousandth of an inch (0.0254 mm). A "1-oz circuit board" is a circuit board with 1 oz of copper per one square foot area, or a thickness of 1.37 mils. 10 mils is 0.254 mm, 1.37 mils is 0.0348 mm. Cross-section area $$\ A = 2.54 \times 10^{-4} \text{m} \times 0.348 \times 10^{-4} \text{m} = 8.84 \times 10^{-9} \text{m}^2\$$.

Thus, the resistance $$\ R = \rho \times \frac{0.1 \text{m}}{8.84 \times 10^{-9} \text{m}^2} = 0.19 \Omega \$$

## When Resistance Can be Ignored

Most of the time, resistance of a wire is too low when you compare it to the resistance of other components and loads, so it's negligible and often safe to ignore. Moreover, $$\ V = IR \$$, the lower the current a load needs to take, the higher its equivalent resistance, so you also ignore the wire resistance if the current delivered by the wire is low, because it's equivalent to connecting a small resistor (a wire) to a large resistor (a device that takes current) - almost no effect.

For example, connect two 1,000 Ω resistors with a 5 cm, AWG-30 copper wire (a thin wire, 0.255 mm in diameter). If we measure the actual resistance between two resistors using an ideal ohmmeter with ideal probes, what would it be?

To calculate its effect, using the formula above for cylindrical wire resistance is often a waste of time, alternatively, we can look up the AWG-30 wire's resistance per unit length from an engineering table on Wikipedia, it says the resistance is "338.6 mΩ/m". In other words, the additional resistance contributed by the wire is $$\ 0.3386 \Omega \times 0.05 \text{m} = 0.01693 \Omega \$$. Ideally, the resistance should be 2000 Ω, but due to the existence of a wire, the measured resistance is 2000.01693 Ω, it's less than 10 parts per million higher, nearly undetectable.

• Remark 3: In non-precision applications, a commonly used type of through-hole resistor is metal film resistor, 5% tolerance, with a temperature coefficient around 50-100 ppm for every 1 °C increase in temperature - the error introduced by the slightest change in temperature is still higher than your wire in this example.

• Remark 4: For even the best general-purpose multimeter, like a Fluke 87, the maximum resolution of resistance measurement is 0.1 Ω, so even measuring the 0.01693 Ω wire resistance is difficult.

Another example is a microcontroller development board, which may require a 5 V DC supply and 50 mA current on average to operate. If you use five meters of AWG-30 to hook the power (positive electrode) and ground (negative electrode), the total resistance is $$\ 0.3386 \Omega \times 5 \text{m} \times 2 = 3.386 \Omega \$$. Total voltage drop across the 5-meter power wire and the 5-meter ground wire, is $$\ 3.386 \Omega \times 0.05 \text{A} = 0.1693 \text{V} \$$. Actual voltage supplied to the microcontroller board is $$\ 5 \text{V} - 0.1693 \text{V} = 4.8307 \text{V} \$$, or 96.6% of the original voltage.

• Remark 5: A common voltage tolerance for digital electronics is +/- 5%.

If the power source itself is error-free, the drop caused by the wire still is well within the limit. Don't forget I used an extreme example here: 10 meters of extremely long and thin wires, which is not really a realistic scenario in most electronics experiments.

As you see, when using wires for interconnection, you can often ignore the wire resistance, and it's likely that you'll never see a mention about wire resistance in schematics. A similar situation occurs when you connect a cable through a socket, a connector, or a clamp - You'll also introduce additional contact resistance, but it's usually insignificant.

• Remark 6: In the industry, the allowed contact resistance introduced by a connector is often 1 Ω. For a high quality connector, sometimes a 0.1 Ω contact resistance is specified.

## When Wire Resistance Should Be Considered

But as the current delivered across a wire goes up, up to a point, you can no longer ignore the additional resistance from the wire. Again, due to Ohm's Law, it also happens when the absolute current is still small, but the resistance of other electrical components around the wire have decreased - it's just two sides of the same coin.

A high wire resistance has three harmful consequences:

1. The voltage drop $$\ V = IR \$$ across the wire becomes excessive and unacceptable, which moves the power supply voltage outside the range of specification. The device may stop working.

2. When the resistance of other electrical components are fairly low, the additional resistance of wire itself is simply too high to ignore.

3. The wire heats up by the current due to its resistance, and the "heater power" is $$\ P = I^{2} R \$$. This represents wasted power. If the wire resistance per unit length is too high, the wire cannot dissipate the heat quickly enough. Temperature will raise to a point when the wire becomes too hot and melt, creating a fire hazard.

## Low voltage DC-power distribution

A common example is power delivered by a USB port. The nominal voltage of USB is 5 V, regulated to +/- 5% as usual. USB 2.0 allows a "low power" device to consume 100 mA, while a "high power" device can receive 500 mA of current. If one use USB as a power source for a charger, the current requirement is even higher, 2000 mA is typical nowadays.

Let's say we have a 1-meter USB cable of questionable quality, which uses two AWG-28 wires (0.361 mm in diameter) for power and ground. Its resistance is 0.42 Ω, when carrying 500 mA of current, we lose 0.21 V due to the cable. To complicate the situation, because the USB power is regulated to +/- 5%, the lowest permissible voltage is, in fact, 4.75 V, the received voltage at the other end of the cable can be as low as 4.54 V - error is much greater than 5% already.

To overcome this problem, the USB 2.0 standard has an additional voltage drop budget for cables.

• The maximum voltage drop (for detachable cables) between the A-series plug and B-series plug on VBUS is 125 mV (VBUSD).

• The maximum voltage drop for all cables between upstream and downstream on GND is 125 mV (VGNDD).

• Functions drawing more than one unit load must operate with a 4.75 V minimum input voltage at the connector end of their upstream cables.

-- Universal Serial Bus Specification Revision 2.0

In other words, for any standard-compliant USB 2.0 high-power device, the manufacturer of this USB device either has to ship the product with a better cable with lower voltage drop, or has to design the device to work down to 4.5 V by any means necessary.

In this case, our device worked. A few days later, someone will find this USB cable and plugg it into a USB wall adapter to charge the smartphone at 2000 mA. Now the voltage drop across the cable is going to be 0.84 V, with only 4.16 V maximum available to the smartphone. The cable either won't work at all, or will charge the smartphone extremely slowly.

• Remark 7: Often in practice, some USB chargers will intentionally regulates the USB to 5.25 V to allow more voltage drop on the cables, even it's strictly a violation of the USB standard.

## Remote Sensing

Cable drop is also a trouble in voltage regulator design. While it's easy to use an adjustable regulator chip to make a power supply and regulate it to +/- 2% or even lower. Unfortunately, just like previous USB example, your regulation only occurs at the output pin of the regulator, not the load.

Source: Remote Sensing is Important for Your Power Supply, by Keysight, fair use.

Additional wire resistance degrades accuracy of a voltage regulator, especially when the load is far away from it, or when the current is high. Typically, one should take special care when laying out the output traces for the regulator: Keep it as short as possible on a PCB.

But the error can never be fully eliminated, especially when the designer has no control over if there's a long cable in between. When it's critical to accurate regulate voltage at the load, one can employ a technique called "remote sensing" to solve the problem. The basic idea is adding two additional wires to "monitor" the "real" voltage at the other side. If the regulator sees a voltage lower than expected, it'll increase its voltage further to overcome the drop.

Source: Remote Sensing is Important for Your Power Supply, by Keysight, fair use.

The remote sensing wires at +s and -s can have the same resistance like the power wires (same thickness), but they are not affected by the voltage drop. It's true even if they have a much higher resistance (thin wires).

One way to think about it, is considering the fact that high current is running through the power wires, producing a $$\ 10 A \times 0.015 \times 2 = 0.3 V\$$ drop, but the sensing wires are only here to transmit a small signal - there's little current running across the sensing wire, so it produces almost no voltage drop across the cable.

Another way is thinking the equivalent input resistance of +s and -s of the sensing input. Ideally, its input resistance should be infinite (i.e. no current goes in, an ideal voltmeter, as if nothing is connected). In practice, a resistance of 1 megaohm (1 MΩ, 1 million ohms) is a realistic expectation. So the equivalent circuit is a small resistor (the wires) connected in series with a huge resistor (the regulator sensing input).

For example, in this schematic, although the sensing wires have total resistance of 200 Ω, but the sensing input resistance is 1 MΩ, many order-of-magnitude higher. The voltage seen by the sensing input is,

$$V_\text{sensed} = 5 \text{V} \times \frac{1,000,000}{1,000,000 + 200}$$

The voltage drop exists, but it's only 0.02%, meanwhile, 99.98% of the voltage from the remote side is measured by the sensing input of the regulator.

## Four-Wire Resistance Measurements

Sometimes it's necessary to measure the resistance of an extremely small resistor (lower than 1 Ω) using an ohmmeter. Resistance of the wires connecting between test probes and your ohmmeter become significant. One solution is to short-circuit the test probes before making a measurement - zeroing out the error. But this requires an additional step, it also introduces an additional source of possible error: the pressure applied between the probes can affect the resistance used for calibration.

A common technique for solving the problem is Four-Wire Resistance Measurement, or Kelvin Measurement.

We can think the output pins of a ohmmeter as a current source and a voltmeter - the current source keeps its output voltage at whatever value it needs for a specific current. Then the output voltage of the current source is measured by the voltmeter. Both the current and voltage are known, so the resistance is determined.

Due to the fact that we are measuring voltage directly across the output terminals of the meter, it cannot distinguish resistance from the resistor-under-test and resistance from the test probes.

Adding two additional wires fixes the problem, we can now measure the voltage at the far end across the resistor-under-test, not the output of our ohmmeter at the near-end. Unaffected by the probe wires, we can make an accurate measurement. It's similar to remote sensing design in voltage regulators.

## Safety Considerations

This is the main consideration that dictates the wire size in utility power installation at homes. When a current passes through a resistor, not only a voltage drop is produced, but this voltage drop heats up the resistor as well. No matter whether the resistor is a resistor component or a wire, we must ensure the dissipated power $$\ P = I^{2} R \$$ does not exceed a maximum limit, otherwise the resistor will overheat.

If it's a wire, the wire can become dangerously hot and melt, creating a fire hazard. To find out the maximum current allowed to carry by a wire, first, the dissipated power in the wire is calculated, next, the flow of heat is identified - what is the ambient temperature of the environment, different materials have different thermal conductivity, etc. Finally, one determine a maximum operating temperature and use it to calculate the maximum permissible current, and finally a safety factor is included.

The actual calculation is fairly complex, and it also needs to follow the Electric Code with approval from regulatory agencies. Rather than calculating it from scratch, an engineering table is used. Again, the table on Wikipedia is a reference.

For example, at 20 °C ambient, a single, unbounded AWG-30 wire in a chassis of an appliance cannot carry more than 0.52 A of current in order to keep its operating temperature under 60 °C.

• Remark 8: If you're designing a product, you must use a reliable handbook with engineering tables calculated according to your local regulatory agency's standards.

The current-handling capacity of traces on a PCB can be found by referring to an engineering table or a calculation program as well.

## Application: Wire-Wound Resistor

Resistance of a wire is not always a nuisance, it has useful applications. Wire-wound resistor is a type of resistor made by winding a metal wire, usually nichrome for its resistivity on a core.

Source: Wirewound resistor, by ResistorGuide, fair use.

1. It's easy to produce highly accurate resistors, as its resistance is proportional to the length of a wire.

2. One can make high power resistors easily from a large wire.

It should be noted that a wirewound resistor has the same shape like an inductor, thus it has the highest inductance in all types of resistors. It should be only used in DC only, and audio-frequency circuit perhaps, but it's unsuitable for any AC circuits at a higher frequency.

## Application: Shunt Resistor

Voltage drop due to resistance of a wire is sometimes helpful as well. An easiest way to obtain current measurement is connecting a low-value shunt resistor in series and measure the voltage drop across it, since $$\ I = \frac{V}{R} \$$.

Using a high-value resistor stops sufficient current from being delivered to a circuit-under-test, it's desirable to make the shunt resistance as low as practical. There will still be a voltage drop, called burden voltage in a multimeter, but low enough to be acceptable.

If you open a multimeter, you'll find a shunt resistor similar to this picture. As you see, it's just a glorified piece of wire.

Source: Open Air Resistor - Metal Element Current Sense, by TT Electronics, fair use.

If high accuracy is not needed, you can make a free shunt resistor by drawing a trace on a circuit board - the wire (trace) itself is your shunt resistor.

Source: Low ohmic shunt resistor direct on PCB copper layer, fair use

• Also, since you are writing so much, may as well include skin effect for AC. Most of what you wrote ignores AC. Even at frequencies as low as 60 Hz, where the depth is $8.5\:\text{mm}$, it still matters. So when carrying something like $1000\:\text{A}$, one might consider something that's 2 inches or more in diameter. But that's six times or more beyond the skin depth. So the wiring is broken down as shown in Fig. 3 of US Patent 1904162 in order to equally distribute the current into the core as well as the surface. (Mostly, I'm teasing you. You've written a lot. And I appreciate it.) – jonk Jan 26 at 8:47
• @jonk Just fixed the wrong number. Well, if skin effect counts, perhaps a mention of wire inductance and its effect on decoupling can also be added on a later date. – 比尔盖子 Jan 26 at 8:53
• Good work. I +1, already. There's so much to write about wiring. You covered a lot. (That patent is interesting, though.) About connectors at the ends of wires? Now, that's a LONG subject that would take a book or two to cover well. – jonk Jan 26 at 8:55
• Thanks for taking the time to post this. The other answers are good, but this one really goes above and beyond! It's going to take a newbie such as myself a bit of time to read and understand all of this, so give me some time before I accept it. The worked basic examples are also a nice addition for someone such as myself, since I was particularly wondering how this is relevant for electronics work! – The Pointer Jan 26 at 11:08
• I have seen the trace resistance be a problem in lab / R&D shops where prototypes are debugged on extender boards. I had a case where a high current board wouldn't work on the extender. Turned out that there was enough drop across the extender that the voltage supervisor kept the thing in reset. – mpdonadio Jan 27 at 4:12

The resistance of a wire, (or more generally, the interconnect) comes into play at all scales of electrical design.

In commercial power distribution systems, conductor resistance causes some of the electrical power to be lost as heat. So the less resistance, the less power is loss. This is why in some limited application superconductors are being considered because they have zero, or near zero resistance.

At the other extreme, silicon integrated circuits used aluminum interconnects for traces on the silicon die. Then IC manufacturers, I think it was IBM, developed a method where they could use copper for the on-chip connections. The lower resistance of copper relative to aluminum allowed higher speeds on the chips.

In between those two extremes (think of server farms or a chassis of boards inside a radar system), delivering hundreds of amps of current from a power supply(ies) to it's various loads with minimal or low loss is a design challenge.

One more example. The Large Hadron Collider (LHC) in Europe uses superconducting magnets to steer the particles around the LHC ring. This is the only way they could provide the high currents the strong magnetic fields needed.

• Yes, it is a rookie mistake I see often. The engineer only considering the current carrying capacity and ignoring the voltage drop. – Mattman944 Jan 26 at 0:45

The text is a little bit sloppy but it is basically correct. There are two concerns when choosing a wire size. First is heating in the wire. If the wire will become hot and cause a burn hazard (or fire hazard) then you must use a larger wire. For further reading use the search term "ampacity table."

Second is voltage drop. This is more likely to be a problem in lower voltage applications, for two reasons. First, if I lose 1 Volt in a mains application, it is no big deal. I may get 119V instead of 120, or 229V instead of 230V. No big deal.

But if I have a 12V battery feeding an inverter, I cannot afford to lose 1V out of 12V in the wire, because it may cause the inverter to shut down prematurely, and because that is a much greater loss on a percent basis.

Low cost USB cables sometimes give rise to excessive voltage drop and may cause problems for devices which attempt to charge at higher currents such as 1.5A or 2.1A.

So voltage drop is likely to be the limiting factor in low voltage power applications. And wire heating is likely to be the limiting factor in mains voltage applications.

Logic or data signals on PCB's will seldom run into voltage drop or overheating problems in normal use. But it may be necessary to consider power loss and trace heating on electronic PCB's if power circuitry is involved.

• Crappy USB charger cables is also a place where normal consumers may have run into this problem when trying to charge modern smartphones or drive a Raspberry Pi. – pipe Jan 26 at 0:36
• I believe that if one translates "copper" into Chinese and then back to English, it will come back as "aluminum". – Sredni Vashtar Jan 26 at 16:34

How common is it to consider the resistance of a wire? Is the resistance of a wire an important consideration when doing electronics work? Does this resistance need to be factored into calculations, as would other components (such as resistors)?

I see many good answers here but I want to add a few extra points not mentioned yet.

Ideal circuits do not exist - that's why they're called ideal. But, once we've got the key goal done with our circuit (say, power conversion or NAND two signals), we can get as many complications into account as you wish; wire resistance is definitely one of them. Though, as others have pointed out, this is not often the case, because it is negligible or still made it as such with the considerations read in the book (to begin with).

But there're many other things to take into account when you're in a certain domain of EE. Let's consider low-resistance sensing e.g. when $$\R_x<1\Omega\$$ - this happens more often than you might think because e.g. a copper rod is in that a range; a length of wire; a conducting bar with very low resistivity like gold.

When you tie up two components anyhow you want (breadboard, PCB, whatever) a contact resistance forms between the two terminals, so two terminals you get two contact resistances. Where you had a 'node' in your circuit, that is a zero resistance point, it no longer is. In general contact resistances are in the order of $$\m\Omega\$$ - your resistance could be so low that when you sense it with your instrumentation, in actuality you're only measuring contact resistances or still making a measurement with very high uncertainty. Contact resistances cannot be measured and are more like random variables and depend on temperature.

But there's more. When you tie up two metals of different electronegativity, a potential difference in between is established. This is called the Volta effect. So you'd have a potential increase or drop at every node, essentially.

Taking into account both of these, can you picture how much your original circuit would change?

And finally there's noise, which is a real thing and can be measured. It can often be ignored because usually SNR is high - but that's not always the case. Noise can effectively be averaged out (but there's still an alternative component) through multiple measurements. You have to consider, moreover, that your circuit is effectively an antenna so it'll catch EM waves - that's another noise contribution, and there are many more.

TL;DR In general, not often. It largely depends on your domain. But what you should always be with you is that it definitely is a real thing and may have to be taken into account.

• "When you tie up two metals of different electronegativity, a potential difference in between is established." - Is this ever relevant, outside of thermocouples and solid state physics? The total difference caused by this in every circuit is zero, because you have to get back to the original metal as you go around the circuit. – user253751 Jan 28 at 11:13
• @user253751: The net current is zero but the potential difference is not. That's what happens in a unbiased p-n junction, to name one. Such voltage may be very small, but still alter your voltmeter readings. When measuring very low resistances though, even such small deviation can matter; that's why there's a particular procedure for these scenarios, which precisely relies on switching current polarity. – edmz Jan 28 at 19:42
• When you measure across a disconnected diode with your voltmeter, it reads zero volts. Why? – user253751 Jan 29 at 10:07
• @user253751: There'd be a few things to say and perhaps you may want to open a question to expand on; but, in very short terms, the potential difference is across the junction, where the electric field exists and very large currents occur, not across the diode. The field is very weak there, almost null and thus way below your voltmeter's resolution. In fact when the diode's I-V curve is derived, one assumes no voltage drop far from the junction (which is in the range of microns in width). – edmz Jan 29 at 12:49
• which is my point: all these potential differences necessarily cancel out when you have a loop. So outside of thermocouples (where they do not cancel because the junctions are at different temperatures) and solid state physics (where you are analyzing behaviour close to the junction) what is the relevance? – user253751 Jan 29 at 12:51

Real devices have non-trivial sample variations in their properties. Circuit designers must allow for these variations. Sample variation can overwhelm certain other considerations.

Consider that commercially available resistors are offered in various tolerance ranges; 10% being one of them. If a designer chooses to employ a 1k Ohm 10% resistor in a given circuit, the resistance of an actual device installed into a real circuit might be anywhere from 900 to 1100 Ohms. If the wire resistance is at most a few milliOhms, it is inconsequential when compared to the allowed sample variation in the resistor value.

On the other hand, if a circuit requires a 10 Ohm 0.1% resistor, the resistance of the wire and contact between wire and device might need to be considered.

When designing for digital circuits and audio analog circuits, wire resistance can generally be ignored because component values make it inconsequential.

When it comes to conductor shape, effects of high frequency or fast switching are usually a larger concern than DC resistance; inductance, ringing, RF coupling, reflections at sharp corners and skin effect, for example.

1. In our product range we connect devices running over distances of upto 300 meters. Hence, when calculating the minimum vorlage requirement compliance for a power receiving device at the remote end.
2. we have established a definition for the cable and we also consider it in our calculations.
3. When there is a need for higher wattage or longer distances, the cable resistance is surely considered.
4. Another example is power track on a PCB, a poorly drawn power net had a drop of about 0.35 V across it in one of the design I had reviewed.

You can add two more ingenious techniques for eliminating the wire resistance to the physical idea of superconductivity and circuit idea of remote sensing above. I have illustrated the written with two pictures from another source dedicated to the same subject. They show the voltage distribution along a real conductor with line resistance Rl.

The first idea - negative resistance, is quite simple and intuitive: to compensate the wire resistance Rl that "creates" a voltage drop I.Rl, insert somewhere in the line a negative "resistor" with resistance -Rl. It will create voltage I.Rl that neutralizes the voltage drop; the result is zero wire resistance (Rl - Rl = 0). This idea has been used for a long time in phone repeaters. The advantage is the compensator is a 2-terminal element, which can be inserted anywhere along the line (in the picture below it is combined with the input voltage source thus making it a source with negative internal resistance). The disadvantage is that it compensates only specific resistance (and only resistance).

The second idea is to drive the load, when possible, by a current source instead, as usual, by a voltage source. Then, if for some reason, the wire acquires some resistance Rl that "creates" a voltage drop I.Rl, the current source will increase its internal voltage with I.Rl and will neutralize the voltage drop. Current interfaces exploit this idea to transmit data over long distances.

In summary, all artificial techniques neutralizing the wire resistance do the same - they insert voltage equivalent to the voltage drop across the line.

Finally, I would like to say a few words about the book of Practical Electronics for Inventors. I bought it in 2000 hoping to find valuable tips on how to understand and invent circuits... but I was disappointed. Although it is useful for a wide range of readers, it is still not a book for inventors... it is rather a book for conventional technicians. If it really was a book for inventors, it would show the ideas behind circuit solutions, clever tricks behind them, their evolution, philosophy... not just describe their structure and operation...

There are a lot of good answers here. Practical cases where wire length and xsectional area are important:

Measuring very small resistances. Using 4-wire measurements removes the resistance of the wires from the measurement.

When connecting solar cells and solar panels together, the resistance of the connecting tinned copper ribbons affects the efficiency of the system.

I'll just add a few additional examples of why this is important and actually taken into consideration pretty much all the time.

• The first is not really electronics, but actually very high-voltage electricity. There's a reason long-distance (and most importantly high power) electricity transmission networks (you know, the big huge towers carrying cables) use very high voltages (measures in hundreds of thousands of volts): that allows to carry the same amount of power using much lower currents. Carrying 1 MW (one megawatt) at 230 V means over 4000 amperes. Carrying the same 1 MW at 400 000 V means 2.5 amperes.

Source: Wikipedia

Since the Joule effect $$\P=R.I^2\$$ means the power dissipated over the power is proportional to the square of the intensity, that's 2.5 million less power lost over the same cables (which would have melted down anyway at the lower voltage).

It's also proportional to the resistance, and thus the wire gauge, which is why they regularly use wire gauges up to... $$\750 mm^2\$$!

Source: Wikipedia

Note that 1 MW is just peanuts, those power lines often carry thousands of MW!

• On a similar note, if you consider rail networks, you'll see that voltages range from a few hundred volts (usually DC) to tens of thousands (usually AC). The lower voltages (often between 300 V and 750 V) are more often used in urban (metro)/suburban networks. They're easier to work with (in terms of safety etc.), but they require "substations" (to convert from the long-distance very high voltage networks quoted above to the voltage used to power the trains) at relatively short intervals.

On the other hand, the higher voltages (usually 15 or 25 kV) are often used for long distance railways, especially high-speed ones given the required power. Those allow substations to be placed at much larger intervals.

There are quite a few long distance railways with voltages "in the middle of the range" (750 V to 3000 V DC) for historical reasons, but new deployments most often use the higher voltages.

• Another example is Power over Ethernet (PoE): using either "phantom power" over the data pairs or using the spare pairs directly, power is "injected" into the Ethernet cable so the device can be powered and connected to the network with a single cable. Ethernet uses quite tiny wires, so their resistance is high, and that means there is a substantial voltage drop between source and powered device.

For that voltage drop to remain acceptable in percentage, while still keeping relatively safe and easy to work with voltage, the PSE (the supplying equipment) supplies between 44 V and 57 V, while the PD (the powered device) will receive anywhere between 37 V and 57 V, depending on the voltage at the PSE and the length of the cable.

Even though most PDs internally use voltages anywhere between 3.3 V and 12 V, if PoE used voltages in that range directly, there would be nothing left (for 3.3 V or 5 V) or definitely not enough (for 12 V) at the end of a 100 m cable run.

• Yet another situation where it matters and which was just alluded to in the other answers: PCB traces. Those work like wires, with their resistance being linked to their width. If you have a lot of current going through a trace (e.g. between a power supply and USB ports that need to be able to provide high currents), and in addition to that it's relatively long, you can't expect to have that run through a standard fine trace.

If you look carefully at existing PCBs, you'll note that there are lots of very fine traces, but a few are quite larger. There's a reason for that!

• @ElliotAlderson Fixed. – jcaron Jan 28 at 15:14