In distribution of electrical power, there are two concerns in choosing the size of the conductors that will carry the electrical current:

- the rating of the conductor for current capacity, and
- the voltage drop that will occur in the conductor.

In systems where the distributed voltage is low, the concern for the voltage drop tends to limit the length of the conductor that can be used. This article examines this relationship in more detail. If you want to skip the technical background that explains the basis for my methods, you can jump right to the final section that present a very simple table-based method.

The resistance of a wire for a certain unit of length is a function of the material used for the wire and its cross section area. For a conductor made of copper and sized according to the American Wire Gauge (AWG standard) we can find the resistance per 1,000-feet of conductor from a table. (See http://www.powerstream.com/Wire_Size.htm for a good listing of this data.) If we use a wire of 10-AWG, we see that 1,000-feet of 10-AWG wire will have a resistance of 1-Ohm. This value is really a physical property of copper, so there is not much variance seen in listings of the resistance of copper wire.

The current carrying capacity or current rating is a value that is more difficult to define with precision. As current flows through a conductor it encounters resistance, which results in a power loss. The lost power is converted to heat. The wire becomes warmer. The amount of temperature increase of the wire above the ambient temperature becomes a factor in rating the maximum current. If the conductor is a single conductor and is in free air space, much more heating can be tolerated. If the conductor is in a bundle with other conductors and enclosed in a conduit with little cooling air, the maximum current allowed will be lower. The type of insulation also affects the current rating. Expensive insulation material like Teflon can tolerate higher temperatures. The result of the many variables affecting wire temperature is the maximum current rating is more variable. For our example conductor of 10-AWG, we see the maximum current for a single conductor in free air is 55-Amperes. For other situations, such as power distribution, the maximum current is derated to only 15-Amperes.

When a current flows in a wire, the voltage drop in the conductor will be equal to the resistance of the wire in Ohms times the current in the wire in Amperes. For wire of a particular gauge, we can compute the resistance by the length of the wire and the resistance of that wire gauge per 1,000-feet. We can say

(1) VoltageDrop = Length-in-Feet x (Resistance/1,000-feet) x Current

Typically in power distribution a limit on the voltage drop to a certain percentage of the system voltage is required. A limit of three-percent is often used. We can set this as our maximum allowed VoltageDrop by

VoltageDrop = SystemVoltage x 0.03

In the case of a 12-Volt system, the result is

VoltageDrop = 0.36-Volts

Now we can calculate the Length-in-Feet of a particular conductor that will create a voltage drop of 0.36. We solve equation (1) for Length-in-Feet:

(2) Length-in-Feet = VoltageDrop / [(Resistance/1,000-feet) x Current]

This rearranges to

(3) Length-in-Feet = VoltageDrop x 1000 / (Resistance x Current)

Now we explore a 10-AWG conductor at its maximum rated current of 55-Amperes in a 12-Volt system

Length-in-Feet = 0.36-Volt / [1-Ohm/1,000-feet) x 55-Ampere]

Length-in-Feet = 6.54-feet

Because this is the total length of the conductors, in a two-way circuit the maximum distance that a 10-AWG conductor could carry its rated 55-Amperes before the voltage drop in a 12-Volt system exceeded 3-percent is only 3.25-feet.

Now we explore a current of only 15-Amperes, which is the rated current for 10-AWG when used in power distribution (according to the table mentioned above). (See the footnotes in the table for the criteria used to obtain this rating.) Recalculating the Length-in-Feet for the 10-AWG conductor before we reach the maximum voltage drop of 3-percent in a 12-Volt system, we have

Length-in-Feet = (0.36-Volt x 1000) / (15-Ampere x 1-Ohm) = 24-feet

Again, for a two-conductor circuit, this distance value is divided in two (because the conductor length will be twice the distance) and we see that a 10-AWG conductor can be used to distribute 15-Amperes in a 12-Volt system for only a maximum distance of 12-feet.

Let's examine a typical situation in a small boat with power distribution from a battery at the stern to a secondary power distribution center at the helm, a distance of about 15-feet. A conductor of 8-AWG is used. The resistance per 1,000-feet for 8-AWG is 0.6282-Ohms. The rated maximum current in power distribution is 24-Amperes. We find the maximum length from

Length-in-Feet = (0.36-Volt x 1000-feet/ (24-Ampere x 0.6282-Ohm)

Length-in-Feet =23.9-feet

For a two-conductor circuit, the maximum length is half that or about 12-feet. Let's try a 4-AWG conductor: the resistance per 1,000-feet is 0.2485 and the maximum current is 60-Amperes. The maximum length for 3-percent drop at 12-Volts is

Length-in-Feet = (0.36-Volt x 1000-feet) / (60-Ampere x 0.0.2485-Ohm)

Length-in-Feet = 24.1-feet

For a two-conductor circuit, the maximum length is half that or about 12-feet.

I see a nice rule of thumb developing here. The maximum length of a conductor that can be used for two-conductor power distribution is equal to the system voltage if the current is to be at the maxim rated current of that conductor for power distribution. In other words, in a 12-Volt system of power distribution, the maximum distance for power distribution at the rated current is 12-feet. Let's test this new rule. We'll use a system voltage of 24-Volts. A 3-percent drop at 24-Volts is 0.72-Volts. Resolving the 4-AWG case above, finds the length to be

Length-in-Feet = (0.72-Volt x 1000-feet/ (60-Ampere x 0.0.2485-Ohm)

Length-in-Feet = 48-feet

For a two-conductor circuit, the maximum length is half that or about 24-feet.

Perfect! The rule of thumb works.

For a system voltage of 12-Volts, and for any distance longer than 12-feet, you must use a larger wire size and derate the current. We can rework the relationship in equation(1) to solve for resistance per 1,000-feet:

(3) Resistance-per-1000-feet = (VoltageDrop x 1000)/(Feet x Amperes)

If we want to distribute 12-Volt power with a 3-percent voltage drop at a distance of 40-feet with a current of 30-amperes, the Resistance-per-1000-feet must be equal or less than

Resistance-per-1000-feet = (0.36 x 1000)/(40 x 30) = 0.3-Ohm

We go to the table and find the conductor size must be 4-AWG (0.2485-Ohm) to meet this requirement. The 40-feet we calculated is twice the two-conductor distance, so the maximum two-conductor distance will be 20-feet. In other words, if 12-Volt power is to be distributed a two-conductor circuit distance of 20-feet and a current capacity of 30-Amperes is desired, the wire used must be 4-AWG in order to maintain the voltage drop at not more than 3-percent.

I have never seen the rule of thumb I proposed above mentioned before. Therefore, I am calling it "Hebert's Rule." The rule says:

In a two-conductor power distribution system where a 3-percent voltage drop is the maximum tolerated, the maximum distance in feet is equal to the system voltage in Volts at the conductor's rated current capacity for power distribution on the basis of 1-Ampere per 700 circular mills of copper conductor.

The corollary rule: in 12-Volt power distribution, the maximum length is 12-feet at the conductor's rated current. This rule is based on the current rating of the wire being calculated on the basis of 1-Ampere per 700 circular mills of copper conductor.

Perhaps I need to demonstrate Hebert's Rule with another example. Let's investigate with a 6-AWG wire.

Wire of 6-AWG has a resistance of 0.3951-Ohms/1,000-Feet, and it is rated to distribute a current of 37-Amperes. According to Hebert's Rule, the length (in feet) of this conductor (in a two-conductor circuit) that will produce a three-percent voltage drop will be equal to the system voltage. For this example we use a system voltage of 48-Volts.

According to Hebert's Rule the maximum length of 6-AWG that can be used to distribute 37-Amperes with less than three-percent voltage drop will be a two-condcutor distance of 48-feet. A two-conductor circuit of 48-feet will have 96-feet of wire. The resistance of 96-feet of 6-AWG conductor will be

(0.3951-Ohms/1000-feet) x 96-feet = 0.0379296-Ohm

If a current of 37-Amperes flows, the voltage drop will be

0.0379296-Ohm x 37-Ampere = 1.40-Volts

A three-percent drop in a 48-Volt system will be

48 x 0.03 = 1.44-Volts

which gives us a margin of 0.04-Volts, or precisely what we required. Hebert's Rule has successfully predicted the maximum length that a 6-AWG conductor can be used to distribute 48-Volts at a current of 37-Ampere: 48-feet (i.e., the system voltage).

Let me repeat Hebert's Rule with slightly more elaboration:

The maximum length (in feet) of a conductor that can be used for two-conductor power distribution is equal to the system voltage (in Volts) if the current is to be at the maxim rated current of that conductor for power distribution as figured according to the 1-Ampere/700-circular mils rating system.

What makes the rule of thumb work is the method used to rate the current capacity of the copper conductor. The copper conductor's current carrying capacity must be rated according to the cross-section area of the copper wire at a ratio of 1-Ampere per every 700-mils of circular cross section. The rated current carrying capacity of conductors according to the 1-Ampere/700-circular mills calculation is as follows:

00-AWG = 190-Amperes 0-AWG = 150-Amperes 1-AWG = 119-Amperes 2-AWG = 94-Amperes 3-AWG = 75-Amperes 4-AWG = 60-Amperes 5-AWG = 47-Amperes 6-AWG = 37-Amperes 7-AWG = 30-Amperes 8-AWG = 24-Amperes 9-AWG = 19-Amperes 10-AWG = 15-Amperes 12-AWG = 9.3-Amperes 14-AWG = 5.9-Amperes 16-AWG = 3.7-Amperes

In a 12-Volt system, the maximum distance those conductors can carry their rated current is 12-feet (or 24-feet of total conductor length) if less than a three-percent voltage drop is desired.

Applying Hebert's Rule to boat electrical power with a system voltage of 12-Volts, we see that the maximum distance that power can be distributed in a two-conductor circuit using a conductor of a particular AWG rating will be 12-feet if the circuit is to carry the current for that AWG rating according to the 1-Ampere/700-circular mils calculation.

Because the voltage drop is a linear function, we can quickly approximate what conductor size we need to use if the distance is longer than 12-feet by looking for the current rating of the next larger gauges and choosing a wire by current rating that is proportional to the distance desired for runs longer than 12-feet.

For example, if 12-Volt power is to be distributed over a distance of 24-feet, we have a ratio of distance of 2:1. We must select the conductor size in the same proportion. If the current to be distributed is to be 15-Amperes, we look at the wire table and find the conductor rated for 30-Amperes. I demonstrate with another example:

To distribute 12-Volts at a distance of 24-feet with a current of 15-Amperes, we must use a conductor rated a 30-amperes. Looking at the wire current table, we see that 7-AWG is rated for 30-Amperes. Now we check our rule. The resistance of 7-AWG is 0.4982-Ohm/1000-feet. An out-and-back distribution of 24-feet will mean 48-feet of wire. The resistance will be

48-feet x 0.4982-Ohm/1000-feet = 0.0239136-Ohm

When a current of 15-Amperes flow in this resistance, the voltage drop will be

0.0239136 x 15 = 0.36-Volt

In a 12-Volt system, a three-percent drop is 12 x 0.03 = 0.36-Volt. Hebert's Rule and its quick length estimation corollary have precisely predicted the proper wire size to use for maintaining the voltage drop at a three-percent tolerance.

You can also easily estimate the distance for a higher voltage drop by multiplying the 12-foot distance (that is, the system voltage) by the proportional increase in voltage drop compared to 3-percent. For example, if a 10-percent voltage drop is allowed, we can multiply the distance by a factor of 10/3 = 3.3-times. That means for the purpose of distributing current with a tolerance for a 10-percent voltage drop, the conductor can be up to 12 x 3.3 = 40-feet long for the rated current.

As an example, if a electrical windlass at the boat can tolerate a 10-percent voltage drop, and the motor typically draws 90-Amperes, we can use 2-AWG conductors up to a distance of 40-feet.

Let's check. A conductor of 2-AWG has a resistance of 0.1563-Ohms/1000-feet. For 80-feet of conductor the resistance will be

80 × 0.1563-Ohms/1000-feet = 0.012504-Ohms

With a current of 90-Amperes, the voltage drop will be

90 × 0.012504 = 1.125-Volts

In a 12-Volt system, a 10-percent voltage drop is 1.2-Volts; we are below that and meet the requirement of the design.

For currents other than the rated current from the table at distances other than the system voltage distance, you can easily interpolate the wire capacity since it is a linear relationship:

Voltage drop limit reached at = current × length

Here the dimension length is really a resistance; the conductor has a certain resistance per foot, so the length of the conductor is its resistance. If we know that a certain conductor can carry a certain current for a certain length, we know the voltage drop limit. Normally we would measure voltage in units of Volts, but for the interpolation we measure it in terms of Ampere-feet. This sounds more confusing than it is. Let me illustrate with an example.

For a conductor of 8-AWG we know from the table and from the concept of the system-voltage length, that the current capacity is 24-Amperes. Since the system voltage is 12-Volts, the length is 12-feet (for a two-conductor circuit). That means the maximum voltage drop limit is going to occur at 24-Amperes times 12-feet or 288-Ampere-feet. If we want to carry more current over a shorter distance, we can select a new current value and then divide the Ampere-feet to find the new distance. For example, if we need the circuit to carry 30-Amperes, how long can it be with 8-AWG? Here is the method:

For 8-AWG Max Voltage Drop = 12-feet × 24-Amperes = 288-Ampere-feet Find maximum length for 30-Ampere: Length = 288-Ampere-feet / 30-Ampere = 9.6-feet

We can also extrapolate to a longer distance with lower current. For example, if we use 8-AWG and only have 18-Amperes in the circuit, how long can the circuit be?

Find length for 18-Amperes Length = 288-Ampere-feet / 18-Ampere = 16-feet

Working with this data in this manner is much simpler than looking up resistance in tables and calculating the actual voltage drop in an iteritive manner as you vary the length.

Here is another very handy table that I have calcuated. It lists the "Ampere-feet" value for various wire gauges that are calculated for a system voltage of 13.2-Volts and a maximum voltage drop of three-percent. With this table you can quickly determine the length of the conductor from the current or vice versa at which a voltage drop of three-percent will occur for a 13.2-Volt power distribution system. The length is the physical span for a two-conductor circuit, so that the total length of wire in the circuit would be twice as much. Since DC power distribution is always a two-conductor circuit, the length parameter is the distance between ends of the circuit, not the total wire of the circuit.

A USEFUL TABLE FOR COMPUTING WIRE SIZE in 13.2-Volt power distribution AWG Ampere-feet 16 = 49.3 14 = 78.4 12 = 124.7 10 = 198.2 08 = 315.2 06 = 501.1 04 = 796.8 02 = 1266.8 01 = 1598.1 0 = 2014.2 00 = 2541.7

For example, to determine a length of conductor:

**Q:** How far can an 8-AWG conductor distribute 15-Amperes of power?

**A:** From above table for 8-AWG, Ampere-feet = 315.2. Thus at 15-Amperes the distance is 315.2 / 15 = 21-feet.

Another example, this time to determine a maximum current of a conductor at a particular length:

**Q:** How much current can be carried by 4-AWG conductor for 18-feet of power distribution?

**A:** From above table, Ampere-feet = 796.8. Thus at 18-feet the current is 796.8 / 18 = 44.3-Amperes

There is one caution: if figuring a maximum current for a short length, if the length in feet is less than the system voltage in volts, you should not exceed the rated maximum current capacity of the wire.

This table is also extremely handy to use for computing conductor size needed for a particular length and current of power distribution. Find the wire gauge necessary by calculating the Ampere-feet; enter the table to find the wire gauge with at least that much capacity. For example:

**Q:** What is a suitable wire gauge to distribute 30-Amperes a distance of 20-feet?

**A:** Multiply 30-Ampere x 20-feet to get 600-Ampere feet. Enter table to find suitable wire with 600-Ampere-feet or more capacity; 4-AWG has 796.8-Ampere-feet, so it can handle this current and length.

The table is based on the resistance per foot value of the conductors for copper wire. Note that the values in the table are ONLY applicable to a 13.2-Volt system. I chose that voltage because that is the nominal charging voltage of a so-called 12-Volt battery. If you wanted to use a different system voltage, you can scale the Ampere-feet values up or down in proportion. If you wanted to calculate on the basis of 12.0-Volts, you would scale the Ampere-feet down by a ratio of 12.0/13.2 = 0.909. If you wanted to calculate on the basis of 24.0-Volts, you would scale the Ampere-feet up by a ratio of 24/13.2 = 1.818. For convenience, I have calculated two more tables, one for 12-Volts as a system voltage and another for 24-Volts:

If you want to be very conservative about power distribution from a 12-Volt battery, you can use 12.0-Volts as the system voltage. Most 12-Volt batteries are actually about 60-percent discharged when their terminal voltage reaches 12.0-Volts, but computing the wire size for power distribution with this figure will tend to limit the voltage drop when you may need it most.

A USEFUL TABLE FOR COMPUTING WIRE SIZE in 12.0-Volt power distribution AWG Ampere-feet 16 = 44.8 14 = 71.3 12 = 113.4 10 = 180.2 08 = 286.5 06 = 455.6 04 = 724.3 02 = 1151.6 01 = 1452.8 0 = 1831.1 00 = 2310.7

The table for 24-Volts demonstrates an important concept: the higher the system voltage, the farther a conductor of a particular gauge can be used to carry current for power distribution (or the more current it can carry for the same distance). This is because a three-percent voltage drop in a 24-Volt system is twice as much voltage as in a 12-Volt system. The voltage drop in the wire is completely independent of the system voltage, as it is a function only of wire resistance and current.

A USEFUL TABLE FOR COMPUTING WIRE SIZE in 24.0-Volt power distribution AWG Ampere-feet 16 = 89.6 14 = 142.6 12 = 226.7 10 = 360.4 08 = 573.1 06 = 911.2 04 = 1448.7 02 = 2303.3 01 = 2905.6 0 = 3662.3 00 = 4621.3

The cross-section area of a circular conductor is measured in a somewhat unusual manner. Because a wire might be made from a number of individual strands, it is not always possible to simply measure the wire's cross-section area as if it were a circle. Instead, wire cross-section is measured in units of circular mils. One circular mil is the area of a circle with a diameter of 1/1000th of an inch, which is then

Area = π × d^{2}
One Circular Mil = 3.1415 × (0.001-inch)^{2}
One Circular Mil = 0.00000314159265-inch^{2}

A conductor of size AWG-10 has a nominal diameter of 0.1019-inch. We calculate its circular mils from

Area = π × d^{2}
Area = π × (0.1019-inch)^{2}
Area = 0.032621072893742-inch^{2}
Equivalent Circular Mils = 0.032621072893742-inch^{2} / 0.00000314159265-inch^{2}
Equivalent Circular Mils =10383.6

One basis for specifying the current carrying capacity of the conductor is to use a rule-of-thumb and allot 700-circular-mils per Ampere. Thus a conductor of 10-AWG would be rated for

Amperes = 10383.6-circular-mills/700-mills-per-Ampere Amperes = 14.8

Here is table of conductor AWG size, diameter, circular mils, calculated Amperes, and rated Amperes

AMPERES AWG Diam. Mils Cal. Rated 1 0.2893 83690 119.5 119 2 0.2576 66360 94.8 94 4 0.2043 41740 59.6 60 6 0.1620 26240 37.5 37 8 0.1285 16510 23.6 24 10 0.1019 10380 14.8 15

The actual maximum current carrying capacity of any conductor must be more carefully calculated to allow for consideration of voltage drop, heating, wire insulation type, ambient temperature, and availability of cooling air. For example a conductor strung individually in open air, as might occur in utility power distribution along pole, and operating at very high system voltage could carry much more current for power distribution than a conductor bundled with other similar conductors, enclosed in a conduit, and carry a very low system voltage.

When choosing a conductor for power distribution in a particular application, there may be specific regulations or recommendations that should be followed. For example, in boat electrical construction compliance with the recommendations of the American Boat and Yacht Council (ABYC) may be necessary. The notion of allotting 1-Ampere for each 700-mils of cross-section in a copper conductor is just a conservative approximation for the current carrying capacity.

When a current flow through a resistance a voltage drop is created. The formula is very simple:

E = I × R where E = Volts I = Amperes R = Ohms

For wire made from pure copper the resistance is well-known and is provided in Ohms-per-1000-feet:

AWG Ohms-per-1000-feet 00 0.0779 0 0.0983 1 0.1239 2 0.1563 4 0.2485 6 0.3951 8 0.6282 10 0.9989 12 1.588 14 3.535 16 6.016 18 6.385

To find the resistance in a particular length, we can just multiply by the distance in feet divided by 1000. To find the voltage drop we multiply by the current in Amperes. For example, to find the resistance of a 6-AWG conductor that is 15-feet long:

0.3951-Ohms/1000-feet × 10-feet = 0.003591-Ohms

To find the voltage drop that occurs if if a current of 10-Amperes flows in that conductor:

E = I × R E = 10 × 0.003591 E = 0.03591-Volts

We have examined how the cross-section area of a conductor sets an upper limit on the current carrying capacity. We have also seen that current carrying capacity will be limited by tne influence of voltage drop in the conductor. We can summarize:

- conductor cross-section in mils sets a maximum current limit
- conductor resistance-pre-foot seta voltage drop limit at that maximum current

We can combine these two to find the voltage drop, E_{drop-at-max-rated-current}, per foot at maximum rated current. As an example, I use 6-AWG.

Example: 6-AWG conductor
I = 3.7 × 10^{1}-Amperes (from cross-section area)
R/feet = 3.951 × 10^{−4} (from resistance of copper)
E_{drop-at-max-rated-current}/feet = I × R/feet
E_{drop-at-max-rated-current}/feet = 1.462 × 10^{−3} Volts/1-feet

If we want the voltage drop to be not more than three-percent of the System Voltage, E_{sys}, then the distance, *d, *at which the voltage drop limits the current will be when:

E_{drop-at-max-rated-current}/feet × *d* = 0.03 × E_{sys}

Solving that relationship for *d* we find:

*d* = 0.03 × E_{sys} / E_{drop-at-max-rated-current}/feet

For the general case where E_{sys} = 1 this simplifies to:

*d* = 0.03-Volts / E_{drop-at-max-rated-current}/feet

Now we substitute the value for E_{drop-at-max-rated-current}/feet for 6-AWG and solve at E_{sys} = 1 to find the distance that a 6-AWG conductor can carry its rated current as a function of the system voltage:

*d* = 0.03-Volts / 1.462 × 10^{−3} Volts/1-feet
*d* = 2-feet

What we have found is a 6-AWG conductor can distribute its rated current of 37-Amperes with less than 3-precent voltage drop in the conductor for a distance of 2-feet for each Volt of the System Voltage being distributed. If the System Voltage being distributed is 120-Volts, then 6-AWG can carry 37-Amperes for 240-feet before a three-precent voltage drop occurs, but if the System Voltage is 12-Volts, 6-AWG can only be used for 24-feet. If the System Voltage is being distributed on two conductors, then the distance will be half. This further simplifies the relationship; the rated current can only be carried for 1-foot distance per 1-Volt of System Voltage.

Looking at the value E_{drop-at-max-rated-current} for a number of conductors of different size, we can see that there is a near constant relationship for the voltage drop at a particular distance (in this case, 1,000-feet):

AWG Amps R E_{drop-at-max-rated-current}
1 119 0.1239 14.47
2 94 0.1563 14.69
4 60 0.2485 14.91
6 37 0.3951 14.62
8 24 0.6282 15.10
10 15 0.9989 14.98
Where
Amps = rated maximum current in Amperes
R = resistance in 1000-feet of the conductor

The table above reveals that the calculations done for 6-AWG really apply to all the conductors, as their rated maximum current in Amperes and their resistance per foot in Ohm are in constant proportion. This means that the value for distance that was found to be 2-feet-per-1-Volt-of-System-Voltage applies to all of these conductors. This relationship is what I have named "Hebert's Rule" earlier in this article. We can now see its derivation, based on the properties of copper conductors.

Any time the length of the conductor is less that the limit imposed by the calculation of the voltage drop, the conductor current should be limited to the maxium rated current. To demonstrate, we consider a 12-Volt system voltage. We know that the conductors can be used up to 12-feet at the rated current. This really means two things: for distances less than 12-feet the conductor should not be used for more than its rated current; and for distances greater than 12-feet the conductor current should be de-rated by calculating the voltage drop so as to be less than the usual three-percent maximum desired.

Copyright © 2015 by James W. Hebert. Unauthorized reproduction prohibited!

Author: James W. Hebert

This article first appeared May 11, 2013 in a previous form and was moved to this form and location on September 12, 2015.

The section on using a table of Ampere-feet was added March 13, 2016, the result of an epiphany that awakened me at 5 a.m.

The addendum was added January 16, 2018.

The CCS was edited to improve math presentation January 17, 2018; please clear browser cache to apply new style sheet if math equations cannot be read.

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Author: James W. Hebert