Electrical power is distributed from the source of power to the load using electrical conductors. There are two principal concerns for the conductors:

- the voltage drop occurring due to resistance in the conductors, and
- the heating of the conductor occurring due to the voltage drop and current flowing in the conductor.

The first concern, voltage drop, will be examined in detail in this article, and a simple method to assess the requirement for conductor size to limit voltage drop will be presented. A separate article (to be written) will examine conductor heating.

The general goal in distribution of electric power is to avoid loss of power in the conductors that distribute the power as much as possible. Since conductors will have some resistance, voltage drop in the conductor is inevitable, but a reasonable goal is to limit the voltage drop to not more than three-percent of the system voltage being distributed.

The voltage drop in a conductor is proportional to the current flowing in the conductor times the resistance of the conductor. The resistance in a conductor is proportional to its length. Unfortunately, the resulting voltage drop is not dependent on the original voltage, so the lower the original voltage--the system voltage--the more critical the need to limit voltage drop becomes.

In the case of a nominal "12-Volt" battery DC system, the system voltage can be seen as being 13.2-Volts, the lead-acid battery terminal voltage at full-charge. Three-percent of 13.2-Volts is 0.396-Volts. In a well-designed system of 13.2-Volt DC power distribution, the overall voltage drop from the source of power to a load consuming that power should be limited to not more than 0.396-Volts. That figure will be used as the allowed voltage drop in the following discussion.

The resistance of a wire for a certain unit of length is a function of the material used for the wire--here we assume pure copper--and its cross section area. For a conductor made of pure copper with a cross section area according to the American Wire Gauge (AWG standard) we can find the resistance per 1,000-feet of conductor as listed below:

TABLE 1 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 4.016 18 6.385

To get the Ohms-per-foot the values in Table 1 must be divided by 1,000. That will result in very small values, more easily handled in scientific notation:

TABLE 2
AWG Ohms-per-foot
00 0.779 x 10^{-4}
0 0.983 x 10^{-4}
1 1.239 x 10^{-4}
2 1.563 x 10^{-4}
4 2.485 x 10^{-4}
6 3.951 x 10^{-4}
8 6.282 x 10^{-4}
10 9.989 x 10^{-4}
12 1.588 x 10^{-3}
14 2.535 x 10^{-3}
16 4.016 x 10^{-3}
18 6.385 x 10^{-3}

When a current flows in a wire, the voltage drop in the conductor will be equal to the current in Amperes times the resistance of the wire in Ohms. We can say this mathematically as:

(1) Volts = Amperes x Ohms

This is the well know property of electricity called *Ohm's Law.*

For wire of a particular cross-section, we can compute the resistance by its length L_{feet} and its resistance Ohms_{per-foot}. The wire resistance can now be stated as:

(2) Ohms = L_{feet} x Ohms_{per-1-foot}

Substituting the expression for Ohms from (2) into (1) gives:

(3) Volts = Amperes x L_{feet} x Ohms_{per-1-foot}

Because we have only one relationship and four variables, we cannot solve equation (3) for just one variable. We can rearrange to solve for Ampere-Feet, an unusual unit but a very useful one (as will be demonstrated):

(4) Amperes x L_{feet} = Volts / Ohms_{per-1-foot}

Equation (4) can now be used with inputs for the value for a particular resistance-per-foot (using the values for AWG standard gauge pure copper conductors) and a particular voltage drop to calculate result in Ampere-feet. For example, we use the resistance-per-foot of 10-AWG conductor from Table 2 (9.989 x 10^{-4} Ohms-per-foot), and a voltage drop of 0.396-Volt (three-percent of 13.2-Volts) to compute the value of Ampere-feet for 10-AWG conductor:

(6) Ampere-feet = 0.396 / 9.989 x 10^{-4}
(6) Ampere-feet = 396.4

In the same manner a value in Ampere-feet for a 0.396-Volt drop can be precomputed for all the AWG wire gauges in Table 2:

TABLE 3 AWG conductor values for Ampere-feet for 0.396-Volt drop: 00 5,083.4-Ampere-feet 0 4,028.5-Ampere-feet 1 3,196.1-Ampere-feet 2 2,533.6-Ampere-feet 4 1,593.6-Ampere-feet 6 1,002.2-Ampere-feet 8 630.4-Ampere-feet 10 396.4-Ampere-feet 12 249.4-Ampere-feet 14 156.8-Ampere-feet 16 98.6-Ampere-feet 18 62.0-Ampere-feet

The calculated values in Table 3 for are not in units of feet, they are in units of Ampere-feet. This is very handy because for any value of current in Amperes, the value for Ampere-feet can be divided by the amount of current to get the physical length of the conductor in feet. For example, if a 10-AWG conductor is to distribute a current of 30-Amperes, the length that causes the 0.396-Volt drop is then:

(7) Length = 396.4-Ampere-feet / 30-Amperes = 13.2-feet

The same method can be used to solve for the maximum current allowed (to limit the voltage drop) for a particular length. For example if a 10-AWG conductor must be 20-feet long, the maximum current before the voltage drop will exceed the limit will be

(8) Current = 396.4-Ampere-feet / 20-Feet = 19.8-Amperes

Table 3 is a very useful collection of data, but it only accounts for one conductor. Direct current electrical power is generally distributed by two conductors, a positive and a negative circuit. Typically the two conductors will be of the same wire gauge. To account for two conductors, the values for length that will result from using the Ampere-feet data must be considered for the total length of the two conductors added together. In terms of the distance between the power source and the load, the length value obtained must be reduced by half to account for the total length of the two conductors.

A simpler method, perhaps, is just to begin with the Ampere-feet value reduced by half; the distance calculated will then account for two conductors. To that end, a new table with half the values from Table 3 is calculated below in Table 4.

In Table 4 (below) are calculated values in Ampere-feet that can be used to quickly assess either the maximum current or the maximum length of a two-conductors circuit distributing 13.2-Volt (nominal "12-Volt" DC power) that will produce a three-percent voltage drop in the conductors:

TABLE 4 AWG conductor values in Ampere-feet for a 13.2-Volt system voltage with two conductors in the circuit and three-percent voltage drop: 00 2,541.7-feet-Ampere 0 2,014.2-feet-Ampere 1 1,598.0-feet-Amperes 2 1,266.8-fee-Amperes 4 796.8-feet-Amperes 6 501.1-feet-Amperes 8 315.2-feet/Amperes 10 396.4-feet/Amperes 12 198.2-feet/Amperes 14 78.4-feet/Amperes 16 49.3-feet/Amperes 18 31.0-feet/Amperes

Table 4 above is very handy for quickly calculating the maximum length of a particular two-conductor circuit for a given current, or for calculating the maximum current for a given length. Here are two examples; in each case the result calculated is specific to a system of 13.2-Volt power that tolerates no more than a three-percent drop in the conductors:

Q: What is the maximum length for 6-AWG wire in a two-conductor circuit to carry 40-Amperes?

A: From the table above we get the Ampere-feet value for 6-AWG: 501.1. Then we divide that value by 40-Amperes, to get the length in feet:

(9) 501.1-Ampere-feet / 40-Amperes = 12.53-feet

Q: What is the maximum current that 6-AWG wire can for carry 20-feet in a two-conductor circuit?

A: From the table above we get the Ampere-feet value for 6-AWG: 501.1. Then we divide that value by 20-feet, to get the current in Amperes:

(9) 501.1-feet/Amperes / 20-feet = 25.0-Amperes

Table 4 can also be used to select a conductor of appropriate Ampere-feet for a specific combination of distance and current. For example:

Q: What is the smallest AWG size for a two-conductor circuit than can carry 40-Amperes a distance of 15-feet?

A: The two values, Feet and Amperes, are multiplied to get a necessary minimum value of Ampere-feet:

(9) 40-Amperes x 15-feet = 600-Amperes-feet

Table 4 is then inspected to find the AWG gauge with at least a 600-Ampere-feet value; in this case a 4-AWG conductor will be suitable.

I have never seen the method of calculating either length or current maxima for power distribution presented in this manner. The concept of pre-computing a value of Ampere-feet for standard AWG copper conductors that is shown here is my own method of solution to a common problem.

Copyright © 2021 by James W. Hebert. Unauthorized reproduction prohibited!

Author: James W. Hebert

This article first appeared April 3, 2021.

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Author: James W. Hebert