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# Wire length tables

## Introduction

Describe wire length, voltage loss, etc. here.

## Tables

The following table shows the American Wire Gauge (AWG) size for several different round trip (RT) wire lengths based on maximum current (A) for a 12V DC system (assuming 3% maximum wire loss):

Table 1. Wire Sizes
RT length 5A 10A 15A 20A 25A 30A 40A 50A
15 ft 16 12 10 10 8 8 6 6
20 ft 14 12 10 8 8 6 6 4
25 ft 14 10 8 8 6 6 4 4
30 ft 12 10 8 6 6 4 4 2
40 ft 12 8 6 6 4 4 2 2
50 ft 10 8 6 4 4 2 2 1
60 ft 10 6 4 4 2 2 1 1/0
70 ft 8 6 4 2 2 2 1/0 2/0
80 ft 8 6 4 2 2 1 1/0 2/0
90 ft 8 4 2 2 1 1/0 2/0 3/0

This table was adapted from Engineering Toolbox  .

## Calculating Wire Sizes

1. copper wire resistivity = 11.2 
2. 3% loss from wires
3. allowable voltage drop = 0.36V
• 12V system, 3% (0.03) wire loss -> 12V x 0.03 = 0.36V

Variable Declarations:

• An: circular mil area
• n: AWG
• m: AWG
• Conductor resistivity: constant, 11.2
• Current: current through wire in Amps, acquired from table
• Roundtrip (RT) wire length: complete length of wire, ft, acquired from table
• Allowable voltage drop: voltage drop through wires, V
Table 2. Calculating Wire Sizes
Eqn. # Equation Step/Explanation
#1 $A_n=(5*92^\frac{36-n}{39})^2$ Eqn. 1 was used, rearranged, and iterated on to produce Table 1. .
#2 $n=36-\frac{39*log\sqrt{\frac{A_n}{5}}}{log(92)}$ Algebraically rearrange Eqn. 1 to solve for n. The resulting equation is Eqn. 2 to the left.
#3 $n=1-m$ If n<0, replace n with 1-m in Eqn. 1  .
#4 $A_n=(5*92^\frac{36-(1-m)}{39})^2$ Substituting n equals 1-m into Eqn. 1 yields this equation (Eqn. 3). Use Eqn. 3 if n<0.
#5 $m=\frac{39*log\frac{\sqrt{A_n}}{5}}{log(92)}-35$ Solve Eqn. 3 for m, the resulting equation (Eqn. 4) is to the left.
#6 $A_n=\frac{(conductor-resistivity)*(current[A])*)roundtrip-wire-length[ft])}{allowable-voltage-drop[V]}$ Eqn. 3 is also a rearrangement of Eqn. 1. Use Eqn. 4 to solve for A_n

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