Wire Size Selection - Trade Offs Between Keeping it Cool and Getting Power to the End of the Wire

Oct 12, 2026

Wire Size Selection explains how to estimate electrical resistance of AWG copper conductors and evaluate resulting voltage drop and power loss as heat. It provides practical resistance approximations for common wire gauges, shows example calculations for a 20AWG loop carrying 7 amps over 15 feet, and interprets the impact on system efficiency. It also connects current-caused temperature rise to insulation limits, discusses derating and a 5% wiring-loss rule of thumb, and demonstrates how upsizing can reduce both voltage drop and temperature rise.

LanguageEnglish

CategoryGeneral

FormatPDF

Pages2

Reading Time1 min

AccessFree to download

Wire Resistance Approximation

Quick AWG-to-resistance rules

Voltage Drop and Energy Loss

Example: 7A through 20AWG over 15 feet

Temperature Rise and Insulation Limits

Free-air vs bundled wiring

Sizing Strategy and Derating

5% loss rule of thumb and upsizing calculations

Path length formula for max allowable run

Bob Nuckolls AeroElectric Connection 6936 Bainbridge Road Wichita, KS 67226-1006 Voice/Fax: 316-685-8617 [E-mail: nuckolls@aeroelectric.com](E-mail: nuckolls@aeroelectric.com) Wire Size Selection Trade Offs Between Keeping it Cool and Getting Power to the End of the Wire . . . . If the material from which wires are made had zero resistance, then any size wire could carry any amount of current! Consider the formula for power where watts = volts x amps. This relationship works whether the watts are used to do something useful (light up a bulb) or worthless (warm up a wire) . First a little background: As a useful rule of thumb, remember that 10AWG wire has a resistance of about 1 milliohm per foot. Every time you step three AWG sizes, you double/half the wire's cross section. So, it follows that 13AWG wire is 2 milliohms/foot, 16AWG is 4 milliohms/foot, 19AWG is 8 milliohms/foot and 22AWG is 16 milliohms/foot. In the other direction, 7AWG is a half milliohm; 4AWG is a quarter milliohm and 1AWG is one-eighth milliohms/ foot. An excellent estimate on intermediate sizes is calculable; just do linear interpolation. For example, 20AWG is about 1/3rd of the way between 8 and 16 milliohms/foot. So, take 1/3 of the difference (8 / 3 = 2.6) or 2.6 milliohms per wire step. Take the 19AWG value of 8 milliohms; add 2.6 milliohms to yield approx 10.6 milliohms/foot. A check of a REAL wire table sez 20AWG is is 10.2 milliohms per foot . . . Not bad for a quick approximation. By memorizing the 1 milliohm/foot value for 10AWG along with the 3AWG steps for doubling/halving resistance allows you to do fast, estimates on wire resistance. Now, knowing the resistance, you can calculate the LOSSES in any particular wire. Suppose we draw a 7-amp load through a piece of 20AWG in a composite airplane and roundtrip from bus bar to load and back to ground is 15 feet. 15 feet times .0102 ohms is .153 ohms. Ohms law sez Volts = Amps x Ohms so the voltage drop in this hunk of wire is 7 x.159 or about 1.113 volts. Hmmmm 1.114 volts/14 .0 volts shows that 8% of the energy intended for the device at the other end of the wire isn't getting there. In the table below, you can look up the resistance of various AWG sizes of copper wire, the current ratings for the wire assuming 35EC and 10EC rise, and feet of wire run for 5% loss at the each temperature-limited current rating. | Wire Data Table | | | | | | | --- | --- | --- | --- | --- | --- | | AWG | Milli
Ohms/
Foot | 35EC Rise | | 10EC Rise | | | | | Amps | Max
Feet | Amps | Max
Feet | | 2 | 0.156 | 100 | 44 | 54 | 83 | | 4 | .249 | 72 | 39 | 40 | 70 | | 6 | .395 | 54 | 33 | 30 | 59 | | 8 | .628 | 40 | 28 | 20 | 55 | | 10 | 1.00 | 30 | 23 | 15 | 47 | | 12 | 1.59 | 20 | 22 | 12.5 | 35 | | 14 | 2.53 | 15 | 18 | 10 | 28 | | 16 | 4.01 | 12.5 | 14 | 7 | 25 | | 18 | 6.39 | 10 | 11 | 5 | 22 | | 20 | 10.2 | 7 | 10 | 4 | 17 | | 22 | 16.1 | 5 | 7 | 3 | 15 | We know further that watts = volts times amps. So a 1.113 volts drop in the wire at a 7 amp load is 7.8 watts. This power is lost to the air and surrounding mass as HEAT. © 1996, 1997, 1998 Bob Nuckolls, Wichita, [Kansas. E-mail: nuckolls@aeroelectric.com. This document may be reproduced mechanically](Kansas. E-mail: nuckolls@aeroelectric.com. This document may be reproduced mechanically)[ ](Kansas. E-mail: nuckolls@aeroelectric.com. This document may be reproduced mechanically)or electronically for non-profit, educational purposes when produced in its entirety and without modification. Referring to the wire table note that 20AWG wire will suffer a 35 degree C temperature rise when loaded with 7 amps. This is a free-air figure. Suppose the wire is buried in a wire bundle? Closed up where air cannot circulate freely around the wire, 7 amps will cause it to get much hotter. Okay, let's take the free air rise and say we're going to run this wire through the tailcone where we expect to see a hot-day soak up to 65EC. With a 35

FAQs

How can you quickly approximate copper wire resistance for common AWG sizes?

Memorize that 10AWG is about 1 milliohm per foot, then step three AWG sizes to roughly double or halve cross-sectional area (and resistance). Intermediate sizes can be estimated using linear interpolation between the bounding milliohm/foot values.

How do you evaluate voltage drop losses for a given wire run and current?

Use the resistance to compute the voltage drop via Ohm’s law (volts = amps × ohms), where the total ohms comes from resistance per foot times the wire length. The resulting voltage drop indicates how much power is not delivered to the load.

How does temperature rise influence wire sizing decisions?

Current through the conductor causes temperature rise, which must be considered relative to insulation limits. Free-air rise is typically lower than bundled or enclosed conditions, so higher temperatures can increase insulation risk even if the copper conductor itself tolerates the heat.

Oct 12, 2026

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