Understanding conductor size and route length

What you should be able to do

Use actual conductor length and applicable resistance-per-length data to predict wire-loop resistance in a simple paired DC circuit.

Core Idea

Resistance increases with length. For the same material and temperature, a larger conducting cross-sectional area reduces resistance. Insulation thickness and overall cable diameter do not by themselves establish conductor area. In AWG sizes, a smaller gauge number generally denotes a larger conductor. Confirm the cable's specification rather than choosing by appearance. Construction, material and temperature also matter.

Read The Four Pictures

The copper cross sections illustrate area, without assigning an AWG size. The cable route illustrates detours rather than a straight line between endpoints. The two-core cable reminds you to include outgoing and return paths. The documentation represents the applicable manufacturer's conductor resistance data. A caliper or visual comparison does not replace that data.

Worked through

A fictional installation sketch specifies this one-way route: Horizontal run: 110 ft. Rises and drops: 20 ft total. Detour around an obstruction: 10 ft additional. Specified service-loop allowance: 10 ft total. Actual one-way conductor route: 150 ft. For two equal conductors following that same route, the combined conductor length is 300 ft. This is 150 ft of two-core cable containing 300 conductor-feet; it is not a direction to order 300 ft of that cable.

Resistance And Drop

The exercise supplies a hypothetical resistance of 0.006 Ω/ft for each conductor under the stated conditions. It is not presented as a universal value for any AWG size. R wire loop = 2 × 150 ft × 0.006 Ω/ft = 1.8 Ω. At the specified 0.50 A: V drop = 0.50 A × 1.8 Ω = 0.90 V. Using only the 110 ft horizontal run would produce 1.32 Ω and 0.66 V drop, understating the actual route's loss.

Compare Two Options On Paper

Option A uses the original hypothetical conductor data: 0.006 Ω/ft. Option B has verified worksheet data of 0.004 Ω/ft at the same conditions. With the same 150 ft one-way route: A: R loop = 1.8 Ω; drop at 0.50 A = 0.90 V. B: R loop = 1.2 Ω; drop at 0.50 A = 0.60 V. This comparison identifies a lower-resistance option. It does not establish that Option B meets every terminal, cable-listing or system requirement.

Unit Check

If the applicable data instead reads 6 Ω per 1,000 ft per conductor: 6 ÷ 1,000 = 0.006 Ω/ft. Use the units consistently. A value already expressed as loop resistance for a cable pair must not be doubled again. If outgoing and return conductors differ, calculate each resistance separately and add them rather than applying the equal-wire shortcut.

Practice With Answers

1. Double the main example's one-way route to 300 ft, keeping everything else fixed.

Answer: R loop = 3.6 Ω and drop = 1.80 V at 0.50 A.

1. Does twice the overall cable diameter guarantee half the resistance?

Answer: No; conductor material and conducting area must be identified.

1. Should the spare coil left installed be ignored?

Answer: No; installed conductor length contributes resistance.

1. Is the shortest straight line on the floor plan sufficient?

Answer: No; account for the actual approved route.

1. Does an acceptable voltage-drop calculation prove cable suitability?

Answer: No.

Application Limits

Check equipment terminal capacity, wiring method, environmental suitability, listing and all applicable system requirements separately. Data transmission limits are also separate from a simple DC voltage-drop result. The worksheet assumes one load and two equal conductors carrying the same current. Distributed loads, shared returns or parallel conductor arrangements require the actual circuit model. Do not replace required route planning with an arbitrary extra percentage. Record the measured or specified route allowances clearly.

Where beginners go wrong

Voice Recap

Use the actual route, the actual conductor data and consistent units. In our example, 150 feet one way means 300 conductor-feet in the paired circuit. At 0.006 ohm per foot, the wire loop is 1.8 ohms and loses 0.9 volt at half an amp.

Sources