11.4 Voltage Drop, Conductor Resistance & Trade Math

11.4 Voltage Drop, Conductor Resistance & Trade Math

Exam Focus: The two voltage-drop formulas, the K constants, Chapter 9 Table 8, solving for circular mils, and the offset/bending arithmetic that appears under general trade knowledge.

Voltage drop is where theory and the code book meet. The formula is memory work; the circular mil area comes out of Chapter 9, Table 8, which you are allowed to have in front of you.

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The Two Formulas

Where:

• K = resistivity constant in ohm-circular mils per foot: 12.9 for copper, 21.2 for aluminum (at 75 °C)
• I = load current in amperes
• D = one-way circuit length in feet — the 2 (or 1.732) already accounts for the return path
• CM = conductor circular mil area from Chapter 9, Table 8

$\%VD = \frac{VD}{E_{source}} \times 100$

[!CAUTION] The most common mistake is doubling the distance twice. If a panel is 150 ft from the load, D = 150. The factor of 2 in the numerator is the return conductor. Entering 300 ft and keeping the 2 doubles your answer.

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Circular Mil Areas You Should Recognize

Values are read from NEC Chapter 9, Table 8, Conductor Properties, which lists size, stranding, overall area in circular mils and square inches, and direct-current resistance per 1,000 ft for coated and uncoated copper and for aluminum.

Two useful patterns: each three AWG sizes roughly doubles the circular mil area, and above 4/0 the sizes are expressed directly in thousands of circular mils (250 kcmil = 250,000 CM).

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Worked Example — Finding the Drop

A 120 V single-phase circuit carries 16 A to a load 100 ft from the panel using 12 AWG copper.

1. $CM = 6{,}530$, $K = 12.9$
2. $VD = \dfrac{2 \times 12.9 \times 16 \times 100}{6{,}530} = \dfrac{41{,}280}{6{,}530} = 6.32\ \text{V}$
3. $%VD = \dfrac{6.32}{120} \times 100 = 5.27%$

That exceeds the 3% branch-circuit guidance, so upsize:

• 10 AWG (10,380 CM): $VD = 41{,}280/10{,}380 = 3.98\ \text{V} = 3.3%$ — still over
• 8 AWG (16,510 CM): $VD = 41{,}280/16{,}510 = 2.50\ \text{V} = 2.08%$ — acceptable

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Worked Example — Solving for Conductor Size

Rearranging for circular mils is faster than guess-and-check:

$CM = \frac{2 \times K \times I \times D}{VD_{allowed}}$

Problem: a 240 V single-phase feeder carries 60 A a distance of 250 ft in copper. Hold the drop to 3%.

1. Allowable drop: $240 \times 0.03 = 7.2\ \text{V}$
2. $CM = \dfrac{2 \times 12.9 \times 60 \times 250}{7.2} = \dfrac{387{,}000}{7.2} = 53{,}750\ \text{CM}$
3. From Table 8, 4 AWG is 41,740 CM (too small) and 3 AWG is 52,620 CM (still short), so 2 AWG at 66,360 CM is the first size that satisfies the target.

Remember: the ampacity requirement and the voltage-drop requirement are two separate tests. Size for ampacity first using Table 310.16 and 110.14(C), then check voltage drop and upsize if needed. The larger of the two results wins.

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Three-Phase Example

A 208 V three-phase feeder carries 100 A a distance of 200 ft in 1/0 AWG copper.

$VD = \frac{1.732 \times 12.9 \times 100 \times 200}{105{,}600} = \frac{446{,}856}{105{,}600} = 4.23\ \text{V}$

$\%VD = \frac{4.23}{208} \times 100 = 2.03\% \quad \text{— acceptable}$

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Why Voltage Drop Matters Even Though It Is Advisory

The NEC's 3% and 5% figures sit in informational notes, which 90.5(C) makes non-enforceable. The physics is not advisory:

• Resistive loads: power falls with the square of voltage. A 5% drop costs about 10% of the heat output.
• Motors: a motor running at reduced voltage draws more current to deliver the same mechanical output, raising winding temperature and shortening insulation life.
• Electronic ballasts and drivers: may drop out or run hot at sustained low voltage.

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Trade Math Under General Electrical Trade Knowledge

Conduit Offsets

The travel of an offset is found from the offset height and the bend angle multiplier:

Example: a 6 in offset using 30° bends. Distance between bend marks = $6 \times 2.0 = 12\ \text{in}$, and the run shrinks by $6 \times 1/4 = 1.5\ \text{in}$.

Other Trade Constants Worth Memorizing

• 1 hp = 746 W
• 1 kW = 1.34 hp
• 1 kWh = 1,000 watts drawn for one hour
• Right triangle: $c = \sqrt{a^2 + b^2}$ — used for saddle bends and diagonal runs
• Circle area = $\pi r^2$ — the basis of every conduit-fill calculation in Chapter 9