2.3 Voltage Drop Calculations

Voltage drop questions sit at the intersection of theory and Code. NEC 2023 does not make a hard 3% or 5% limit a mandatory requirement for most branch circuits; those figures appear in Informational Notes (for example, near 210.19(A) and 215.2(A)) as fine-print recommendations for efficiency and equipment performance. Arkansas exam writers still expect you to calculate drop and sometimes to compare your result to the recommended percentages.

Why Electricians Care

Long runs to barns, shops, outbuildings, and large residences are common in Arkansas work. Excess drop causes dim lights, sluggish motors, and nuisance issues long before ampacity is violated. The Prov General Knowledge area tests whether you can quantify that drop.

The Approximate Formulas

Trade schools and most open-book exams use these forms (D = one-way circuit length in feet; CM = conductor area in circular mils):

Single-phase (or DC):

VD = (2 × K × I × D) / CM

Three-phase:

VD = (1.732 × K × I × D) / CM

The factor 2 accounts for the outgoing and return conductors in a single-phase circuit. Three-phase uses √3 (1.732) with one-way length.

K Values

Some references use 11–12.9 for copper depending on temperature assumption. Unless the stem gives another constant, 12.9 Cu / 21.2 Al is the expected exam pair.

Circular Mils — Chapter 9 Table 8

You need CM from Chapter 9, Table 8 (Conductor Properties) in NEC 2023. Memorize the common sizes; look up the rest during the open-book exam:

(Values match Table 8; confirm on exam day if an item hinges on an exact digit.)

Worked Example — Single-Phase Copper Drop

A 20 A load is supplied by 12 AWG copper over a 150 ft one-way run on a 120 V circuit. Find voltage drop and percent drop.

VD = (2 × 12.9 × 20 × 150) / 6,530

= (77,400) / 6,530 ≈ 11.9 V

%VD = (11.9 / 120) × 100 ≈ 9.9%

That far exceeds the recommended 3% branch guidance. Upsize the conductor or shorten the run for good practice — even though ampacity of 12 AWG copper might still be acceptable for a 20 A breaker under other rules.

Worked Example — Meeting 3% on 240 V

Same 20 A load, 240 V, still 150 ft one-way. What minimum copper CM keeps drop at or below 3%?

Allowed VD = 0.03 × 240 = 7.2 V

Rearrange: CM = (2 × K × I × D) / VD

CM = (2 × 12.9 × 20 × 150) / 7.2 = 77,400 / 7.2 = 10,750 CM

10 AWG is 10,380 CM (slightly under); 8 AWG at 16,510 CM clears the target. Exam answers often ask for the next standard size that satisfies the calculation.

Worked Example — Three-Phase Aluminum Feeder

A 60 A continuous-duty load (use 60 A in the VD formula when the stem gives load current) on a 208 V three-phase feeder, 200 ft one-way, 1/0 AWG aluminum.

VD = (1.732 × 21.2 × 60 × 200) / 105,600

= (440,064) / 105,600 ≈ 4.17 V

%VD = (4.17 / 208) × 100 ≈ 2.0% — within a 3% feeder recommendation.

Alternate Method — Using Table 8 AC Resistance

Some problems give or expect use of ohms per 1,000 ft from Table 8:

VD ≈ I × R_path, where R_path for single-phase ≈ 2 × (ohms/1,000 ft) × (D/1,000)

Example: 10 AWG Cu uncoated, about 1.21 Ω/1,000 ft (confirm Table 8 value on exam), 100 ft one-way, 15 A, 120 V:

R_loop ≈ 2 × 1.21 × 0.1 = 0.242 Ω

VD ≈ 15 × 0.242 ≈ 3.63 V → about 3.0%

K-method and Table 8 resistance method should be close; use whichever the question's data supports.

Feeder + Branch Combined Guidance

Informational Notes commonly frame:

• Feeder: aim ≤ 3%
• Branch: aim ≤ 3%
• Feeder + branch together: aim ≤ 5%

If a question gives separate feeder and branch drops of 2.5% and 2.0%, combined 4.5% meets the 5% note even though each piece is under 3%.

Exam Strategy for Arkansas Prov Items

1. Identify single-phase vs three-phase (watch for 208Y/120 wording).
2. Confirm copper vs aluminum for K.
3. Use one-way feet for D — do not double the distance yourself when the formula already has the 2 or 1.732 factor.
4. Convert VD to percent before comparing to 3% / 5%.
5. When sizing up, compute required CM, then pick the next Table 8 size that meets or exceeds it.

Worked Example — Choosing Between Copper and Aluminum

Suppose a 40 A single-phase load, 180 ft one-way, 240 V, target 3% (allowed VD = 7.2 V).

Required CM = (2 × K × I × D) / VD.

Copper (K = 12.9): CM = (2 × 12.9 × 40 × 180) / 7.2 = 185,760 / 7.2 = 25,800 CM → 6 AWG Cu (26,240 CM) meets it.

Aluminum (K = 21.2): CM = (2 × 21.2 × 40 × 180) / 7.2 = 305,280 / 7.2 = 42,400 CM → 4 AWG Al (41,740 CM) is barely under; 3 AWG or 2 AWG aluminum is the safe exam pick if 4 AWG is not listed as acceptable. Higher K for aluminum is why Al feeders often jump one or two sizes versus copper for the same drop target.

Voltage-drop math is a reliable point bank on the nine-item General Knowledge slice if you keep the formulas on a scratch sheet at the start of the exam.