Worked example
A 20 A load at the end of a 100 ft run of 12 AWG copper on a 120 V circuit:
- VD = 2 × 12.9 × 20 × 100 ÷ 6,530 = 7.90 V.
- 7.90 ÷ 120 = 6.6%, over the 3% guideline.
- Moving up to 10 AWG (10,380 circular mils) gives 4.97 V, or 4.1%.
Circular mil areas
| Size | Circular mils |
|---|---|
| 1 AWG | 83,690 |
| 2 AWG | 66,360 |
| 3 AWG | 52,620 |
| 4 AWG | 41,740 |
| 6 AWG | 26,240 |
| 8 AWG | 16,510 |
| 10 AWG | 10,380 |
| 12 AWG | 6,530 |
| 14 AWG | 4,110 |
| 1/0 AWG | 105,600 |
| 2/0 AWG | 133,100 |
| 3/0 AWG | 167,800 |
| 4/0 AWG | 211,600 |
Try 5 electrician exam practice questions
A 100-amp subpanel is located 300 feet from the main panel. The load is 80 amps at 240 volts single-phase. Using 2/0 AWG copper conductors, what is the approximate voltage drop?
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D. 1.9% Single-phase voltage drop = 2 × K × I × D ÷ CM, with K = 12.9 for copper, I = 80 A, D = 300 ft and 2/0 AWG = 133,100 circular mils: 2 × 12.9 × 80 × 300 ÷ 133,100 ≈ 4.65 volts. 4.65 ÷ 240 = 1.9%, within the 3% recommended for feeders in the informational note to 215.2.
A 40-amp, 240-volt branch circuit feeds a wall oven. What is the minimum conductor size for copper conductors with 75°C insulation?
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A. 8 AWG Table 310.16: 8 AWG copper is rated 50 amps at 75°C, which is sufficient for a 40-amp circuit; 10 AWG (35 A at 75°C) is too small. 6 AWG would work but is not the minimum. The 60°C column (40 A for 8 AWG) only governs where terminations are not rated 75°C.
A dwelling unit has a 240-volt, 12-kW electric range. Using the standard calculation method, what is the demand load for the range?
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C. 8.0 kW Table 220.55, Column C: for one household range rated not over 12 kW, the standard-method demand load is 8 kW. The 12 kW nameplate is not used directly, and Note 1 (5% per kW over 12 kW) does not apply to a 12 kW range.
A kitchen dishwasher is connected to a dedicated branch circuit. The manufacturer specifies a 15-amp breaker. Can the electrician install a 20-amp overcurrent device instead?
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C. No, the manufacturer's listed overcurrent protection must be followed Manufacturer listing requirements take precedence. The NEC requires that equipment be installed per its listing, which includes the specified overcurrent protection rating.
A small commercial office has 120 general-use receptacles at 180 VA each, 40 fluorescent luminaires at 150 VA each, and one 208V, 3-phase air-conditioning unit with a nameplate of 7,000 VA. What is the calculated service load using the standard method?
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C. 28.8 kVA Receptacles: 120 × 180 VA = 21,600 VA; Table 220.44 allows the first 10 kVA at 100% and the remainder at 50%, so 10,000 + 5,800 = 15,800 VA. Lighting: 40 × 150 VA = 6,000 VA at 100%. Air conditioning: 7,000 VA at 100%. Total = 15,800 + 6,000 + 7,000 = 28,800 VA, or 28.8 kVA.
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Common questions
For a single-phase circuit, VD = 2 × K × I × L ÷ CM, where K is the conductor's resistivity (about 12.9 for copper and 21.2 for aluminum), I is the current in amps, L is the one-way length in feet and CM is the wire's area in circular mils. For three phase, replace the 2 with 1.732.
The NEC suggests, in informational notes, no more than 3% on a branch circuit and 5% in total for the feeder and branch together. Informational notes are advice rather than enforceable rules, though some codes (energy codes, for example) and specifications make them mandatory.
One way. The 2 in the single-phase formula already accounts for the current flowing out and back.
Aluminum has higher resistivity than copper, so the same size carries more voltage loss. That is why aluminum runs are usually one or two sizes larger.
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Sources, checked 2026-10-29: NFPA 70, National Electrical Code (2023): Chapter 9, Table 8 (circular mil areas), and the informational notes on voltage drop for branch circuits (210.19) and feeders (215.2). K values of 12.9 (copper) and 21.2 (aluminum) are the commonly taught approximations at 75°C. QuizCram is not affiliated with the agencies or standards bodies named here.
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