Voltage Drop: Why It Matters and How to Calculate It

The 30-second version

Wire has resistance. Current through resistance drops voltage (E = I × R) and makes heat. Too much drop means dim lights, hot motors, nuisance tripping, and equipment that dies young. The NEC's informational notes recommend limiting drop to 3% on a branch circuit or feeder, 5% total from service to the last outlet.

Try the voltage drop calculator to run the numbers instantly.

The formulas

Single-phase:

VD = (2 × K × I × D) / CM

Three-phase:

VD = (1.732 × K × I × D) / CM

Where:

• VD = voltage drop (volts)
• K = resistivity constant — approximately 12.9 for copper, 21.2 for aluminum (ohms-cmil/ft at 75°C)
• I = load current (amps)
• D = one-way circuit length (feet)
• CM = conductor area in circular mils (Chapter 9, Table 8)

Percent drop = VD ÷ source voltage × 100.

Worked example

120 V circuit, 16 A load, 150 ft one-way, #12 copper (6,530 CM):

VD = (2 × 12.9 × 16 × 150) ÷ 6,530 = 9.5 volts — about 7.9%. Way over the recommendation.

Same load on #8 copper (16,510 CM): VD = 3.75 V = 3.1%. That's the difference wire size makes on a long run: two sizes, not one, is the usual answer.

Field rules of thumb

• Every wire size up cuts drop roughly in half per two sizes (CM roughly doubles every two AWG sizes).
• On 120 V, drop hurts twice as fast as on 240 V for the same load and distance — one more reason to run 240 V to outbuildings.
• Size for the actual load, but think about what the circuit will feed in ten years. Nobody ever complained the wire was too big.
• Aluminum needs roughly two sizes larger than copper for the same drop.

Common mistakes

Using round-trip distance in the formula (the 2 and 1.732 already account for it — D is one-way), forgetting that the drop percentage applies at the load's operating voltage, and ignoring the starting current on motor loads at the end of long runs.