How much voltage will you lose?

How do I calculate voltage drop?

For a single-phase circuit, voltage drop = 2 × K × I × L ÷ CM, where K is 12.9 for copper (21.2 for aluminum), I is the current, L is the one-way length in feet and CM is the wire's area in circular mils. For 20 A over 100 feet of 12 AWG copper (6,530 circular mils): 2 × 12.9 × 20 × 100 ÷ 6,530 = 7.9 volts, or 6.6% of 120 V.

How much voltage drop is acceptable?

The NEC recommends no more than 3% on a branch circuit and 5% for the feeder and branch together. At 120 V, 3% is 3.6 volts. Motors run hot and lights dim when the drop is much larger.

How do I reduce voltage drop?

Use a bigger wire, shorten the run, or run the load at 240 V instead of 120 V — the same power draws half the current, so the drop in percent falls to a quarter. The table in the result shows the drop for the next sizes up.

Is the length one way or round trip?

Enter the one-way distance from the panel to the load. The formula already doubles it for the return path on a single-phase circuit; three-phase circuits use √3 instead of 2.

Why is aluminum wire drop higher?

Aluminum has about 64% more resistance than copper for the same size — K of 21.2 against 12.9 — so it loses more voltage. That is why aluminum feeders are usually one or two sizes larger than copper.

Why does voltage drop matter more at 12 volts?

Because the same drop in volts is a much larger share of a small voltage. A 10 A load 20 feet from the source on 10 AWG copper loses 0.50 V, which is 4.1% of 12 V, while the same current and wire at 120 V would lose the same 0.50 V, only 0.41%. On 14 AWG the 12 V circuit loses 10.5%. That is why low-voltage lighting, RV and solar wiring uses much larger conductors than household circuits for the same current and length.

Does a three-phase circuit lose less voltage?

For the same current, length and wire, the formula uses √3 = 1.73 instead of 2, so a three-phase circuit loses about 13% less voltage than a single-phase one. A 30 A load 150 feet away on 10 AWG copper drops 9.69 V on a 208 V three-phase supply (4.66%).

What the formula assumes

The calculation VD = 2 × K × I × L ÷ CM treats the wire as a resistor whose resistance is set by K, the resistivity in ohm-circular-mils per foot, and by the cross-section in circular mils (one circular mil is the area of a circle one thousandth of an inch in diameter). K of 12.9 for copper and 21.2 for aluminum are common design values for wire at working temperature, so the result is an estimate: the real resistance depends on temperature and on the wire's construction. It considers a resistive load; for long AC runs with motors, the inductance of the conductors adds to the drop, and starting current is much larger than running current. The percentage is the drop divided by the source voltage.

Worked example: a 240 V, 30 A circuit 150 feet away

With 10 AWG copper (10,380 circular mils), VD = 2 × 12.9 × 30 × 150 ÷ 10,380 = 11.18 V, or 4.66% of 240 V, with 228.8 V remaining at the load. That is above the 3% guideline. Going up one size to 8 AWG (16,510 circular mils) cuts it to 7.03 V, 2.93%. The same run in 10 AWG aluminum drops 18.38 V, 7.66%, because aluminum's K is larger. The same current on a 120 V circuit, with the same wire and distance, would have the same 11.18 V loss, but that is twice the percentage.

What drop is acceptable

The National Electrical Code does not set a mandatory limit in its general rules; informational notes under the branch circuit and feeder sections suggest that a drop of 3% on a branch circuit or feeder, and 5% for both together, gives reasonable efficiency. Specific installations may have their own rules, and equipment such as motors, electronics and LED drivers has voltage tolerances in its documentation. A conductor that is large enough to carry the current safely (its ampacity) can still produce too much drop on a long run, and the size must satisfy both. The wire size calculator on this site picks a conductor for both limits.

Common mistakes

Entering the total round-trip length instead of the one-way distance doubles the drop twice, since the formula already includes the return path. Using the wrong voltage in the percentage (120 V for a 240 V load) halves or doubles the answer. Entering the breaker size as the current is conservative when the real load is smaller, so use the actual running current if you know it. Finally, a copper-clad aluminum conductor and a plain aluminum one are not the same: select the metal that matches the wire in the cable.

Sources: National Electrical Code (NFPA 70), informational notes to 210.19(A)(1) and 215.2(A)(1) on voltage drop (3% and 5%); conductor properties in NEC Chapter 9, Table 8 and Table 9 for circular mils; the formula with K = 12.9 (copper) and 21.2 (aluminum) as commonly published in electrical references. Related: wire size calculator, wire resistance calculator and Ohm's law calculator.