Voltage Drop Calculator

What voltage drop is and why it matters

Every conductor has resistance, and current flowing through resistance loses voltage. The supply might be 120 volts at the panel, but the far end of a long run sees less, and the difference is voltage drop.

The effect is easy to underestimate. A 6 percent drop on a 120 volt circuit leaves 112.3 volts at the outlet, which most electronics tolerate. The same 6 percent on a 240 volt motor leaves 225.6 volts, and a motor that is designed for 230 volts will draw more current to produce the same torque, run hotter, and fail earlier. On a 12 volt lighting circuit, 6 percent is 0.72 volts and the lamps visibly dim toward the end of the run.

Two separate questions decide wire size, and they are often confused:

• Ampacity: can the conductor carry the current without overheating? This is a safety question and the tables in the electrical code answer it.
• Voltage drop: will the equipment at the far end receive enough voltage to work? This is a performance question, and it depends on length in a way ampacity does not.

A 20 amp circuit can be wired in 12 AWG and still deliver poor voltage at 200 feet. The wire is safe; the run is too long for the load.

The formula for single phase and three phase

The drop depends on current, conductor resistance and length. Resistance tables list ohms per 1,000 feet, so the length is divided by 1,000 before it is used.

Single phase: volts lost = 2 x current x (ohms per 1,000 ft x one-way feet / 1,000)

The 2 accounts for the return conductor. Current goes out on the hot leg and back on the neutral, so the loop is twice the one-way distance. Leaving the 2 out halves the answer and is the most common error in hand calculations.

Three phase: volts lost = 1.732 x current x (ohms per 1,000 ft x one-way feet / 1,000)

Three phase uses 1.732, the square root of 3, instead of 2. The factor comes from the way the three phases combine, and it assumes a balanced load. An unbalanced load needs a more careful calculation than this page does.

Percent drop = volts lost / supply voltage x 100. That last step matters because 7.72 volts is trivial on a 480 volt feeder and serious on a 120 volt branch circuit.

The other form of the same formula uses circular mils: volts lost = 2 x K x current x length / circular mils, with K around 12.9 for copper and 21.2 for aluminum. For the 12 AWG example that gives 7.90 volts against the 7.72 from the resistance table, about 2 percent higher, because K is a rounded constant rather than a table value.

Worked example: 20 amps on a 120 volt circuit

A workshop wants a 20 amp outlet 100 feet from the panel, wired in 12 AWG copper. The code table gives 12 AWG copper a resistance of 1.93 ohms per 1,000 feet at 75 degrees Celsius.

• One-way resistance: 1.93 x 100 / 1,000 = 0.193 ohms.
• Loop resistance: 2 x 0.193 = 0.386 ohms.
• Volts lost: 20 x 0.386 = 7.72 volts.
• Percent: 7.72 / 120 = 6.43 percent.
• Voltage at the outlet: 120 - 7.72 = 112.28 volts.

That is more than twice the 3 percent design target, so the next sizes are worth checking. Each step up in gauge cuts the resistance by about a third.

The 10 AWG step removes almost 3 volts and still misses the 3 percent target at 4.03 percent. Only 8 AWG gets under it, and 6 AWG leaves a comfortable margin. Copper prices climb fast at that size, which is why long runs often move the panel closer instead.

Worked example: a 240 volt tool and a three phase run

A 240 volt, 40 amp welder sits 150 feet from the panel, wired in 8 AWG copper. The resistance is 0.764 ohms per 1,000 feet.

• One-way resistance: 0.764 x 150 / 1,000 = 0.1146 ohms.
• Volts lost: 2 x 40 x 0.1146 = 9.17 volts.
• Percent: 9.17 / 240 = 3.82 percent.

That fails a 3 percent branch circuit target and passes the 5 percent total. Dropping to 6 AWG gives 0.491 x 150 / 1,000 = 0.0737 ohms, 2 x 40 x 0.0737 = 5.90 volts, and 5.90 / 240 = 2.46 percent. The higher supply voltage is doing most of the work here: the same 9.17 volts on a 120 volt circuit would be 7.6 percent.

Now a three phase case. A 208 volt, 30 amp motor load is 250 feet away on 8 AWG copper.

• One-way resistance: 0.764 x 250 / 1,000 = 0.191 ohms.
• Volts lost: 1.732 x 30 x 0.191 = 9.92 volts.
• Percent: 9.92 / 208 = 4.77 percent.

6 AWG gives 0.491 x 250 / 1,000 = 0.1228 ohms, then 1.732 x 30 x 0.1228 = 6.38 volts, which is 3.07 percent, still a hair over. 4 AWG lands at 4.00 volts and 1.92 percent. Three phase saves conductor compared with single phase at the same power, because the 1.732 factor replaces the 2 and each conductor carries less current for the same total watts.

Wire resistance by gauge

These values come from NEC Chapter 9 Table 8 for uncoated conductors at 75 degrees Celsius, the temperature the code uses for most circuit calculations. Resistance rises with temperature, so a conductor running hot has slightly more drop than the table predicts.

Aluminum carries about 61 percent of the conductivity of copper for the same size, so the same gauge drops roughly 1.65 times as many volts. That is not a reason to avoid aluminum: it is lighter and cheaper for service entrance and feeder work, and it is widely used there. It is a reason to size aluminum two gauges larger when matching a copper design, and to use the aluminum column rather than the copper one.

An 8 AWG aluminum run of 100 feet at 20 amps on 120 volts gives 1.26 x 100 / 1,000 = 0.126 ohms, then 2 x 20 x 0.126 = 5.04 volts, which is 4.20 percent. The same run in 8 AWG copper is 2.55 percent, so the material choice alone changes the verdict.

Where the 3 percent and 5 percent numbers come from

The familiar targets are 3 percent on a branch circuit and 5 percent for the feeder plus branch combined. They are recommendations in informational notes to the National Electrical Code, at 210.19(A) for branch circuits and 215.2(A) for feeders, not enforceable rules. The NFPA 70 page is the place to confirm the current edition and the exact wording.

Informational notes are explanatory text. Section 90.5 of the code says as much, which means an inspector cannot fail an installation for a 4 percent drop on the basis of that note alone. What can bind you is a local amendment, a project specification, or the listing instructions of the equipment, and some jurisdictions do adopt the 3 percent figure as a rule.

The 5 percent budget is shared, not stacked. If the feeder eats 2 percent, the branch circuit has 3 percent left. Designers often give the feeder 1 to 2 percent and leave the rest for the branch, which is why a long service run can force larger conductors even when the branch circuit is short.

OSHA's electrical wiring rules in 29 CFR 1910 Subpart S cover conductor sizing and overcurrent protection for workplaces, and they point back to the same code framework rather than setting a separate voltage drop number.

Low voltage runs and other mistakes

The lower the supply voltage, the more a given number of lost volts matters. A 12 volt landscape lighting run of 50 feet at 5 amps on 12 AWG copper shows why.

• One-way resistance: 1.93 x 50 / 1,000 = 0.0965 ohms.
• Volts lost: 2 x 5 x 0.0965 = 0.965 volts.
• Percent: 0.965 / 12 = 8.04 percent.

8 AWG brings it to 0.38 volts and 3.18 percent, and 6 AWG to 0.25 volts and 2.05 percent. The usual fix on landscape lighting is not bigger wire but a transformer tap one step higher, or shorter runs with fewer fixtures on each.

One more: a load that cycles, such as a compressor or a welder, draws a starting current several times its running current. The drop during that inrush is brief, but it can trip a contactor or dim every light on the circuit. Sizing the run for the running current and then checking the start is the practical order.

This page estimates drop from resistance alone and ignores reactance, which matters on large feeders and on long runs at high current. For a service entrance or a feeder over roughly 100 amps, use the full calculation from the code rather than this estimate.