Voltage drop

Why it matters

Every conductor has resistance, so current flowing through it loses some voltage along the way. On a long run, a motor at the far end may get noticeably less than 120 V or 240 V: it runs hot, lights dim, and electronics misbehave. Voltage drop is why a conductor that passes the ampacity check can still be too small for a long run, and it is a common calculation on the job and in exams.

Awareness level only. This is paper and calculator practice. Choosing conductors for real installations is done under a licensed electrician's supervision.

Is it a code rule?

Mostly, no. For general branch circuits and feeders, the NEC's voltage-drop guidance appears in informational notes (for example in Article 210 for branch circuits and Article 215 for feeders). Those notes suggest a maximum of about 3% on a branch circuit and 5% total for feeder plus branch circuit. Because they are informational notes, they are not enforceable on their own. Some specific articles, such as those for certain sensitive equipment and fire pumps, and many engineering specifications and energy codes, do make limits mandatory. Good practice is to design within the 3% / 5% guidance anyway.

The formula

The standard trade formula for a single-phase circuit is:

VD = (2 × K × I × L) / CM

• VD: voltage drop in volts.
• K: the resistance of a one-mil-foot of conductor in ohms (about 12.9 for copper and 21.2 for aluminium at typical operating temperature; these are trade approximations).
• I: the load current in amps.
• L: the one-way length of the run in feet. The 2 accounts for the current going out and coming back.
• CM: the conductor's area in circular mils. Chapter 9 of the NEC lists conductor properties; for example 12 AWG is about 6,530 circular mils and 10 AWG is about 10,380.

For three-phase circuits, replace the 2 with √3 (about 1.732).

Percent drop is:

% VD = VD / source voltage × 100

Worked example

A 120 V, 16 A load sits 100 ft (one way) from the panel, on 12 AWG copper.

VD = (2 × 12.9 × 16 × 100) / 6,530
   = 41,280 / 6,530
   ≈ 6.32 V
% VD = 6.32 / 120 × 100 ≈ 5.3 %

That is well over the 3% suggestion. Try 10 AWG:

VD = 41,280 / 10,380 ≈ 3.98 V
% VD ≈ 3.3 %

Closer, but still above 3%. You might go up to 8 AWG, shorten the run or split the load. Note that the ampacity check alone would have accepted 12 AWG; voltage drop is what made the difference.

Solving for the conductor size

Rearranged, the same formula gives the minimum circular mils for a target drop:

CM = (2 × K × I × L) / allowed VD

For a 3% limit on 120 V, the allowed drop is 3.6 V. With the example above: CM = 41,280 / 3.6 ≈ 11,467 circular mils. 10 AWG (10,380) is just short, so the next size up meets it.

Why it also matters for safety

Voltage drop is not just about performance. A motor starved of voltage draws more current to make the same power, which heats its windings. Long runs also raise the impedance of the fault path, which slows how quickly a breaker trips on a ground fault. That is why long runs often need a larger equipment grounding conductor too (Article 250 has a rule for increasing it in proportion).

Common mistakes

• Using the round-trip length for L and then also multiplying by 2.
• Using 2 instead of 1.732 on three-phase.
• Mixing copper and aluminium values of K.
• Forgetting to convert the volt drop to a percentage of the source voltage.
• Treating the 3% / 5% guidance as optional good-to-have on long motor or pump circuits, where it really matters.

Quick recap

Single-phase VD = 2KIL / CM; three-phase uses 1.732 instead of 2. Divide by the source voltage for percent. The NEC's 3% / 5% guidance lives in informational notes, but good design follows it, and long runs often need a bigger conductor than the ampacity check alone suggests. Use conductor areas from Chapter 9 of your edition (NFPA 70; checked on 2026-10-23).

Try it

Open the Do tab. The calculator has every function but one: write the single-phase voltage-drop formula.