Voltage Drop Calculator (NEC)

What is a voltage drop calculator?

A voltage drop calculator tells you how many volts a circuit loses on its way from the panel to the load. No conductor is a perfect conductor: every foot of wire has resistance, and pushing current through that resistance burns off a slice of your supply voltage as heat. By the time the electricity reaches the far end of a long run, the motor, light, or receptacle may be seeing noticeably less voltage than the breaker is feeding.

This tool applies the standard approximation used by electricians in the field. You pick the system (DC, single-phase AC, or three-phase AC), the conductor material, the wire size in AWG, the one-way length of the run, the load current, and the source voltage. It returns the volts lost, that loss as a percentage of the supply, the voltage actually arriving at the load, and a plain verdict on whether the run meets the recommended limit.

Why voltage drop matters

Excessive voltage drop is not a code violation in most cases — it is a performance and efficiency problem, and the National Electrical Code addresses it in informational notes rather than hard rules. The consequences are real regardless:

• Motors run hot. An induction motor starved of voltage draws more current to deliver the same torque, which heats the windings and shortens its life.
• Lights dim and flicker. Incandescent output falls off steeply with voltage, and LED drivers can behave erratically at the low end.
• Heaters underperform. Resistive heat output scales with the square of voltage, so a 5% voltage loss costs nearly 10% of the heat.
• Energy is wasted. The volts you lose are dissipated as heat in the wire itself — you pay for them, and they warm your walls instead of your load.

The NEC’s guidance (in the informational notes to 210.19(A) and 215.2(A)) is to keep the drop at or below 3% on a branch circuit, and no more than 5% total across the feeder and the branch circuit combined. This calculator flags your run against that 3% target.

How does the calculator work?

The calculation rests on the circular-mil form of Ohm’s law, using the conductor properties published in NEC Chapter 9, Table 8.

Conductor resistivity. The constant $K$ is the resistance of one circular mil of conductor, one foot long, expressed in ohm-circular mils per foot:

$$K_{\text{copper}} = 12.9 \qquad K_{\text{aluminum}} = 21.2$$

Aluminum’s $K$ is roughly 64% higher than copper’s, which is exactly why an aluminum run of the same gauge drops noticeably more voltage.

Conductor area. Each AWG size has a fixed cross-sectional area in circular mils ($CM$), also from Table 8 — for example 12 AWG is $6{,}530$ CM and 8 AWG is $16{,}510$ CM. A bigger number means a fatter wire, less resistance, and less drop.

Voltage drop, DC and single-phase AC. Current has to travel out to the load and back along the neutral, so the conductor length counts twice:

$$V_{\text{drop}} = \frac{2 \times K \times I \times L}{CM}$$

Voltage drop, three-phase AC. In a balanced three-phase system the return currents partially cancel, so the multiplier is $\sqrt{3} \approx 1.732$ instead of 2:

$$V_{\text{drop}} = \frac{1.732 \times K \times I \times L}{CM}$$

Here $I$ is the load current in amps and $L$ is the one-way length of the run in feet (enter it in meters if you prefer — the calculator converts for you).

Percent drop and voltage at the load. The drop is then compared against the supply and subtracted from it:

$$\%\ \text{drop} = \frac{V_{\text{drop}}}{V_{\text{source}}} \times 100$$ $$V_{\text{load}} = V_{\text{source}} - V_{\text{drop}}$$

Worked example

A 20 A single-phase branch circuit is wired with 12 AWG copper and runs 100 ft (30.48 m) one way from a 120 V panel.

• 12 AWG has an area of $6{,}530$ CM, and copper’s constant is $K = 12.9$.
• Single-phase, so the multiplier is 2:

$$V_{\text{drop}} = \frac{2 \times 12.9 \times 20 \times 100}{6{,}530} = 7.90 \text{ V}$$

• As a percentage of the supply:

$$\%\ \text{drop} = \frac{7.90}{120} \times 100 = 6.58\%$$

• Voltage arriving at the load:

$$V_{\text{load}} = 120 - 7.90 = 112.10 \text{ V}$$

At 6.58% the drop is more than double the recommended 3%, so this run fails the target: the load only sees 112.10 V.

Fixing it by upsizing the conductor

Keep everything the same but pull 8 AWG copper ($16{,}510$ CM) instead:

$$V_{\text{drop}} = \frac{2 \times 12.9 \times 20 \times 100}{16{,}510} = 3.13 \text{ V}$$

That is 2.60% of 120 V, leaving 116.87 V at the load — comfortably inside the 3% target. Two gauge steps up bought back nearly 5 V.

Practical notes

• Enter the one-way distance, not the round trip. The formula already doubles the length for DC and single-phase circuits. Measuring the wire as it actually routes — up walls, along joists, around obstacles — matters more than the straight-line distance between the panel and the load.
• The multiplier is the whole difference between the systems. DC and single-phase both use 2 because the current makes a full round trip on two conductors. Balanced three-phase uses 1.732 because the three line currents are 120° apart and partially cancel in the return path.
• This is the classic NEC approximation. It uses DC resistance only and ignores conductor reactance, temperature rise, and power factor. That is accurate enough for the great majority of branch circuits and short feeders. For long runs, large conductors, or a poor power factor, use the AC impedance figures in NEC Chapter 9, Table 9 instead. Treat the result here as a solid design estimate, not a substitute for an engineered study.
• Aluminum needs roughly two gauge sizes to match copper. Because $K$ is 21.2 rather than 12.9, aluminum drops about 64% more voltage at the same gauge and current — factor that in before switching materials to save money.
• Voltage drop and ampacity are different questions. A wire can be perfectly legal for the current it carries yet still drop too much voltage over a long run. Size for ampacity first with an Ohm’s law calculator and a watts-to-amps calculator, then check voltage drop and upsize if needed. Once the gauge is settled, confirm the wires fit the raceway with a conduit fill calculator.

Frequently asked questions

Is 3% a legal limit? No. In the NEC it appears as an informational note — guidance, not an enforceable rule. Local amendments, utility requirements, or a specific piece of equipment may impose a stricter limit, and some jurisdictions do make it binding. Design to 3% on the branch circuit and 5% overall and you will satisfy nearly everyone.

Why does length count twice? Current has to make a complete circuit. On a 100 ft one-way run, the electricity travels 100 ft out on the hot conductor and 100 ft back on the neutral — 200 ft of resistance in total. The factor of 2 in the formula accounts for that automatically, so you only enter the one-way distance.

What is a circular mil? It is the area of a circle one mil (one thousandth of an inch) in diameter. Wire tables use it because the area in circular mils is simply the diameter in mils squared — no $\pi$ required — which makes conductor comparisons easy.

How do I reduce an excessive drop? In order of practicality: use a larger conductor (the most common fix), shorten the run, split the load across two circuits, or raise the supply voltage — a 240 V circuit carrying the same power draws half the current and therefore drops half the volts.