Voltage drop happens along the cable, not at the equipment using the electricity. Every foot of conductor adds resistance, and current flowing through that resistance consumes part of the available voltage before it reaches the load.
A meter at the panel can therefore show the full supply voltage while the far end of a long circuit reads low. The panel measurement describes the starting point, not what survives the journey through the conductors.
The calculation is Ohm's law applied to the cable instead of the load, and two multipliers control the result. The current travels both ways, which makes the electrical path twice the run length you measured. And for the same load, moving from 120 V to 240 V improves the percentage drop by a factor of four rather than two.
The current goes out and comes back
Cable runs are normally described by their one-way physical distance. A load 100 ft from the panel has 100 ft of outgoing conductor and another 100 ft carrying the current back, so the calculation has to account for 200 ft of copper. That return path is the factor of two at the front of the equation, and leaving it out models half the circuit and reports half the real loss.
voltage drop = 2 × one-way distance × current × ohms per foot
ohms per foot = ohms per 1,000 ft ÷ 1,000
percent drop = voltage drop ÷ supply voltage × 100
voltage at load = supply voltage − voltage drop
The default case uses a 20 A load, a 100 ft one-way run, 12 AWG copper and a 120 V supply. The resistance is 1.93 ohms per 1,000 ft, taken from NEC Chapter 9, Table 8 for uncoated copper at 75 C.
ohms per foot = 1.93 ÷ 1,000
ohms per foot = 0.00193
voltage drop = 2 × 100 × 20 × 0.00193
voltage drop = 7.72 V
percent drop = 7.72 ÷ 120 × 100
percent drop = 6.43%
voltage at load = 120 − 7.72
voltage at load = 112.28 V
The load receives 112.28 V after 7.72 V has been lost in the conductors, and at 6.43 percent that is graded HIGH because it is past both the 3 and the 5 percent figures. Those verdict labels belong to the voltage drop calculator, which grades 3 percent or less as GOOD, over 3 through 5 as ACCEPTABLE, and over 5 as HIGH. The bands are not themselves code requirements.
The 7.72 V is spread along the whole current path rather than lost at any one point. A measurement beside the panel is taken before that resistance has accumulated, so a normal reading there establishes nothing about the voltage available at the far end.
The same load at 240 volts
The comparison that matters keeps the load, the cable and the distance unchanged. A 2,400 W load draws 20 A at 120 V but only 10 A at 240 V, which is the same relationship the watts to amps calculator works in.
| Supply | Load | Current | Voltage lost | Drop | At the load | Verdict |
|---|---|---|---|---|---|---|
| 120 V | 2,400 W | 20 A | 7.72 V | 6.4333% | 112.28 V | HIGH |
| 240 V | 2,400 W | 10 A | 3.86 V | 1.6083% | 236.14 V | GOOD |
120 V current = 2,400 W ÷ 120 V = 20 A
240 V current = 2,400 W ÷ 240 V = 10 A
Current creates the first improvement. Voltage drop across a fixed resistance is proportional to current, so halving the current halves the absolute loss from 7.72 V to 3.86 V. That change on its own is worth a factor of two.
Supply voltage creates the second, and it is a different mechanism rather than more of the same one. The percentage compares the lost voltage against the starting voltage, and that reference has doubled, so the already smaller loss is halved again as a proportion.
absolute-loss improvement = 7.72 ÷ 3.86 = 2
percentage improvement = 6.4333 ÷ 1.6083 = 4
The two effects are independent, so they compound. Half the current produces half the voltage loss, and twice the supply voltage then makes that smaller loss half as large again in percentage terms, which leaves the combined figure at one quarter of the original.
A second comparison exposes the half-measure and is worth running deliberately. If the current stays at 20 A while only the reference voltage changes, the cable still loses the full 7.72 V, and the percentage falls only because the same loss is divided by twice the supply.
| Supply | Current | Implied load | Voltage lost | Drop | Verdict |
|---|---|---|---|---|---|
| 120 V | 20 A | 2,400 W | 7.72 V | 6.43% | HIGH |
| 240 V | 20 A | 4,800 W | 7.72 V | 3.22% | ACCEPTABLE |
That is not the same-load comparison, because holding 20 A at 240 V means the load has doubled to 4,800 W. Changing the reference voltage alone is worth two; halving the current as well is what makes the improvement four.
For the unchanged 2,400 W load the verdict moves from HIGH to GOOD without touching the 12 AWG cable or the 100 ft route. That compounding is the electrical reason long feeds to outbuildings are so often run at 240 V.
Turn the formula around
The usual equation starts with a distance and returns a percentage. Rearranged, it answers the question people actually have when planning: the longest one-way run that stays inside a chosen percentage at a given voltage, current and conductor resistance.
percent = 2 × L × current × ohms per foot ÷ voltage × 100
percent × voltage = 200 × L × current × ohms per foot
L = percent × voltage ÷ (200 × current × ohms per foot)
At 20 A on 120 V, these are the longest one-way runs that stay inside 3 percent. Each figure is cut down to whole feet rather than rounded, because rounding up would cross the limit the number exists to respect.
| Copper gauge | Resistance per 1,000 ft | Maximum one-way length |
|---|---|---|
| 14 AWG | 3.07 ohms | 29 ft |
| 12 AWG | 1.93 ohms | 46 ft |
| 10 AWG | 1.21 ohms | 74 ft |
| 8 AWG | 0.764 ohms | 117 ft |
| 6 AWG | 0.491 ohms | 183 ft |
| 4 AWG | 0.308 ohms | 292 ft |
This is a voltage-drop comparison and not an installation recommendation. It shows how resistance changes the distance available under one electrical target, with current and supply voltage held fixed.
For 12 AWG the default 100 ft run is well past the 46 ft that limit allows at 20 A on 120 V. Moving the same 2,400 W load to 240 V changes the current and the percentage reference together.
120 V reach = 3 × 120 ÷ (200 × 20 × 0.00193) = 46.63 ft
240 V reach = 3 × 240 ÷ (200 × 10 × 0.00193) = 186.53 ft
The whole-foot figures are 46 ft and 186 ft. Before that cut the formula gives exactly four times the reach, for the same reason the percentage fell to a quarter: the current halved while the supply voltage doubled. A circuit losing a quarter of the percentage per foot can run four times as far before it meets the same boundary.
The current is the load, not the breaker
Voltage drop is calculated on the current the circuit actually carries. A breaker rating sets an overcurrent-protection boundary; it does not mean the circuit sits at that current, and entering it when the load never approaches it makes the run look worse than it is.
| Actual current | Voltage lost | Drop | Verdict |
|---|---|---|---|
| 20 A | 7.72 V | 6.43% | HIGH |
| 16 A | 6.18 V | 5.15% | HIGH |
| 12 A | 4.63 V | 3.86% | ACCEPTABLE |
| 8 A | 3.09 V | 2.57% | GOOD |
At 20 A the circuit is firmly in the HIGH band, at 16 A it is still just above 5 percent, 12 A moves it into ACCEPTABLE, and 8 A brings it comfortably inside GOOD. The same cable crosses two verdict boundaries across that range with nothing about the conductor or the distance changed.
Sizing to a breaker rating the load never reaches therefore points at heavier cable than the situation calls for. The current to use comes from the load you expect, subject to the separate rules that govern circuit sizing.
What a step up in gauge buys
Larger copper has lower resistance, so every step up the gauge scale cuts the loss for the same current over the same distance. On the default 20 A, 100 ft, 120 V run the progression looks like this across six common sizes.
| Copper gauge | Voltage lost | Drop | Voltage at load | Verdict |
|---|---|---|---|---|
| 14 AWG | 12.28 V | 10.23% | 107.72 V | HIGH |
| 12 AWG | 7.72 V | 6.43% | 112.28 V | HIGH |
| 10 AWG | 4.84 V | 4.03% | 115.16 V | ACCEPTABLE |
| 8 AWG | 3.06 V | 2.55% | 116.94 V | GOOD |
| 6 AWG | 1.96 V | 1.64% | 118.04 V | GOOD |
| 4 AWG | 1.23 V | 1.03% | 118.77 V | GOOD |
Each step cuts resistance by close to the same proportion: 37.1 percent from 14 to 12 AWG, 37.3 from 12 to 10, 36.9 from 10 to 8, and 35.7 from 8 to 6. Because the reduction is proportional rather than fixed, the same fraction of a large loss saves more volts than the same fraction of an already small one.
The first step, from 14 to 12 AWG, saves 4.56 V on this run. The last, from 6 to 4 AWG, saves 0.73 V. The ladder flattens as it descends, so the early steps carry nearly all the benefit, while moving a suitable fixed-power load to 240 V attacks the current and the percentage reference at once instead.
What this does not decide
NEC 210.19(A) carries an Informational Note recommending that branch-circuit voltage drop at the farthest outlet not exceed 3 percent, with feeder and branch-circuit drop combined kept to 5 percent, and NEC 215.2(A) carries a parallel note for feeders.
NEC 90.5 states that Informational Notes are informational only and are not enforceable as requirements. Those percentages are therefore recommendations rather than code, and the GOOD, ACCEPTABLE and HIGH labels are this calculator's own grading of them.
Energy codes and local amendments often make the 3 percent figure binding anyway, and staying inside it is defensible engineering regardless, because equipment at the far end then receives voltage closer to what it was designed for.
The figures here are copper only, and aluminium needs its own conductor data. The engine uses the NEC Chapter 9, Table 8 DC resistance for uncoated copper at 75 C rather than modelling AC impedance or the operating conditions of a particular installation.
Voltage drop settles one constraint out of several. Ampacity, temperature correction, conduit fill, overcurrent protection and applicable local rules are separate questions that a favourable percentage cannot answer or override, so treat this as a way to compare options rather than as a specification. The circuit has to be sized and installed by a licensed electrician.