Voltage Drop Calculator delivers fast voltage loss, load-end voltage, power loss, conductor sizing, and maximum run length for copper or aluminum electrical circuits with accuracy.
Voltage drop is the portion of a circuit’s supply voltage that gets used up overcoming the resistance of the conductor itself, instead of reaching the load. The numbers above come from the same calculation laid out below — the formula, the resistivity constants, and the 3%/5% thresholds all trace back to standard conductor resistivity values and the NEC’s own guidance on branch-circuit and feeder sizing.
How It’s Calculated
For single-phase, two-wire circuits, voltage drop is:
$$VD = \frac{2 \times K \times I \times L}{CM}$$
For three-phase circuits, the 2 is replaced with √3 (about 1.732):
$$VD = \frac{\sqrt{3} \times K \times I \times L}{CM}$$
$VD$ is the voltage lost across the run, in volts. $K$ is the conductor’s resistivity constant — 12.9 for copper, 21.2 for aluminum, based on standard conductor resistivities at typical operating temperature. $I$ is the load current in amps.
$L$ is the one-way conductor length in feet (the calculator also accepts meters, converting at 1 m = 3.28084 ft). $CM$ is the conductor’s cross-sectional area in circular mils — 6,530 cmil for 12 AWG copper, for example.
The 2 in the single-phase formula is there because current makes a round trip — out through the hot conductor, back through the neutral or second conductor — so a 100-foot run is actually 200 feet of current-carrying wire. Three-phase circuits use √3 instead, since the voltage relationship between three conductors 120° out of phase isn’t a simple there-and-back loop; √3 is the constant that ties line-to-line voltage to the drop on each conductor.
From that single $VD$ figure, everything else follows. Voltage drop as a percentage is $(VD / V) \times 100$. Voltage delivered to the load is $V – VD$. Power lost in the conductor is $VD \times I$ for single-phase circuits, or $$\sqrt{3} \times VD \times I$$ for three-phase — the same three-phase power relationship applied to the voltage that’s lost rather than the voltage that’s delivered.
The calculator also checks that result against NEC guidance. NEC 210.19(A), in its informational note on branch-circuit sizing, recommends keeping voltage drop under 3% for reasonable operating efficiency, and the related feeder guidance in 215.2(A) puts the combined feeder-plus-branch-circuit total at 5%. Those percentages run in both directions in the tool: solving for the conductor size needed to stay under 3% at a given current and length,
$$CM_{3\%} = \frac{2 \times K \times I \times L}{0.03 \times V}$$
and solving for the maximum run length a given conductor can support before crossing the 3% or 5% mark,
$$L_{max} = \frac{V \times p \times CM}{2 \times K \times I}$$
where $p$ is 0.03 or 0.05 depending on which limit is being checked.
The formula only runs on positive, non-zero values for voltage, current, and length — a circuit with no current or no length has no voltage drop to compute in the first place.
Worked Example
Take a 120V, single-phase branch circuit carrying 15A over a 100-foot run of 12 AWG copper wire (6,530 circular mils) — a common setup for a long kitchen or garage receptacle circuit.
Plugging those numbers into the single-phase formula:
$$VD = \frac{2 \times 12.9 \times 15 \times 100}{6530} = \frac{38700}{6530} \approx 5.93\text{ V}$$
That’s the voltage lost to the conductor’s resistance. As a share of the 120V source, that’s $(5.93 / 120) \times 100 \approx 4.94\%$, and the load itself only sees $120 – 5.93 = 114.07$ V.
The effective resistance behind that number is $(2 \times 12.9 \times 100) / 6530 \approx 0.395\ \Omega$, and at 15A that works out to $5.93 \times 15 \approx 88.90$ W dissipated in the wire itself before any of it reaches the load.
Checking that against the 3% threshold: 3% of 120V is 3.60V, so this run is already 2.33V past that target. Hitting 3% exactly at this current and length would take $(2 \times 12.9 \times 15 \times 100) / 3.60 \approx 10{,}750$ circular mils — about 4,220 cmil more than 12 AWG has.
In practical terms that lands between standard sizes: 10 AWG (10,380 cmil) is still a touch short of clearing 3% here, so getting under it at this length and current would mean stepping up to 8 AWG (16,510 cmil).
Run the formula the other direction and 12 AWG copper at 15A stays under 3% only out to about 60.74 feet, and under the 5% ceiling out to about 101.24 feet — at exactly 100 feet, that’s 1.24 feet of room left before hitting the 5% limit.
| Metric | Result |
|---|---|
| Voltage drop | 5.93 V |
| Voltage drop (% of source) | 4.94% |
| Voltage at the load | 114.07 V |
| Power lost in the conductor | 88.90 W |
| Loop resistance | 0.395 Ω |
| Conductor size needed for 3% | 10,750 cmil |
| Max length at 3% | 60.74 ft |
| Max length at 5% | 101.24 ft |
What the Result Means
NEC 210.19(A) recommends keeping branch-circuit voltage drop under 3%, but nothing prevents a circuit from working correctly above that number — it’s a performance guideline, not a hard safety limit the way ampacity is. The calculator tracks two thresholds pulled straight from that guidance and its feeder counterpart:
| Voltage drop | What the tool shows | What it means |
|---|---|---|
| 3% or less | Within 3% Target | Meets the NEC 210.19(A) informational note for branch circuits |
| Above 3%, under 5% | Above 3% Target, distance still under the 5% max | Exceeds the branch-circuit guideline on its own, but stays inside the combined feeder-plus-branch 5% ceiling from 215.2(A) |
| Above 5% | Above 3% Target, 5% max distance exceeded | Beyond both informational notes — the conductor is undersized for the run at that current |
The cmil and distance figures work the same way in reverse. A negative “cmil available” figure means the chosen conductor is smaller than what’s needed to hit 3% at that current and length — the number shown is how many circular mils short it is. A negative “ft available” figure at the 5% mark means the run has already gone past the distance where even the looser 5% guidance holds.
The wattage figure is heat, not a code violation on its own — it’s power the conductor’s resistance converts to heat (I²R loss) instead of delivering to the load. On longer or higher-current runs that shows up as equipment running warm and slightly under-voltage, which is an efficiency and performance issue long before it becomes a safety one.
What Changes the Result
Conductor material matters directly, since $K$ sits right in the numerator: copper’s constant of 12.9 versus aluminum’s 21.2 means an aluminum conductor of the same size, length, and current carries about 64% more voltage drop than copper.
Conductor size ($CM$) sits in the denominator, so it’s an inverse relationship — doubling the circular-mil area roughly halves the voltage drop for the same current and length, which is why undersized voltage-drop runs get fixed by upsizing the wire rather than changing anything else about the circuit.
Circuit configuration changes the multiplier out front: 2 for single-phase versus √3 (≈1.732) for three-phase, a difference of roughly 13%. It also changes how power loss gets calculated — $VD \times I$ for single-phase versus $\sqrt{3} \times VD \times I$ for three-phase.
Current and length both sit in the numerator and scale the result directly — double either one and voltage drop doubles right along with it.
Source voltage doesn’t change the voltage drop itself — $VD$ only depends on $K$, $I$, $L$, and $CM$ — but it changes the percentage. The same 5.93V drop on a 240V circuit instead of 120V would come out to about 2.47% instead of 4.94%, because the same number of volts lost is a smaller share of a larger base voltage.
FAQs
What is a good voltage drop percentage for a circuit?
NEC 210.19(A)’s informational note points to 3% for a branch circuit on its own, and 5% total when a feeder and branch circuit are combined. Staying under those numbers isn’t mandatory the way ampacity limits are, but it keeps equipment running at close to its rated voltage.
How do you calculate voltage drop on a 120V circuit?
Multiply 2 (for single-phase) by the conductor’s K constant, the current in amps, and the one-way length in feet, then divide by the conductor’s circular-mil area. A 15A, 100-foot run on 12 AWG copper at 120V comes out to about 5.93V, or roughly 4.94% of the source voltage.
Does aluminum wire have more voltage drop than copper?
Yes. Aluminum’s resistivity constant (21.2) is significantly higher than copper’s (12.9), so for identical size, length, and current, an aluminum conductor loses about 64% more voltage than copper.
How far can I run 12 AWG wire on a 15A, 120V circuit?
Solving the formula for length shows 12 AWG copper at 15A and 120V stays under the 3% guideline out to about 60.74 feet, and under the 5% combined limit out to about 101.24 feet. Past that, either upsize the conductor or expect the drop to climb above the NEC’s informational note thresholds.
Is voltage drop different for three-phase circuits?
Yes. Single-phase circuits use a factor of 2 in the formula, for the round trip through hot and neutral. Three-phase circuits use √3 (about 1.732) instead, reflecting the line-to-line voltage relationship between three conductors carrying current 120° apart.
Does the NEC require voltage drop calculations?
No. The 3% and 5% figures come from informational notes in 210.19(A) and 215.2(A), which are recommendations for efficient operation, not enforceable requirements the way ampacity and overcurrent protection sizing are.
What’s the difference between voltage drop and wire ampacity?
Ampacity is the maximum current a conductor can carry without overheating — a hard NEC limit tied to conductor insulation and safety. Voltage drop is a separate performance measure covered only by informational notes, so a wire can sit well within its ampacity rating and still have voltage drop well above 3% or 5% on a long enough run.
Does running a circuit at a higher voltage reduce voltage drop?
Not the voltage drop itself — that number only depends on current, length, conductor size, and material. What changes is the percentage: the same volts lost make up a smaller share of a higher source voltage, which is part of why 240V circuits tolerate longer runs than 120V circuits before crossing the 3% or 5% marks.
These results are for planning and estimation purposes only. Final conductor sizing, breaker sizing, and any actual wiring work should be verified against your local electrical code and performed or inspected by a licensed electrician.