Voltage Drop Calculator
Enter the load, the voltage, the conductor and the one-way distance, and this calculator returns the drop in volts, the drop as a percentage, the voltage arriving at the load, and whether the run clears the usual 3% and 5% marks. When it does not, it names the next size up that does.
Panel to load, measured along the route the cable actually takes.
- Voltage at the load
- 228.1V
- Against the 3% branch-circuit figureNEC 210.19(A) informational note — a recommendation, not a rule
- Over
- Against 5% totalFeeder plus branch circuit combined
- Passes
- Smallest size under 3%At this length and load
- 8 AWG copper
- Total conductor in the circuitUseful for pricing the wire
- 300ft
What this figure includes
Uses the DC resistance figures from NEC Chapter 9, Table 8 at 75 °C, which is the standard approximation for branch circuits and short feeders. It ignores reactance, so it is not accurate for very large conductors in steel conduit at high current — use the Table 9 impedance method there. It does not check ampacity; size for ampacity first, then check drop.
How this is calculated
The formula
Voltage drop is Ohm's law applied to the wire itself. The wire has resistance, current flows through it, and the product is voltage that never reaches the load.
Single phase: VD = (2 × R × L × I) ÷ 1000 Three phase: VD = (1.732 × R × L × I) ÷ 1000
Where R is the conductor's resistance in ohms per 1,000 ft, L is the one-way length in feet, and I is the current in amps. Percentage drop is VD ÷ system voltage × 100.
The factor of 2 is the part people drop. Current has to get back to the source, so a 150 ft run is 300 ft of copper. On three phase the return paths partly cancel, which is where 1.732 — the square root of 3 — comes from.
Worked example
A 20 A, 240 V single-phase circuit on 12 AWG copper, 150 ft out to a detached garage.
1. R for 12 AWG copper = 1.98 Ω per 1,000 ft 2. VD = (2 × 1.98 × 150 × 20) ÷ 1000 = 11.9 V 3. Percent = 11.9 ÷ 240 = 4.95% 4. Voltage at the load = 240 − 11.9 = 228 V
That is nearly 5% — legal, but poor. Step to 10 AWG copper (1.24 Ω) and the drop falls to 7.4 V, or 3.1%. Step to 8 AWG and it is 4.7 V, under 2%.
What the NEC actually says
This is the point most calculators get wrong, so it is worth being precise.
The 3% branch-circuit and 5% total figures come from informational notes — 210.19(A) for branch circuits and 215.2(A) for feeders. Informational notes are explanatory. They are not enforceable requirements, and an inspector cannot fail a job for a 4% branch circuit on that basis alone.
What *is* enforceable is 110.3(B): equipment must be installed according to its listing and labelling. If a piece of equipment specifies a minimum operating voltage, and your run does not deliver it, that is a violation with teeth. Certain articles also carry hard drop limits — sensitive electronic equipment (647.4(D)), fire pumps, and some elevator and industrial installations.
So treat 3% as good practice rather than law, and treat the equipment nameplate as the real requirement. In practice most jurisdictions and most engineers design to 3% anyway, because the consequences of ignoring it are not code violations, they are callbacks.
Why it matters at the load
Undervoltage is hardest on motors. A motor is a constant-power device: give it less voltage and it draws more current to do the same work. More current means more heat in the windings, and heat is what kills motor insulation. A well pump or a compressor at the far end of a long, undersized run will run hot and fail years early, and nothing about it looks like a wiring fault.
Resistive loads simply underperform — a water heater takes longer, an incandescent lamp dims. Electronics with switching power supplies mostly do not care until they hit their dropout voltage, then they cut out entirely, which shows up as intermittent faults nobody can reproduce.
Size for ampacity first
Voltage drop is the second check, never the first. Start with NEC 310.16 ampacity, apply the temperature and bundling adjustments, confirm the overcurrent device, and only then run the drop calculation. If the drop pushes you up a size, the larger conductor is still governed by the original breaker — you are upsizing the wire, not the protection. On a circuit with equipment grounding conductors, 250.122(B) requires you to increase the EGC proportionally when you upsize the ungrounded conductors for drop.
Frequently asked questions
- What is an acceptable voltage drop?
- 3% on a branch circuit and 5% for feeder plus branch combined is the widely used design target. Both come from informational notes in the NEC, so they are recommendations rather than enforceable limits.
- Is the 3% rule enforceable?
- Not on its own. What is enforceable is NEC 110.3(B) — equipment has to get the voltage its listing requires — plus specific articles like 647.4(D). Most designers hold to 3% regardless.
- Why is the distance doubled?
- Because current travels out on one conductor and back on the other. A 150 ft run is 300 ft of conductor, and both legs drop voltage.
- How far can I run 12 AWG at 20 amps?
- About 60 ft on a 120 V circuit and about 120 ft at 240 V before you hit 3%. At 120 V the same load hits the limit in half the distance, because the drop is the same volts against half the voltage.
- Does aluminium wire drop more voltage?
- Yes, roughly 60% more for the same AWG, because aluminium has higher resistivity. The usual answer is to go up two sizes — 2/0 aluminium in place of 1/0 copper, for instance.
- If I upsize the wire, do I upsize the breaker?
- No. The breaker protects the load and stays the same. But if you upsize the ungrounded conductors for drop, NEC 250.122(B) requires the equipment grounding conductor to be increased in the same proportion.
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