Voltage Drop Calculator
Work out how much voltage disappears between the panel and the load. Enter the current, the one-way distance, the conductor size and material, and the calculator returns the volts lost, the percentage, and the smallest size that keeps the drop at or under the 3% the NEC suggests.
The 3% limit is an NEC informational note (210.19(A)), not an enforceable rule — your inspector, AHJ or equipment spec may want something tighter. Resistance here is the DC value at 75 °C; long runs on large conductors also see some AC reactance, so treat the result as the usual design estimate.
Formula and calculation
Worked example
A 30 A load on a 240 V single-phase circuit 100 ft from the panel, wired in 8 AWG copper: VD = 2 × 12.9 × 30 × 100 ÷ 16,510 = 4.69, which is 1.95% of 240 V, so the load sees 235.31. Within the 3% recommendation for a branch circuit.
What drives voltage drop
Two numbers drive the drop: the resistance of the conductor and the current through it. Doubling the run doubles the drop, and so does doubling the current. Changing from 14 AWG to 10 AWG cuts the resistance to about 40% of what it was, which is why upsizing is usually the cheapest fix.
As a rule of thumb, 14 AWG and 12 AWG branch circuits are fine at their rated 15 A and 20 A out to about 50 ft. Beyond roughly 100 ft, run the numbers: the voltage at the load is what the equipment actually runs on, and motors, compressors and LED drivers all dislike a sagging supply.
What the results mean
- Voltage drop
- The volts lost between the source and the load, from VD = 2 × K × I × L ÷ CM (1.732 instead of 2 for three-phase). This is the number that eats into what your equipment actually sees.
- Drop as a share of supply
- The drop as a share of the source voltage. The NEC suggests 3% or less on a branch circuit and 5% or less over the feeder plus branch. Those are informational notes, not rules, but they are the usual design target.
- Voltage at the load
- What the equipment at the far end actually gets: source voltage minus the drop. Motors, compressors and electronics all run hotter and less reliably as this falls.
- Against the 3% guideline
- A pass or fail read against the 3% suggestion, in plain words, for the conductor you selected.
- Smallest conductor for 3%
- The smallest conductor on the size list that keeps the drop at or under 3% for this current and distance, with the circular mils needed — or a note that no listed size is enough.
- Substituting your numbers
- The same formula with your numbers substituted, so you can check the arithmetic by hand or show it on a permit drawing.
Frequently asked questions
What is an acceptable voltage drop?
The NEC suggests 3% or less on a branch circuit and 5% or less from the service to the farthest outlet. Those figures appear in informational notes, which are recommendations rather than enforceable rules, but they are what most designers and inspectors work to. On a 120 V circuit 3% is 3.6 V; on 240 V it is 7.2 V.
Does voltage drop change with the supply voltage?
The volts lost do not — they depend on current, conductor and length only. The percentage does. A 6 V drop is 5% of a 120 V circuit but just 2.5% of a 240 V circuit, which is why higher-voltage circuits reach farther on the same wire and why 240 V runs are preferred for long or heavy loads.
How do I fix excessive voltage drop?
Four levers, in rough order of cost: upsize the conductor, shorten the run, reduce the current, or raise the supply voltage. Going up one or two AWG sizes fixes most residential runs. For very long feeders, parallel conductors or a transformer closer to the load can be cheaper than one enormous cable.
Why does the formula use 2 for single-phase?
In a two-wire circuit the current flows out along one conductor and back along the other, so the resistance of the whole loop drops the voltage, and the loop is twice the one-way length. Three-phase circuits share the return path between phases, and the multiplier becomes 1.732 instead of 2.
Does this match what an electrician would calculate?
Yes — it is the K method taught in US trade textbooks and apprentice exam prep, with K = 12.9 for copper and 21.2 for aluminum taken from the 75 °C DC resistance in NEC Chapter 9, Table 8. Real installations also have terminations, bundled cable heating and AC reactance, so treat the answer as the design estimate it is.