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Calculation details
Planning estimate only, not financial, tax, or legal advice. Verify assumptions and current rules before making decisions.
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What does this calculator estimate?
Voltage drop calculator: find wire voltage drop and end of run voltage from length, current, wire gauge, and material, AC or DC. Free, private, runs locally.
How this calculator works
Formula
Voltage drop = 2 x length (ft) x amps x resistance per 1,000 ft / 1,000 (round-trip run); drop % = drop / volts x 100.
Worked example
120 V circuit, 15 A, 100 ft of AWG 0 wire: 4.76 V dropped across the round trip, a 3.97% drop, under the 5% rule of thumb.
Cite this tool
BoringToolsKit. “Voltage Drop Calculator.” boringtoolskit.com/voltage-drop-calculator/ (reviewed August 2026). Free to reference in articles, syllabi, and answer posts with a link.
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Worked example: a 12V battery feed to a 10A inverter
Take a 12V system feeding a 10A load over a 50 ft run of 12 AWG copper, using this calculator's resistance table (1.588 ohms per 1000 ft). The formula first converts the table value to the actual run length: 1.588 x 50 / 1000 = 0.0794 ohms one way. The wire carries current out and back, so the length factor doubles: 0.0794 x 2 = 0.1588 ohms of total round-trip resistance. The calculator shows this as its 'Wire resistance' figure, displayed as 0.16 ohms after rounding.
Ohm's law turns that resistance into lost volts: 10 A x 0.1588 ohms = 1.588 V, which the result hero shows as 1.59 V. As a percentage of the supply, 1.588 / 12 x 100 = 13.23%, so the load sees about 10.41 V instead of 12 V. That sits well above the common 3% target for sensitive loads, and the fix is visible in the table itself: repeating the same math at 6 AWG (0.395 ohms per 1000 ft) gives 0.0395 ohms round-trip, a 0.395 V drop, and 3.29%. Note the small oddity in the numbers: 6 AWG alone lands just over 3% here, so a run like this needs either a shorter run, a lower current, or the next size down.
One check on the inputs: the calculator refuses to run on non-positive values. Volts of 0, amps of 0, or length of 0 all return the 'Enter positive values' message, because a drop percentage divides by volts and any zero or negative supply voltage would make the percentage meaningless.
Where the 2x multiplier and the resistance table bend
The built-in doubling is only correct when the load sits at the far end of the cable. If you tap loads along the run, the effective electrical length is shorter. A 100 ft run with the load at the midpoint behaves like a 50 ft run in this formula, so the calculator overstates its drop. Conversely, a motor that draws its full rating only at startup sees much higher momentary drop: a 10A motor pulling 40A inrush on the 50 ft 12 AWG example drops 6.35 V, over half a 12 V bank, for the fraction of a second the inrush lasts.
The resistance values are DC numbers for standard annealed copper at roughly 20 degrees C. Two real-world effects push actual drop higher. First, resistance rises with temperature, on the order of 0.4% per degree C, so a conductor at 60 degrees C carries about 16% more resistance than the table value; the 1.59 V example becomes roughly 1.84 V. Second, alternating current adds skin effect and inductive reactance, which matter at longer AC runs and are not modeled here.
Practically, the 3% and 5% conventions are limits you apply, not rules the calculator enforces. It reports the raw percentage and leaves the judgment to you. It also does not check ampacity: 12 AWG copper is typically rated near 20-25 A depending on the standard applied, and the table tops out at 6 AWG. A request for 2 AWG or a 200 A feed silently falls back to the 12 AWG value, which understates drop badly, so keep entries inside the listed gauges and verify conductor sizing separately.