Voltage Loss Calculator

How much voltage are you losing in your wire run?

Voltage loss in a conductor equals twice the one-way cable length multiplied by current and resistivity, divided by the wire cross-sectional area; a drop exceeding 3 of supply voltage is generally considered excessive for branch circuits. Voltage loss across a conductor depends on wire gauge, cable length, current draw, and conductor material. Enter your circuit details to find out whether your installation falls within acceptable limits and what percentage of supply voltage you are losing to resistance.

Updated August 2026 · How this works

Example calculation — edit any field to use your own numbers

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Worth knowing
How It Works
The formula, explained simply

Think of a garden hose: the longer and narrower the hose, the more pressure you lose by the time water reaches the sprinkler. Electrical wire works the same way — every conductor has resistance, and resistance steals voltage before it reaches your load. A light bulb, motor, or appliance at the end of a long run gets slightly less voltage than the panel delivers, and that deficit is voltage drop.

The calculator uses Ohm's Law (voltage equals current multiplied by resistance) applied to the full round-trip path of the circuit. You enter the one-way cable run in feet; the tool doubles it because current must travel to the load and return through the return conductor, both of which add resistance. The resistance per foot depends on the American Wire Gauge (AWG) number and whether the conductor is copper or aluminum. Lower AWG numbers mean physically thicker wire and lower resistance per foot — 4 AWG is far thicker than 14 AWG, which is why large motors use large conductors.

The percent voltage drop is the calculated drop divided by your supply voltage, expressed as a percentage. This percentage is what really matters to your equipment: a 5-volt drop on a 480 V circuit is trivial, but the same 5-volt drop on a 120 V circuit takes away over four percent of the supply. The compliance verdict shown below the result tells you directly whether your configuration meets, approaches, or exceeds the commonly used 3 percent guideline for branch circuits.

When To Use This
Right tool, right situation

Use this tool whenever you are planning a new circuit run, troubleshooting unexplained equipment faults, or checking whether an existing wire can handle a load increase. It is directly applicable to residential branch circuits, outbuilding subpanel feeds, EV charger installations, and commercial lighting or HVAC feeds where the panel is far from the load. The tool is also useful for solar and battery system DC wiring, where voltage drop at low voltages has a proportionally larger impact on system efficiency.

The tool is not appropriate as a substitute for a full load analysis by a licensed electrician when the circuit feeds life-safety equipment, medical devices, or large industrial machinery. It also does not account for derating due to conduit fill, high ambient temperatures, or continuous loads (which the NEC requires to be derated to 80 percent of breaker capacity). For those scenarios, calculate the derated current first and then use this tool with the derated figure.

A good rule of thumb: if this tool shows you at or above the 3 percent guideline, go up one AWG size and recalculate before ordering wire. The cost difference between 12 AWG and 10 AWG on a typical run is small compared to the cost of rewiring after the fact.

Common Mistakes
Why results sometimes look wrong

Mistake 1 — measuring one-way and forgetting to double. The most common error is entering the total round-trip distance as the one-way run, which doubles the calculated drop. The calculator already doubles the value you enter. If you measure 85 feet from panel to outlet, enter 85 — not 170. Entering the round-trip length will make a compliant circuit look like it fails.

Mistake 2 — using nameplate watts instead of actual running amps. A motor's rated wattage represents its output power, not its input current demand. A 1-horsepower motor draws more amps than its output wattage implies because of efficiency losses. Always use the measured or nameplate amperage, not a watt-to-amp conversion from output power, or the voltage drop calculation will understate the true drop during operation.

Mistake 3 — ignoring starting current on motor loads. The tool calculates drop at the steady-state running current you enter. Induction motors draw five to seven times running current during startup, briefly deepening the voltage drop for every other load on the same circuit. If the calculated drop is near the guideline at running current, starting transients will push it well above it. Size the conductors to hold drop well below the guideline if the circuit feeds motors.

The Math
Worked examples and deeper derivation

The core formula is: Voltage Drop = 2 × L × I × Rft, where L is the one-way cable length in feet, I is the current in amps, and Rft is the resistance per foot of one conductor in ohms per foot. The factor of 170 feet (for the example run of 170 ft total) accounts for the complete circuit path. The result is a voltage in volts.

For the example circuit — copper 12 AWG, 120 V supply, 15 A, 170 ft total path — the resistance of the full run is 0.3366 Ω. Multiplying by 15 A gives a drop of 5.05 V. Dividing by the 120 V supply and multiplying by 100 yields 4.21% percent drop. The voltage remaining at the load is 114.95 V.

Percent drop = (Voltage Drop / Supply Voltage) × 100. This normalisation is what makes the 3 percent guideline universal across 120 V, 240 V, and 480 V systems. The absolute drop number varies widely with supply voltage; the percent captures how much of the available pressure you are losing regardless of system voltage level.

Garage subpanel feed — copper 12 AWG, 120 V circuit
120 V supply, 15 A load, 85 ft one-way run, 12 AWG copper
The round-trip path is 170 ft. At 15 A through 12 AWG copper, the total resistance is 0.3366 Ω, producing a voltage drop of 5.05 V. That represents 4.21% of the 120 V supply, and the load sees 114.95 V. The compliance verdict is: Marginal (3–5%) — consider upsizing conductor. For a standard branch circuit this result is marginal — the run is long enough that upsizing to 10 AWG would bring the drop comfortably within the 3 percent guideline.
Commercial 480 V motor circuit — large aluminum conductor
480 V supply, 100 A load, 300 ft one-way run, 4/0 AWG aluminum
With a 600 ft round trip and 100 A, the circuit resistance is 0.0600 Ω. The resulting voltage drop is 6.00 V, leaving 474.00 V at the motor terminals. Percent drop is 1.25%. The verdict: Within guideline (≤ 3%) — acceptable for branch circuits. For a motor load, even a drop below the 3 percent guideline matters — starting current can be five to seven times the running current, temporarily deepening the drop during startup.
Short low-power lighting circuit — copper 14 AWG
120 V supply, 5 A load, 20 ft one-way run, 14 AWG copper
Even the thinner 14 AWG wire performs well over a 40 ft round trip. Resistance is 0.1257 Ω, and the drop is only 0.63 V — just 0.52% of supply. The load receives 119.37 V, and compliance reads: Within guideline (≤ 3%) — acceptable for branch circuits. This illustrates why short runs rarely need oversize wire: resistance accumulates with length, not with gauge alone.
Expert Unlock
The thing most explanations skip

The resistivity values in this tool are for conductors at 75 degrees Celsius — the standard temperature for NEC ampacity tables. A conductor running near its ampacity limit will be hotter, and resistance rises with temperature (roughly 0.4 percent per degree C for copper), so the actual drop at a loaded conductor will exceed the calculated value. For circuits loaded above sixty percent of ampacity in high-ambient environments, apply a temperature correction factor before trusting the result. Additionally, the formula assumes DC resistance; for AC circuits at power frequency (50 or 60 Hz), skin effect and proximity effect add a small impedance increment that becomes relevant for conductors larger than 2/0 AWG — the Chapter Nine of the NEC AC impedance values supersede the DC figures at those sizes.

Why does my voltage drop exceed 3 percent and what can I do about it?

What is the 3 percent voltage drop guideline and where does it come from?
The 3 percent figure is a widely used rule of thumb for branch circuits — it keeps voltage at the load close enough to nominal that equipment performs within its design window. The figure appears in general electrical engineering guidance and in recommendations from several code bodies as a practical target for branch-circuit wiring. It is not a hard legal ceiling in all jurisdictions, but going above it risks equipment running outside its rated voltage tolerance, which can shorten motor life, cause lighting flicker, and produce heat in sensitive electronics.
Why does a longer wire run cause more voltage drop even if the gauge is the same?
Resistance is proportional to length: double the run, double the resistance, double the drop. Voltage drop follows Ohm's law — drop equals current multiplied by resistance — so every extra foot of wire adds a small but real resistance in series with the load. The calculator doubles your one-way entry because current flows out through one conductor and returns through the other, so the effective resistive path is the full round trip. A 100-foot run actually means 200 feet of current-carrying conductor.
Should I use copper or aluminum wire to reduce voltage drop?
Copper has roughly 61 percent of aluminum's resistivity, which means a copper conductor of the same gauge carries the same current with significantly less voltage drop. Aluminum wiring is common in service entrances and large feeders because it is lighter and cheaper per foot for high-ampacity runs, but it requires larger gauge for the same performance as copper. If drop is your primary concern and weight or cost is secondary, copper wins at equivalent gauge. If you must use aluminum, step up at least two AWG sizes to approximate the drop performance of copper.

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