Voltage Drop Calculator

How to use the voltage drop calculator

Enter the current the circuit carries in amps, the one-way distance from the panel to the load in feet, the copper conductor size, and the supply voltage, and the tool returns the volts lost along the run, that loss as a percentage of the supply, the voltage actually arriving at the load, and a verdict on whether the run is sized sensibly. The defaults model a very common case: 20 A over 100 ft of 12 AWG copper on a 120 V circuit. Copper is assumed throughout, which is what most branch wiring uses. Enter the current the load genuinely draws rather than the breaker rating, since drop scales directly with amps and sizing to a breaker you never approach will send you to needlessly heavy cable.

Wire is not a perfect conductor. It has resistance, and every foot of it converts a little of your supply voltage into heat instead of delivering it to the load. Thicker wire has lower resistance per foot, which makes the gauge selector the main lever you have. The values behind it come from NEC Chapter 9, Table 8, which lists conductor DC resistance at 75 C: 12 AWG solid uncoated copper is 1.93 ohms per 1,000 ft, 10 AWG is 1.21, and 8 AWG is 0.764. Each step up the gauge scale cuts the resistance by roughly a third, so moving two sizes from 12 to 8 more than halves the loss on an identical run without changing anything else.

This is the tool for the question that starts why does my shed, garage or detached workshop run tools weakly. A circuit can be entirely safe and still deliver soft voltage at the far end when the run is long and the wire is thin, because the loss happens along the cable rather than at the source, and a meter at the panel reads perfectly normal. Motors run hot and lose torque on low voltage, incandescent lighting dims noticeably, and some electronics behave strangely. Two things drive the loss and both sit in the formula: length and current. Double the distance and the drop doubles; double the amps and it doubles again. A long feed to an outbuilding is the classic problem case precisely because it maximises both at once.

Read the percentage against the recognised targets. NEC 210.19(A) carries an Informational Note recommending that branch-circuit voltage drop at the farthest outlet not exceed 3 percent, with the combined feeder and branch-circuit drop kept to 5 percent, and NEC 215.2(A) carries a parallel note for feeders. Both are recommendations rather than code requirements, because NEC 90.5 states plainly that Informational Notes are informational only and not enforceable as requirements. In practice energy codes and local amendments often make the 3 percent figure binding anyway, and it is defensible engineering regardless. Use this to plan a run before you buy cable or to diagnose one already misbehaving, and have a licensed electrician size and install anything permanent.

The formula

This is Ohm law applied to the cable rather than the load. Current flowing through resistance produces a voltage across that resistance, and here the resistance belongs to the conductors themselves. The only detail that catches people out is the factor of two: current leaves the panel along one conductor and returns along the other, so the resistance you are fighting covers twice the one-way distance you measured.

Conductor resistance is tabulated per 1,000 ft, so it is divided by 1,000 to give ohms per foot before being multiplied by the round-trip length. The percentage is then that loss compared with the supply voltage you started from, which is why the identical absolute drop is only half as serious on a 240 V circuit as on a 120 V one, and why long feeds to outbuildings are so often run at 240 V.

voltage drop = 2 × distance × current × (ohms per 1000 ft ÷ 1000)
percent drop = voltage drop ÷ supply voltage × 100
voltage at the load = supply voltage − voltage drop
2 × 100 ft × 20 A × 0.00193 = 7.72 V

Worked example with the defaults: 20 A over 100 ft of 12 AWG copper at 1.93 ohms per 1,000 ft gives 2 × 100 × 20 × 0.00193 = 7.72 V of loss. Against a 120 V supply that is 7.72 ÷ 120 × 100 = 6.4 percent, leaving 112.28 V at the load. That is over the 5 percent figure the NEC notes give for the whole path, so the verdict says to upsize. Step to 8 AWG copper at 0.764 ohms per 1,000 ft and the same run drops just 3.06 V, or 2.5 percent, comfortably inside the 3 percent branch-circuit recommendation.

Change the supply voltage instead of the wire and watch what happens. The same 20 A over the same 100 ft of 12 AWG still loses 7.72 V in absolute terms, because nothing about the conductor changed, but on a 240 V circuit that is only 3.2 percent rather than 6.4. The absolute loss is identical; its significance halves. That is the real reason 240 V is preferred for long feeds, alongside the fact that a given amount of power at 240 V needs half the current, which halves the absolute drop as well and compounds the advantage.

Three limits worth naming. This calculator assumes copper: NEC Chapter 9, Table 8 gives 12 AWG aluminium at about 3.18 ohms per 1,000 ft against copper 1.93, roughly 65 percent higher, so an aluminium run of the same gauge drops correspondingly more. It uses tabulated DC resistance at 75 C, which is close enough for ordinary branch circuits but not the AC impedance figure a long high-current feed really wants. And voltage drop is only one of several constraints on conductor size, alongside ampacity, temperature correction, conduit fill and overcurrent protection. This is a planning number, never a substitute for a licensed electrician sizing and installing the circuit.

Frequently asked questions