Voltage Drop Formula

The one formula

Every voltage drop calculation on this site, and in every textbook, reduces to Ohm’s law applied along a conductor:

Everything below is this equation with something substituted in. Pick the right k for your system, get Z right for your conductor and load, and measure L one way rather than there and back, and the answer follows.

DC circuits

DC is the simplest case. There is no reactance, no power factor and no phase angle: the only thing opposing current is resistance, so Z is just R. This makes the DC formula exact rather than approximate. Whatever error there is in the answer comes from not knowing the current, the length or the conductor temperature precisely, never from the formula itself.

Because low-voltage DC systems draw high current for modest power and are compared against a small supply voltage, the percentage drop climbs fast. A 12 V circuit losing half a volt has already lost 4.2%. See the DC voltage drop calculator for the sizes and distances that go with common battery and solar voltages.

Single-phase AC

The multiplier is the same 2 as DC, for the same reason: a two-wire circuit carries the full load current out along one conductor and back along the other. What changes is that Z is now an impedance rather than a plain resistance, because an alternating current sets up a magnetic field around the conductor and that field opposes changes in the current.

At unity power factor (a heater, an incandescent lamp), the reactive part contributes nothing and Z_e collapses back to R. The further the load is from unity, the more the reactance matters.

Three-phase AC

This gives the line-to-line voltage drop, which is what you compare against a line-to-line supply voltage like 208 V, 400 V or 480 V. If you need the line-to-neutral drop on a balanced system, divide the result by √3, or calculate with the line-to-neutral voltage and a multiplier of 1. Mixing the two is a common way to produce an answer that is out by 73%.

Where 2 and √3 come from

This is the part most explanations skip, and it is the part worth understanding because it stops you ever picking the wrong one.

The 2

In a two-wire circuit, whether DC or single-phase, the current has to complete a loop. It leaves the source, travels the length of the run through one conductor, passes through the load, and returns the length of the run through the other. So the current passes through two run-lengths of conductor, and each contributes I × R × L of drop. Total: 2 × I × R × L.

The √3

In a balanced three-phase circuit there is no dedicated return conductor. Each of the three line currents is 120° out of phase with the others, and at every instant they sum to zero: each conductor acts as the return path for the other two. Current therefore traverses only one run-length per phase.

That gives a line-to-neutral drop of I × Z × L. But three-phase supplies are quoted line-to-line, and a line-to-line voltage is √3 times the line-to-neutral voltage. Scaling the drop the same way gives √3 × I × Z × L.

Impedance and power factor

For AC, the conductor opposes current with resistance R and inductive reactance X_L. The two act at right angles to each other, but voltage drop is not the magnitude of that combination. It is the component of the conductor impedance lying in the direction of the load current:

This is stated in the notes to NEC Chapter 9, Table 9, which also publishes an effective-impedance column computed at 0.85 power factor. Recomputing that column from the table’s own R and X_L values is one of the checks the calculation engine on this site is tested against.

Note what happens at the extremes. At PF = 1, cos θ = 1 and sin θ = 0, so Z_e = R: reactance drops out entirely. At PF = 0.7, sin θ is about 0.71, so the reactance term is contributing nearly as much as it possibly can. This is why a resistance-only estimate is exactly right for a resistive load and progressively optimistic for a motor.

The AC calculator shows how much a resistance-only figure understates the drop across the whole power factor range.

The circular-mil K method

The form most often taught in the trade avoids looking anything up. Instead of a tabulated resistance it uses the conductor’s cross-sectional area in circular mils and a constant that carries the resistivity of the metal:

Note that L here is in plain feet, not thousands of feet: the K constant absorbs the factor of 1000. That difference between the two forms is worth watching for.

How the two methods compare

Take 20 A on 12 AWG copper over 100 ft, single-phase. 12 AWG is 6530 circular mils.

A 0.3% difference, which is the rounding in the K constant. Both are fine for design work. The table method wins when you want the actual published resistance for a specific stranding and temperature; the K method wins when you are working from a conductor area and have no table to hand.

The metric form

Outside North America, resistance is quoted in ohms per kilometre and conductor areas in mm². The formula takes the same shape with resistivity in place of a constant:

Worked through: 20 A on 4 mm² copper over 30 m, single-phase, gives (2 × 0.0209 × 30 × 20) ÷ 4 = 6.27 V. On a 230 V supply that is 2.7%.

Where resistance comes from

Resistance is not a property of a wire alone: it depends on the metal, the cross-sectional area, the length and the temperature:

At 20 °C, copper is 1.7241 × 10⁻⁸ Ω·m. This value is the definition of 100% IACS, the conductivity standard everything else is measured against. Aluminium used in conductors is 61% IACS, which works out to 2.8264 × 10⁻⁸ Ω·m.

Resistivity rises with temperature, and the correction that goes with the standard reference tables is:

Going from 75 °C to 90 °C raises copper resistance, and therefore voltage drop, by about 4.8%. Running cool works the other way: at 25 °C, resistance is about 16% lower than the tabulated figure.

Percentage and load voltage

The percentage is what design limits are written against, and it is why the same volts lost matter far more at 12 V than at 480 V. Half a volt is 4.2% of a 12 V supply and 0.1% of a 480 V one.

Rearranged for size and length

The formula has one unknown appearing once and linearly, so it rearranges exactly for whichever quantity you actually need.

Take the smallest listed conductor whose impedance is at or below this figure. That is what the wire size calculator does.

The answer is in the same length unit as Z is quoted per: thousands of feet, if Z is in Ω per 1000 ft. This is what the maximum cable length calculator computes.

Every variable

Questions

To see any of these formulas run with your own numbers, and with each substitution written out, use the voltage drop calculator. For five complete examples worked start to finish, see voltage drop calculation.