How to Calculate Voltage Drop

What you need first

Five things. If any of them is a guess, the answer inherits the guess.

For AC below unity power factor you also want the power factor, and for large conductors the raceway type, because both feed into the reactance term.

The five steps

1. Look up the conductor resistance

From the resistance table, in ohms per 1000 ft at 75 °C. 12 AWG stranded copper is 1.98. 4/0 copper is 0.0608. Aluminium is roughly 1.64 times the copper figure at the same size.

If the conductor will run substantially hotter or cooler than 75 °C, correct it: R₂ = R₁ × [1 + α × (T₂ − 75)], with α of 0.00323 for copper and 0.00330 for aluminium.

2. Reduce the run to a one-way length

In thousands of feet, because that is what the resistance is quoted per. A 150 ft run is 0.15. If you measured total cable length, halve it first.

3. Pick the multiplier

2 for DC and single-phase, because the current goes out and comes back. √3, which is 1.732, for balanced three-phase, because no conductor carries the full return current. The derivation is on the formula page.

4. Multiply through

5. Convert to a percentage

The percentage is what design targets are written against, and what makes results comparable between a 12 V circuit and a 480 V one.

Worked through

A 30 A single-phase load 175 ft from a 240 V panel on 8 AWG copper.

Step 1  R = 0.778 Ω per 1000 ft      (8 AWG stranded copper, 75 °C)
Step 2  L = 175 ÷ 1000 = 0.175
Step 3  multiplier = 2               (single-phase)
Step 4  V_drop = 2 × 30 × 0.778 × 0.175
        V_drop = 8.17 V
Step 5  %V_drop = (8.17 ÷ 240) × 100 = 3.40%
        V_load  = 240 − 8.17 = 231.83 V

3.40% is just over a 3% target. One size up to 6 AWG (0.491 Ω per 1000 ft) gives 5.15 V and 2.15%, which clears it with margin. That is usually the shape of the decision: the first size that passes, plus a look at whether the next one up is worth the money.

The three big mistakes

1. Using the round-trip length as if it were one-way

This doubles the answer. The multiplier already accounts for the return conductor. That is the entire reason the multiplier exists. If the load is 100 ft away, the one-way length is 100 ft, regardless of the fact that you pulled 200 ft of cable to reach it.

The reverse mistake, halving a genuine one-way measurement because "the current goes both ways", halves the answer, and is worse, because it makes an inadequate conductor look adequate.

2. Getting the units of L wrong

Resistance per 1000 ft needs length in thousands of feet. Substituting 175 instead of 0.175 gives an answer a thousand times too large. This error is at least self-announcing: if your 240 V circuit reports an 8000 V drop, you have found it.

The circular-mil K method uses plain feet instead, because the constant absorbs the factor of 1000. Mixing the two conventions is where this usually starts.

3. Using the wrong multiplier

Choosing single-phase for a three-phase circuit overstates the drop by 15%, so you buy conductor you did not need. Choosing three-phase for a single-phase circuit understates it by the same 15%, which is the dangerous direction.

Checking your answer

Four checks that catch most errors before they cost anything.

• Is the percentage plausible? Branch circuits usually land between 0.5% and 8%. A result of 0.02% or 400% means a units error.
• Does doubling the length double the drop? It must. If it does not, something is wrong with the arithmetic.
• Cross-check with the K method.V_drop = (2 × 12.9 × I × L_feet) ÷ cmil for single-phase copper. The two methods should agree within about 1%, and they use completely different inputs, so agreement is real evidence.
• Compare against the chart.The voltage drop chart lists percentage drop by size and distance; find the nearest row and column and see whether your answer sits between them.

Two useful shortcuts

Volts per amp per 100 ft

Multiply the conductor resistance in Ω per 1000 ft by 0.2 and you have volts per amp per 100 ft on a two-wire circuit. It turns every subsequent calculation into one multiplication.

For three-phase, multiply the result by √3 ÷ 2 = 0.866.

Scaling a known result

Voltage drop is linear in current and in length, and inversely proportional to conductor area. So once you have one answer you can move it around without starting over:

• Double the current → double the drop
• Halve the length → halve the drop
• Go up one AWG size → drop falls to about 79% (each size is ~1.26× the area)
• Go up three AWG sizes → drop roughly halves (area roughly doubles)
• Copper to aluminium at the same size → drop rises about 64%
• Two conductors in parallel → drop halves

Questions

Five full examples are worked out on the voltage drop calculation page, and the calculator shows the same substitutions live as you type.