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Drawing No. EH–EE–028 // Electrical Engineering

Voltage Drop Calculator

Reviewed October 2026

Calculate voltage drop, percentage loss, load-end voltage and conductor loss for DC, single-phase and balanced three-phase circuits using AWG, kcmil or mm² conductors.

Scope: Preliminary calculation only. Verify important quantities and local requirements before purchase, construction or installation.

How to use this calculator

Enter the circuit voltage, current, one-way length and conductor size. Select copper or aluminium and the circuit type.

1 — EnterUse the dimensions or values you know.
2 — ReviewCheck units and assumptions shown beside each field.
3 — ApplyAdd a practical allowance before ordering or building.

Inputs

Results

Voltage drop
—
Estimated conductor drop
Percentage drop
—
Relative to supply
Load-end voltage
—
Supply minus drop
Conductor loss
—
Approximate I²R loss

The equation, and the two terms this page leaves out

What is being computed, and the size of what is not.

Voltage drop along a conductor is an impedance problem, not a resistance problem. The general form for a balanced circuit is

ΔV = k · I · L · (R′·cosφ + X′·sinφ)  k = 2 for DC and single-phase, √3 for balanced three-phase

where R′ and X′ are resistance and reactance per unit length and L is the one-way route length. The factor of 2 on a single-phase circuit accounts for the return conductor; the √3 on a three-phase circuit is the line-to-line relationship, not a third conductor — a frequent source of error when the two are mixed up.

Conductor resistance is temperature dependent:

R₂₀ = ρ₂₀ / A  R(T) = R₂₀ · [1 + α(T − 20)]

with ρ₂₀ = 1.724 × 10⁻⁸ Ω·m and α = 0.00393 /K for annealed copper, 2.826 × 10⁻⁸ Ω·m and 0.00403 /K for aluminium. Running a copper conductor at its 70 °C rated temperature rather than 20 °C raises its resistance by 20%; at 90 °C, by 28%.

What the calculator above does: it applies temperature correction and computes ΔV = k·I·L·R(T)·cosφ — the resistive term only. It has no reactance input. For small conductors that is a good approximation; for large ones it is not, and the next two sections quantify when the difference starts to matter.

Worked example: a 120 m three-phase submain

400 V, 80 A, copper, power factor 0.85, conductors at 70 °C.

A 400 V three-phase submain feeds an 80 A load 120 m away. Trial size 35 mm² copper, power factor 0.85 (so sinφ = 0.527), conductors assumed at their 70 °C operating temperature. For a 35 mm² multicore LV cable the reactance is about 0.08 Ω/km.

R₂₀ = 1.724×10⁻⁸ / 35×10⁻⁶ = 0.493 Ω/km → R₇₀ = 0.493 × 1.197 = 0.589 Ω/km ΔV = √3 × 80 × 0.120 × (0.589×0.85 + 0.080×0.527) = 9.03 V → 2.26%

Now the same circuit with each refinement removed in turn:

MethodΔV% of 400 VRelative to full calculation
R at 70 °C, with reactance9.03 V2.26%100%
R at 70 °C, no reactance (this calculator)8.33 V2.08%92%
R at 20 °C, with reactance7.66 V1.92%85%
R at 20 °C, no reactance6.96 V1.74%77%

Every omission runs the same way: it makes the cable look better than it is. Taking the bottom row and comparing it against a 2% internal design target would pass a circuit that actually sits at 2.26%. Neither error is large on its own; together they are 23%, and they are both non-conservative.

Two further points this example does not show. Real stranded cable resistance runs roughly 5–10% above the ρ/A figure because of the lay of the strands — manufacturer or standard tabulated values are preferable to the ideal calculation where the margin is tight. And the load current here is the design current; for a continuous load a code may require the circuit to be assessed at 125% of it.

When reactance stops being negligible

Cable reactance is roughly constant with size. Resistance is not.

For LV multicore cables the reactance per unit length sits near 0.07–0.09 Ω/km almost regardless of conductor area — it is set by conductor spacing and geometry, not by cross-section. Resistance, meanwhile, falls in inverse proportion to area. So the X/R ratio climbs steadily with cable size:

Copper conductorR at 70 °CX (typical multicore)X / RReactance contribution at pf 0.85
2.5 mm²8.25 Ω/km0.080.01negligible
16 mm²1.29 Ω/km0.080.06under 4%
35 mm²0.589 Ω/km0.080.14about 8%
95 mm²0.217 Ω/km0.080.37about 19%
185 mm²0.112 Ω/km0.080.72about 32%
240 mm²0.086 Ω/km0.080.93about 37%
400 mm²0.052 Ω/km0.081.55about 49%

Reactance overtakes resistance somewhere around 240–300 mm². Above that point a resistance-only calculation is not an approximation, it is half the answer — and the error grows worse as power factor falls, because the reactive term scales with sinφ. At pf 0.7 the reactive contribution on a 240 mm² run is larger again.

Practical boundary: use the resistance-only result up to roughly 95 mm² / 3/0 AWG and at power factors above 0.9, where the omission is a few percent. Beyond that, take R and X from the cable manufacturer's data or from the tabulated (r, x) values in IEC 60364-5-52 Annex G or NEC Chapter 9 Table 9, and compute the impedance drop properly. Note also that increasing conductor size above the crossover buys progressively less voltage-drop improvement — paralleling smaller cables or raising the distribution voltage becomes the better move.

What counts as too much drop

The headline figures, and the widely misunderstood status of the American ones.

FrameworkLightingOther usesStatus
NEC (US) — branch circuit3%3%Informational note, not a requirement
NEC (US) — feeder + branch total5%5%Informational note, not a requirement
BS 7671 (UK), public LV supply3%5%Appendix 4, from origin of installation
IEC 60364-5-52, public LV supply3%5%Annex G recommendation
IEC 60364-5-52, private supply6%8%Annex G recommendation

The NEC point is worth dwelling on, because it is misquoted constantly. NEC 210.19(A) and 215.2(A)(1) place their voltage-drop figures in informational notes, which Article 90 states are not enforceable as Code requirements. They are good engineering practice and most specifications adopt them, but a circuit exceeding 3% is not by itself a Code violation in general-purpose work. Some articles do impose mandatory limits for specific applications — sensitive electronic equipment, fire pumps and certain PV circuits among them — and a project specification or local amendment can make the 3%/5% figures contractual. Check which applies before arguing either way.

BS 7671 and IEC 60364 are stated differently: the limit applies from the origin of the installation to the point of use, so the distribution drop and the final-circuit drop share one budget. Spending 4% on a long submain leaves 1% for everything downstream of it.

Why the limit exists at all

What this calculation does not check

Frequently Asked Questions

Practical questions about assumptions, inputs, interpretation and limitations.

Where this sits in a cable calculation

Voltage drop is one of four checks. A conductor has to pass all of them.

  1. Step 01Design current Convert the connected load to a design current, apply diversity and the continuous-load uplift, and establish the protective device rating.
  2. Step 02Ampacity and derating Base ampacity for the conductor and insulation, corrected for ambient temperature, grouping and installation method. The derated value must exceed the device rating.
  3. Step 03 — this pageVoltage drop Check the drop at the design current over the route length against the applicable limit. On long runs this usually governs and forces a size above the ampacity answer.
  4. Step 04Short-circuit withstand and final size Adiabatic check against prospective fault current and clearing time, then the governing size across all four checks.

Working the steps out of order wastes effort: sizing for voltage drop before the ampacity answer is known often produces a conductor that then fails the fault check anyway, and the final size is whichever of the four is largest.

Sources