What is the voltage drop formula for a single-phase circuit?

V_drop = 2 × L × I × R / 1000, where L is the one-way length in metres, I is the load current in amps, and R is the conductor resistance in Ω/km. The factor of 2 accounts for current flowing out on the active and back on the neutral — you always double the one-way length on a single-phase circuit.

What is the NEC voltage drop limit?

The NEC recommends (but does not generally require) a maximum of 3% on the branch circuit alone and 5% combined feeder plus branch circuit at the furthest outlet, per the informational notes to 210.19(A) and 215.2(A). It becomes a hard limit for sensitive-load articles like 695.7 (fire pumps) and 647.4 (sensitive electronic equipment).

What is the BS 7671 voltage drop limit?

BS 7671 Appendix 4 mandates 3% for lighting circuits and 5% for all other circuits, measured end-to-end from the origin of the installation to the furthest point of utilisation. Unlike the NEC, submain and final circuit drops are summed. AS/NZS 3000 aligns with this at 5% end-to-end.

When do I need to include cable reactance in voltage drop?

For cables over about 16 mm² (6 AWG) inductive reactance stops being negligible, and the full formula V_drop = k × L × I × (R·cosφ + X·sinφ) / 1000 should be used, where k is 2 for single-phase or √3 for three-phase and φ is the load power factor. For smaller cables and resistive loads, cosφ ≈ 1 and the shorter resistance-only formulas are accurate enough.

Voltage Drop Calculations: NEC & BS 7671 Guide

Voltage-drop formulas for apprentices: single-phase 2LIR, three-phase √3LIR, worked examples, and NEC 3%/5% vs BS 7671 limits.

Voltage drop is the single calculation an apprentice will do more than any other on site, and the one that catches most people out in the licence exam. Get it wrong on paper and your feeder overheats; get it wrong on the exam and you lose an easy mark. This guide walks through the maths — single-phase and three-phase, NEC and BS 7671 — with worked examples you can copy straight into your notes.

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What voltage drop actually is

Every conductor has resistance. When current flows through it, some of the supply voltage is "spent" pushing current through that resistance before it reaches the load. That lost portion is voltage drop, and it shows up as a hot cable, a dim light at the end of a long run, and a motor that struggles to start.

Two numbers matter: the absolute drop in volts, and the percentage drop relative to nominal supply voltage. Standards are written around the percentage, because a 6V drop is fine on a 480V feeder and catastrophic on a 12V DC circuit.

The core formulas

There are two forms every electrician needs to know cold.

Single-phase (two-wire) — the workhorse

V_drop = 2 × L × I × R / 1000

Where L is the one-way length in metres (or feet), I is the load current in amps, and R is the conductor resistance in ohms per kilometre (or per 1000 ft). The factor of 2 accounts for the current flowing out on the active and back on the neutral — so you always double the one-way length.

Three-phase (balanced) — the commercial one

V_drop = √3 × L × I × R / 1000

On a balanced three-phase circuit the neutral carries no net current, so instead of doubling the length you multiply by √3 (≈ 1.732). Same L, I and R as before, just a different geometric factor. This is the formula that turns up on almost every commercial feeder problem in the exam.

The impedance version — long runs and big cables

For cables over about 16 mm² (6 AWG), inductive reactance starts to matter and you can no longer treat the cable as pure resistance. The full form is:

V_drop = k × L × I × (R·cosφ + X·sinφ) / 1000

Where k is 2 for single-phase or √3 for three-phase, X is the reactance in Ω/km, and φ is the load power factor. For resistive loads (heaters, incandescent lights) cosφ ≈ 1 and the reactance term drops out — which is why the shorter formulas above are exact for small resistive circuits and a good approximation for everything else up to ~16 mm².

Percentage drop and the standards

Every code writes its voltage-drop rules as a percentage of nominal supply voltage, but they don't agree on the number. Know both if you're studying for a licence in a jurisdiction that references either.

NEC (US) — 210.19(A) and 215.2(A) informational notes

The NEC is famously not a design manual and does not require a voltage-drop limit for general branch circuits. It recommends:

3% maximum on the branch circuit alone.
5% maximum combined feeder + branch circuit at the furthest outlet.

Sensitive-load exceptions (fire pumps in 695.7, sensitive electronic equipment in 647.4) do turn these recommendations into hard limits — worth knowing for the exam.

BS 7671 (UK / Ireland) — Appendix 4, section 6

BS 7671 is stricter and mandatory:

3% for lighting circuits.
5% for other uses (power, heating).
Measured from the origin of the installation (the incoming supply terminals) to the furthest point of utilisation — so unlike the NEC, submains and final circuits are summed.

AS/NZS 3000 is essentially aligned with BS 7671 here: 5% end-to-end, measured from the point of supply.

Worked example 1 — single-phase branch circuit (NEC)

A 20A 120V branch feeds a workshop socket 25 m away. Cable is 12 AWG copper, resistance 5.2 Ω/km. Is the drop within NEC's 3% recommendation?

V_drop = 2 × 25 × 20 × 5.2 / 1000 = 5.2 V

5.2 / 120 = 4.3%. That's over the 3% branch-circuit recommendation. Options: shorten the run, drop the load, or upsize the cable to 10 AWG (3.3 Ω/km) which gives 3.3 V or 2.75%.

Worked example 2 — three-phase feeder (BS 7671)

A 100A three-phase 400V feeder runs 60 m to a distribution board. Cable is 25 mm² copper, resistance 0.87 Ω/km. Power factor 0.9 lagging, negligible reactance. What's the drop, and is it within BS 7671's 5%?

V_drop = √3 × 60 × 100 × 0.87 / 1000 = 9.04 V

9.04 / 400 = 2.26%. Comfortably inside 5%, and it still leaves headroom for whatever final circuits the board feeds — remember BS 7671 measures end-to-end, so you need to add the drop of the worst-case final circuit before signing off.

Worked example 3 — long DC / EV circuit

A 32A EV charger on a 230V single-phase supply, 40 m from the DB, cable 6 mm² copper (3.08 Ω/km).

V_drop = 2 × 40 × 32 × 3.08 / 1000 = 7.88 V

7.88 / 230 = 3.43%. Under the 5% BS 7671 limit for a power circuit — but if you already used 1.5% on the submain, you're at 4.9% and the next apprentice adds a socket and busts the limit. Upsize to 10 mm² and the run drops to 2.06%, which is what a careful designer will spec.

The unit conversion that catches everyone out

The formula gives you volts when L is in metres and R is in Ω/km — because the /1000 cancels metres into kilometres. If you're working in imperial you need L in feet and R in Ω per 1000 ft, and the /1000 still lines up. Mixing units is the #1 wrong answer on exam questions: metres with Ω per 1000 ft gives you a drop about 3.28× smaller than reality.

Fast-recall rules for the exam

Single-phase: "two L I R over a thousand." The 2 is the round trip.
Three-phase: "root-three L I R over a thousand." The √3 replaces the 2, not multiplies it.
NEC: 3% branch, 5% total. Recommendation not requirement (except for the sensitive-load articles).
BS 7671 / AS-NZS 3000: 3% lighting, 5% everything else. Mandatory. Measured end-to-end.
Double-check units. Metres go with Ω/km, feet go with Ω per 1000 ft. Never mix.

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