Voltage drop calculation: formula, limits and ready-made tables

Short answer: Voltage drop is the loss of voltage along a cable caused by its resistance, so the voltage at the end of the run is lower than at the start. Single-phase: %e = (2 × L × P × 100) ÷ (k × S × U²); three-phase: %e = (L × P × 100) ÷ (k × S × U²) (L in metres, P in watts, S in mm², k = 56 for copper, U = 230 or 400 V). IEC 60364-5-52 recommends 3% for lighting and 5% for other loads; Turkish wiring regulations allow 1.5% for lighting and 3% for power circuits after the meter. The tables below show how far you can go with each cable size without doing the maths.

What is voltage drop and what causes it?

Every cable has a small resistance. When current flows, some voltage is lost across it and turns into heat (Ohm's law: ΔU = I × R). Resistance grows with length and falls with cross-section, so voltage drop depends on three things: run length, current and cable size. Aluminium conducts less well than copper and drops more on the same run.

Symptoms: lights at the end of the run are dim or flicker, LED drivers switch off and on, motors and compressors struggle to start and run hot, and EV chargers reduce power or show an undervoltage error. A loose terminal or corroded joint can cause a large drop on its own; such points heat up and are a fire risk.

Voltage drop formula

For copper cable and resistive loads (LED, heaters, EV chargers; cos φ ≈ 1):

• Single-phase (230 V): %e = (2 × L × P × 100) ÷ (k × S × U²)
• Three-phase (400 V): %e = (L × P × 100) ÷ (k × S × U²)
• In volts, single-phase: ΔU = (2 × L × I) ÷ (k × S)
• In volts, three-phase: ΔU = (√3 × L × I) ÷ (k × S)

L is the one-way length (m), P watts, I amps, S cross-section (mm²), k conductivity (copper 56, aluminium 35). The 2 in the single-phase formula is there because current flows out and back. For inductive loads such as motors, multiply by cos φ (usually 0.8–0.85).

Worked example: 230 V, 16 A (3,680 W) socket circuit, 25 m, 2.5 mm². %e = (2 × 25 × 3,680 × 100) ÷ (56 × 2.5 × 230²) = 18,400,000 ÷ 7,406,000 ≈ 2.5%. In volts: 2 × 25 × 16 ÷ (56 × 2.5) ≈ 5.7 V.

What is the maximum allowed voltage drop?

• IEC 60364-5-52 (installations supplied from the public low-voltage network): 3% for lighting, 5% for other uses. This is the "3%/5% rule".
• Turkey (Internal Electrical Installations Regulation), after the meter: at most 1.5% on lighting circuits and 3% on power circuits.
• EV charging runs at full load for hours, so keeping the circuit below 3%, and around 2% on long runs, is good practice.

Ready-made table: how far can each cable size go? (single-phase 230 V)

Calculated for copper at the breaker's full current. If the real load is lower, the distance grows in proportion: at half the current you can go twice as far.

Maximum length on three-phase (400 V) circuits

Carrying the same power on three phases cuts voltage drop to about one sixth of single-phase. That is the main advantage of three-phase when feeding a distant building or installing an 11–22 kW charger. See single-phase vs three-phase.

Voltage drop on 12 V and 24 V LED

At low voltage the current for the same power is much higher, and the percentage drop rises sharply: a 0.36 V loss is 3% at 12 V but hardly noticeable at 230 V. Maximum length of the cable from driver to LED strip (copper, 3% drop):

At 12 V, 2.5 A is about 30 W, 5 A 60 W and 8.3 A 100 W; at 24 V the same currents carry twice the power. The strip itself also drops voltage, so the far end of a long run turns yellowish and dim. Re-feed a 12 V strip at least every 5 m and a 24 V strip every 10 m, or feed it from both ends or the middle. On long, powerful runs choosing 24 V (or 48 V) is more effective than a bigger cable. For driver sizing see LED strip driver calculation.

Worked examples

Detached outbuilding (garage, garden house), 50 m, single-phase 25 A: with 6 mm², ΔU = 2 × 50 × 25 ÷ (56 × 6) ≈ 7.4 V, or 3.2% – over the limit. With 10 mm² it falls to 4.5 V, 1.9%. Fed three-phase 3 × 16 A on 4 mm², the same building stays at 1.6%.

Home EV charger, 7.4 kW single-phase, 40 m: 3.3% on 6 mm², 2% on 10 mm². An 11 kW three-phase charger at the same distance stays at about 1.2% on 4 mm². For the full charger circuit calculation see cable size calculation.

24 V LED strip, 10 m × 14.4 W = 144 W (6 A), driver 5 m away: on 1.5 mm², 2 × 5 × 6 ÷ (56 × 1.5) ≈ 0.71 V, i.e. 3%. Moving the driver closer or using 2.5 mm² brings it down to 1.8%.

Garden lighting, 20 × 10 W LED fittings spread along 80 m: because the load is distributed, calculate with half the length (40 m); on 2.5 mm² the drop is only about 0.2%. LED fittings draw little power, so on garden circuits the real issues are usually moisture and joint quality, not cable size.

How to reduce voltage drop

• Increase the cable size: doubling the cross-section halves the drop.
• Use three-phase for long runs and high power.
• Move the board or a sub-distribution box closer to the load; place the LED driver near the strip.
• For low-voltage LED choose 24 V or 48 V instead of 12 V and feed long strips at several points.
• Use copper; aluminium needs about 1.6 times the cross-section for the same drop.
• Tighten terminals and use proper connectors; a loose joint causes both drop and heat.

How to measure voltage drop

With a multimeter, measure the voltage at the end of the circuit first with no load, then at full load (for example while the charger or a heater is running). The difference is the drop on the circuit. Mains voltage itself varies between about 220 and 240 V during the day, so take the two readings one after the other. Measuring inside a distribution board is dangerous; leave it to a qualified electrician.

For breaker sizing see which breaker for which circuit. In Bursa we design and install new circuits, EV chargers and garden lines as an electrical contractor and offer EV charger installation.

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