Three-Phase Voltage Drop Calculator: BS 7671 Method, Balanced and Unbalanced Loads, Worked Examples

Quick Answer: Three-phase voltage drop is calculated using the BS 7671 method: Vd = (mV/A/m × Ib × L) / 1000. For balanced three-phase loads, maximum voltage drop is 5% from the point of supply to the load (per IET Guidance Note 1). For 400V three-phase, 5% = 20V. The mV/A/m values for three-phase are lower than single-phase for the same cable (typically 0.55× the single-phase value) because of the relationship between line and phase voltages.

Summary

Voltage drop is a critical design check in BS 7671. It ensures that equipment at the end of a cable receives adequate voltage for correct operation. For three-phase circuits — supplying motors, HVAC units, large process equipment, EV chargers, or distribution boards — the calculation method follows the same principle as single-phase but uses different values from the BS 7671 tables.

The distinction between single-phase and three-phase voltage drop causes frequent errors. The mV/A/m values in BS 7671 Appendix 4 are given separately for single-phase and three-phase, and they are not the same. Using the wrong column underestimates or overestimates the actual voltage drop. Additionally, unbalanced three-phase loads require a modified approach that accounts for the asymmetric neutral current.

This article covers the standard balanced three-phase calculation, the unbalanced case, how to handle resistive vs reactive loads, and practical guidance for common three-phase circuits.

Key Facts

• BS 7671 voltage drop limit — 5% from origin of installation to load (3% for lighting circuits)
• Three-phase voltage — Line-to-line: 400V; line-to-neutral: 230V; the 5% limit applies to line voltage in most commercial calculations
• 5% of 400V — Maximum 20V voltage drop for a 400V three-phase circuit
• mV/A/m — Millivolts per ampere per metre; from BS 7671 Appendix 4; different values for single-phase and three-phase
• Three-phase mV/A/m — Approximately 0.55× the single-phase value for the same cable (for resistive loads at unity power factor)
• Balanced load — Equal current in all three phases; neutral carries minimal or zero current; standard assumption
• Unbalanced load — Different currents in some phases; neutral carries residual current; affects voltage drop calculation
• Power factor — Inductive loads (motors) have leading/lagging current; affects reactive voltage drop; BS 7671 Table 4D5 gives (mV/A/m)r and (mV/A/m)x values for power factor correction
• Design current (Ib) — The actual current the circuit will carry; use as the basis for voltage drop calculation
• Worst-case single phase — For unbalanced loads, check the phase with the highest current
• Volt drop formula — Vd (V) = (mV/A/m × Ib × L) / 1000

Quick Reference Table: mV/A/m Values (Approximate, Copper Cable, PVC Insulation)

Values are approximate; use BS 7671:2018+A2:2022 Appendix 4 Tables 4D2B–4D5B for precise values. Values vary with temperature and installation method.

Detailed Guidance

Balanced Three-Phase Calculation: Step by Step

Scenario: A 400V three-phase distribution board at the far end of a 75m run of 16mm² copper cable (SWA, clipped direct). Design current is 40A per phase (balanced load).

Step 1: Find mV/A/m from BS 7671 tables

• Cable: 16mm² copper, SWA
• Installation method: clipped direct (Method C in BS 7671)
• Three-phase column: 2.4 mV/A/m (from table above or BS 7671 App 4)
• Note: use the value for operating temperature; for circuits running close to maximum, use the value at maximum conductor temperature

Step 2: Apply formula Vd = (mV/A/m × Ib × L) / 1000 Vd = (2.4 × 40 × 75) / 1000 = 7,200 / 1000 = 7.2V

Step 3: Check against limit

• Maximum allowable: 5% of 400V = 20V
• Actual: 7.2V = 1.8%
• Result: compliant ✓

Step 4: Express as percentage

• 7.2 / 400 × 100 = 1.8% — well within 5% limit

What the mV/A/m Three-Phase Values Mean

The relationship between single-phase and three-phase voltage drop values:

For a balanced three-phase system, the line-to-line voltage drop is √3 × the voltage drop per phase conductor. However, because a three-phase system has three current-carrying conductors (not two as in single-phase), the cable resistance contributes differently.

The three-phase mV/A/m value in BS 7671 gives the line-to-line voltage drop directly, taking account of the three-phase relationship. So you use it exactly like the single-phase value: multiply by design current and cable length.

Do not multiply three-phase mV/A/m values by 2 (as you would need to do for some bespoke calculations). BS 7671 Appendix 4 values for three-phase already represent the line-to-line drop per ampere per metre.

Power Factor Correction for Inductive Loads

For purely resistive loads (heaters, incandescent lighting), the calculation above is accurate. For inductive loads (motors, transformers, HVAC compressors), the actual voltage drop is slightly different due to the phase angle.

BS 7671 Table 4D5 (and equivalent tables for other cable types) provides:

• (mV/A/m)r — resistive component
• (mV/A/m)x — reactive component

Corrected voltage drop formula: Vd = [(mV/A/m)r × cos φ + (mV/A/m)x × sin φ] × Ib × L / 1000

Where:

• cos φ = power factor of the load
• sin φ = reactive component (sin φ = √(1 − cos²φ))

Example (motor at power factor 0.85):

• (mV/A/m)r for 16mm² three-phase = 2.35
• (mV/A/m)x for 16mm² three-phase = 0.175 (inductive reactance)
• cos φ = 0.85; sin φ = √(1 − 0.7225) = √0.2775 = 0.527
• Corrected mV/A/m = (2.35 × 0.85) + (0.175 × 0.527) = 1.998 + 0.092 = 2.09 mV/A/m

For most practical calculations, the uncorrected values give a conservative (higher) result and the power factor correction is not necessary. Use power factor correction when optimising cable size to avoid going up a size unnecessarily.

Unbalanced Three-Phase Loads

When three-phase circuits supply a mix of single-phase and three-phase loads (typical in commercial TPN distribution), the phases may carry different currents.

Approach:

1. Calculate the current on each phase separately
2. Find the phase with the highest current
3. Use the single-phase mV/A/m value for that cable size
4. Apply the single-phase formula: Vd = (mV/A/m_single × I_max × L) / 1000
5. This gives the voltage drop for the worst-case phase

Neutral conductor: For unbalanced loads, the neutral carries the vector sum of the three phase currents minus any harmonic cancellation. For single-phase loads on a three-phase supply: if loads are unequal, the neutral current = √(I_L1² + I_L2² + I_L3² − I_L1×I_L2 − I_L2×I_L3 − I_L3×I_L1) [simplified for 120° displacement].

In practice: for sizing the neutral conductor, if the system has significant harmonic loading (switch-mode power supplies, VFDs, electronic ballasts), the neutral may carry up to 1.73× the phase current due to triplen harmonics. Size the neutral accordingly.

Cascaded Circuits: Adding Voltage Drops

When equipment is supplied through a chain of cables (mains → main distribution board → sub-board → circuit), the voltage drops accumulate:

Example:

• Main feeder: 25m of 95mm², 200A balanced three-phase → Vd = (0.42 × 200 × 25) / 1000 = 2.1V
• Sub-distribution cable: 30m of 35mm², 80A → Vd = (1.10 × 80 × 30) / 1000 = 2.64V
• Final circuit: 40m of 16mm², 32A → Vd = (2.4 × 32 × 40) / 1000 = 3.07V

Total voltage drop: 2.1 + 2.64 + 3.07 = 7.81V As percentage: 7.81 / 400 × 100 = 1.95% — compliant within 5%

In a building with many distribution levels, this cascade effect can consume the budget quickly. Early-stage design should allocate voltage drop budget across the distribution system.

Practical Sizing Check: Three-Phase Motor Supply

Scenario: A 15kW, 400V, three-phase motor. Power factor 0.85, efficiency 93%. Cable run 45m.

Step 1: Calculate design current

• Electrical input = 15kW / 0.93 = 16.1kW
• Ib = P / (√3 × U × cos φ) = 16,100 / (1.732 × 400 × 0.85) = 16,100 / 589 = 27.3A

Step 2: Select cable size

• Start with 6mm² copper (current capacity: ~38A for Method C)
• mV/A/m three-phase (6mm²): 6.4 (from table)
• Power factor correction: (6.35 × 0.85) + (0.195 × 0.527) = 5.4 + 0.10 = 5.5 mV/A/m
• Vd = (5.5 × 27.3 × 45) / 1000 = 6.75V = 1.69% — compliant ✓

Step 3: Verify current capacity 6mm² at method C: 46A (BS 7671 Table 4D1A); design current 27.3A — fine.

Starting Current and Voltage Drop

Motor starting current can be 5–7× rated current. This causes a large momentary voltage drop that may:

• Trip other circuits on the same sub-board due to undervoltage
• Cause interference with sensitive equipment (PLCs, variable speed drives)
• Flicker lighting circuits

For critical motor installations, check voltage drop at starting current:

• Starting current: assume 6 × Ib = 6 × 27.3 = 163.8A
• Vd_start = (5.5 × 163.8 × 45) / 1000 = 40.5V = 10.1%

This level of transient voltage drop is often acceptable (loads are tolerant of brief dips), but if other equipment on the circuit is affected, you need a larger cable or a star-delta or soft-starter motor arrangement.

Frequently Asked Questions

Why is the three-phase mV/A/m value different from single-phase for the same cable?

In single-phase, current flows out on the live and returns on the neutral. The voltage drop is the sum of both conductors' resistive drop: 2 × I × R per metre. In three-phase balanced, the line voltage drop results from the geometry of the three-phase vectors. The relationship means the three-phase voltage drop (line-to-line) for a balanced load uses approximately 0.87× the single-phase factor for the same cable. BS 7671 Appendix 4 calculates this for you and presents separate columns — use the right column.

Can I use a single-phase voltage drop calculator for three-phase circuits?

Only if you manually apply the three-phase correction factor. Using the single-phase mV/A/m value for a three-phase circuit will overestimate the voltage drop by approximately 15–20%, giving a conservative result. This is safe (may result in selecting a larger cable than necessary) but may be uneconomic. For accurate three-phase calculations, always use the three-phase mV/A/m values.

What happens if a cable exceeds the 5% voltage drop limit?

If a calculated design exceeds 5%, the circuit is non-compliant with BS 7671 Regulation 525. Options: increase cable CSA; reduce circuit length (add intermediate boards); accept a temporary derogation where equipment tolerance confirms compliance. Note that voltage drop is a maximum at full design current — at part load the actual drop will be lower.

Regulations & Standards

• BS 7671:2018+A2:2022 — Requirements for Electrical Installations, particularly Regulation 525 and Appendix 4

• IET Guidance Note 1 — Selection and Erection of Equipment: voltage drop interpretation

• IET On-Site Guide — Simplified tables for common circuit types

• IET BS 7671 Wiring Regulations — 18th Edition with Appendix 4 tables

• IET Guidance Notes series — Guidance Note 1 covers voltage drop in detail

• cable sizing — Single-phase cable sizing and current-carrying capacity

• consumer units — Distribution board design considerations