Volt Drop Calculation: Formula, 3% & 5% Limits (BS 7671)

Voltage drop is calculated with the formula VD = (mV/A/m × Ib × L) / 1000, and BS 7671 limits it to 3% (6.9V) on lighting circuits and 5% (11.5V) on all other circuits supplied from the public network. This guide explains the formula, where the limits come from, and works through real cable-sizing examples.

When current flows through a cable, the resistance of the conductor causes a reduction in voltage along the length of the cable. The supply voltage at the origin of the installation (230V nominal for a single-phase UK supply) is reduced by the time it reaches the load at the far end of the circuit. This reduction is called voltage drop.

Every cable has resistance, and that resistance increases with length. The longer the cable run and the higher the current flowing through it, the greater the voltage drop. Excessive voltage drop causes equipment to underperform — lights dim, motors run inefficiently, and sensitive electronic equipment may malfunction. Voltage drop is one of several factors alongside maximum demand that determines correct cable sizing.

BS 7671 (the 18th Edition wiring regulations) sets maximum permitted voltage drop limits to ensure that equipment connected at the far end of a circuit receives sufficient voltage to operate correctly and safely.

The voltage drop limits are set out in Appendix 4 of BS 7671. They apply from the origin of the installation to the most distant point of the circuit. For a low voltage installation supplied from the public network, the limits are:

Lighting Circuits

The maximum voltage drop for lighting circuits is 3% of the nominal voltage. For a 230V single-phase supply, this equates to 6.9V. This means the voltage at the most distant luminaire must not fall below 223.1V under design load conditions.

Other Circuits (Power)

The maximum voltage drop for all other circuits (socket outlets, cookers, showers, fixed appliances) is 5% of the nominal voltage. For a 230V single-phase supply, this equates to 11.5V. The voltage at the load must not fall below 218.5V under design load conditions.

• ✓Lighting circuits: 3% maximum = 6.9V (on a 230V supply)
• ✓Power circuits: 5% maximum = 11.5V (on a 230V supply)
• ✓Limits apply from the origin of the installation to the load
• ✓These limits assume a supply from the public distribution network
• ✓For private supplies or installations fed from a generator, different limits may apply

The standard formula for calculating voltage drop in single-phase circuits is:

VD = (mV/A/m x Ib x L) / 1000

Where:

• ✓VD = voltage drop in volts
• ✓mV/A/m = millivolts per amp per metre, obtained from the cable data tables in BS 7671 Appendix 4
• ✓Ib = design current of the circuit in amps
• ✓L = route length of the cable in metres (one way, not the total conductor length)
• ✓1000 = conversion factor from millivolts to volts

The mV/A/m value is the key figure. It represents how many millivolts of drop occur for every amp of current flowing through every metre of cable. It varies depending on the cable cross-sectional area, conductor material (copper or aluminium), insulation type, and installation method. Larger cables have lower mV/A/m values because they have less resistance per metre.

A ring final circuit is wired in 2.5mm2 thermoplastic (PVC) twin and earth cable, clipped direct. The design current is 32A and the total route length of the ring is 50m.

Step 1: Find the mV/A/m Value

From BS 7671 Appendix 4, the mV/A/m value for 2.5mm2 copper thermoplastic cable at 70 degrees C conductor operating temperature is 18 mV/A/m.

Step 2: The Common Mistake — Treating the Ring as a Radial

A frequent error is to apply the radial formula directly, using the full design current over the full ring length:

VD = (18 x 32 x 50) / 1000 = 28.8V — wrong

This figure is wrong for a ring final circuit. Both legs of the ring feed every load, so no conductor ever carries the full 32A over the full 50m. Using 28.8V would wrongly condemn a perfectly compliant circuit.

Step 3: Apply the Ring Method

The worst case for voltage drop on a ring is the full load connected at the midpoint of the ring — the point furthest from the board along both legs. The current splits between the two legs, so each leg is half the ring length (50 / 2 = 25m) and carries half the design current (32 / 2 = 16A). Applying the formula to one leg:

VD = (18 x 16 x 25) / 1000 = 7.2V

Step 4: Check Against the Limit

This is a power circuit, so the limit is 5% of 230V = 11.5V. The calculated voltage drop of 7.2V is within the 11.5V limit. This design complies with BS 7671.

A lighting circuit is wired in 1.5mm2 thermoplastic (PVC) twin and earth cable, clipped direct. The design current is 10A and the route length to the furthest luminaire is 20m.

Step 1: Find the mV/A/m Value

From BS 7671 Appendix 4, the mV/A/m value for 1.5mm2 copper thermoplastic cable at 70 degrees C conductor operating temperature is 29 mV/A/m.

Step 2: Apply the Formula

VD = (29 x 10 x 20) / 1000 = 5.8V

Step 3: Check Against the Limit

This is a lighting circuit, so the limit is 3% of 230V = 6.9V. The calculated voltage drop of 5.8V is within the 6.9V limit. This design complies with BS 7671.

However, note that 5.8V is close to the 6.9V limit. If the cable route were extended or additional luminaires increased the design current, the circuit could exceed the limit. It is good practice to allow a margin of safety in your designs.

When your voltage drop calculation exceeds the BS 7671 limits, there are several practical strategies to bring the design into compliance.

Increase Cable Size

The most common solution is to select a larger cable cross-sectional area. Larger cables have lower resistance per metre and therefore a lower mV/A/m value. For example, moving from 2.5mm2 (18 mV/A/m) to 4mm2 (11 mV/A/m) reduces the voltage drop by approximately 39%.

Shorten Cable Routes

Since voltage drop is directly proportional to cable length, shorter routes mean less voltage drop. Correct routing through cable safe zones can also help minimise unnecessary cable lengths. During the design stage, plan cable routes to minimise unnecessary diversions and keep total lengths as short as practicable.

Reposition the Distribution Board

If the distribution board can be positioned closer to the loads with the highest demand, the cable lengths to those loads are reduced. This is particularly relevant on larger domestic or commercial installations where cable runs can be long.

Consider Sub-Distribution

On larger installations, a sub-distribution board fed by a larger cable from the main board can reduce the voltage drop on individual final circuits. The larger cable to the sub-board carries the combined load at a lower mV/A/m value, and the final circuits from the sub-board have shorter route lengths.

• ✓Larger cable = lower mV/A/m = less voltage drop
• ✓Shorter route = less cable length = less voltage drop
• ✓DB closer to loads = shorter cable runs to high-demand circuits
• ✓Sub-distribution boards reduce final circuit lengths on larger installations
• ✓Always recalculate after making changes to confirm compliance

Related Course

Inspection & Testing (2391)

The 2391 course covers verifying voltage drop during inspection and testing of completed installations.

View Course

The mV/A/m values tabulated in BS 7671 Appendix 4 are quoted at the maximum conductor operating temperature for the cable insulation type. For thermoplastic (PVC) insulated cables, this is 70 degrees C. For thermosetting (XLPE or LSF) cables, this is 90 degrees C.

In practice, if the circuit is not fully loaded (i.e. the actual current is less than the cable's rated current-carrying capacity), the conductor temperature will be lower than the tabulated maximum. At lower temperatures, conductor resistance is lower, and the actual voltage drop will be less than the calculated value.

BS 7671 Appendix 4 provides a correction method. The tabulated mV/A/m value has two components: a resistive component (r) and a reactive component (x). Only the resistive component varies with temperature. For most domestic circuits using small cables at power frequency, the reactive component is negligible and the resistive component dominates.