**Section 525 of BS 7671 requires the voltage drop in the circuits of an installation to be limited so that the safe and satisfactory operation of current-using equipment is not impaired.**

**Section 525 of BS 7671 requires the voltage drop in the circuits of an installation to be limited so that the safe and satisfactory operation of current-using equipment is not impaired.**

To achieve this circuit conductors larger than that needed to carry the load current may have to be installed to compensate for voltage drop.

This article looks at one method, commonly referred to as the ‘basic’ method, for satisfying the voltage drop requirements of BS 7671.

**Calculating voltage drop using the ‘basic’ method**

When this method is used to determine voltage drop the following assumptions apply:

- the circuit conductors are always assumed to be at their maximum permitted operating temperature.
- the load power factor (pf) is always assumed to be the worst possible value from the point of view of voltage drop, and the actual power factor is not taken into account.

This approach errs on the safe side which may, in some cases, lead to the selection of a larger conductor size. Therefore, a more detailed calculation method (which is not covered in this article) which includes corrections for conductor operating temperature and load power factor is given in section 6 of Appendix 4 of BS 7671.

Note: The basic method takes no account of harmonic content, for which more detailed voltage drop calculations would be necessary.

**Requirements relating to maximum permitted voltage drop**

Regulation 525.1 requires that, in the absence of any other considerations, under normal operating conditions the voltage at the terminals of fixed current‑using equipment shall be greater than the lower limit corresponding to the product standard for the equipment.

Where the fixed current-using equipment is not subject to a product standard, Regulation 525.201 requires the voltage at the terminals to be such that the safe functioning of the equipment is not impaired.

For a low voltage installation supplied directly from the public network these requirements are deemed to be met if the voltage drop between the origin of the installation (usually the supply terminals) and the terminals of fixed current-using equipment or socket-outlets does not exceed the following limits stated in section 6.4 of Appendix 4 of BS 7671 (Regulation 525.202).

- Lighting 3 %
- Other uses 5 %.

Note: A greater voltage drop than stated may be accepted for a motor during starting periods and for other equipment with high inrush currents (Regulation 525.102).

Appendix 4 of BS 7671 includes tables of voltage drop in mV/A/m (millivolts per ampere per metre). The tabulated voltage drop values are for a current of one ampere for a one metre run along the route taken by the cables. The voltage drop for a run of cable(s) is calculated using the following formula:

Where:

mV/A/m is the tabulated mV/A/m value obtained from Appendix 4 of BS 7671

L is the length of the run of the cables, in metres

Ib is the design current, in amperes.

For AC circuits using conductor sizes up to and including 16 mm2 and DC circuits with conductors of any csa, inductance may be ignored and the tabulated voltage drop values take account of resistance only.

For cables having a csa of greater than 16 mm2, separate tabulated voltage drop values are given for the resistance, inductance and impedance of the conductors.

These are identified in the tables by the symbols r, x and z, respectively. For AC circuit conductors having a csa of 25 mm2 or larger the value of mV/A/m to be used in the above equation is the ‘impedance’ value (mV/A/m)z, from the appropriate column of the table.

**Calculating the total voltage drop if a circuit is NOT directly supplied from the origin of the installation**

Where a final circuit is supplied from a distribution board that is not at the origin of the installation, as shown in Fig 1, the total voltage drop between the origin of the installation and the load using equipment is equal to the sum of the voltage drops in the distribution circuit and the final circuit.

Care is needed when adding together the voltage drop of a three-phase distribution circuit and a single-phase final circuit, as the former voltage drop relates to the line-to-line voltage and the latter relates to the line-to-neutral voltage.

As illustrated by the following worked example, expressing all the voltage drops as percentages (of the nominal line-to-line voltage or line-to-neutral voltage, as applicable) before adding them together helps to reduce the likelihood of errors.

**Worked example**

In a low voltage (400/230 V) installation supplied directly from a public low voltage distribution system, a three-phase and neutral distribution board is supplied from the origin by a 50 m run of 4-core 25 mm2 thermosetting insulated armoured cable to *BS 5467 *having copper conductors.

Supplied from the distribution board is a single‑phase circuit consisting of a 20 m run of 2-core 4 mm2 thermosetting insulated armoured cable to *BS 5467 *having copper conductors, supplying a load current of 21 A.

Assuming the current in the distribution circuit is 80 A, as shown in Fig 1, determine the total voltage drop between the origin of the installation and the load under normal operating conditions, expressed as a percentage of the nominal supply voltage.

**The voltage drop ( V**

*D***) in the distribution cable**

The *V**D *in the distribution circuit cable is calculated as follows.

From Fig 2 the tabulated (mV/A/m)z value for the cable is 1.65 mV/A/m, while (L) is 50 m and *I**b *is 80 A. Substituting these values gives a voltage drop of 6.6 V. This is equivalent to 1.65 % of the nominal line-to-line voltage of the supply;

(6.6 V ÷ 400 V) x 100 = 1.65 %.

**The voltage drop ( V**

*D***) in the final circuit cable**

The *V**D *in the final circuit cable supplying the equipment is calculated as follows.

The tabulated mV/A/m value for the 2-core 4.0 mm2 cable is 12 mV/A/m, obtained from column 3 of Table 4E4B in

Appendix 4, L is 20 m and *I**b *is 21 A.

Substituting these values into the formula gives a voltage drop of 5.04 V which is equivalent to 2.19 % of the nominal line-to-neutral voltage of the supply; (5.04 V ÷ 230 V) x 100 = 2.19 %.

Therefore, the total percentage voltage drop between the origin of the installation and the load is equal to the sum of the distribution circuit (1.65 %) and the final circuit (2.19 %), which is 3.84 %. The total voltage drop of 3.84 % falls within the applicable maximum of 5 %

The total voltage drop expressed in volts, line-to-neutral, is 8.83 V given by; (3.84 x 230 V) ÷ 100 = 8.83 V.

**Guidance on other installation matters are covered in the NICEIC and ELECSA Site Guide, newly revised to the 18th Edition of BS 7671.**

For information about the NICEIC Approved Contractor or Domestic Installers schemes, visit www.niceic.com or call 0333 015 6626