Understanding Voltage Drop in Electrical Design

Every electrical cable possesses inherent resistance. As current flows through a conductor over a distance, some of the electrical energy is lost as heat, resulting in a lower voltage arriving at the load than what originated at the supply. If the voltage drop is too severe, equipment will operate inefficiently, motors will draw excess current and overheat, and lighting will dim.

The BS 7671 Statutory Limits

To ensure safe and efficient operation of equipment, BS 7671:2018+A4:2026 mandates strict limits on the maximum allowable voltage drop from the origin of the installation to any load (Regulation 525). The limits are based on the public low voltage distribution system:

• Lighting Circuits: Maximum 3% drop. (e.g., 6.9V on a standard 230V single-phase supply).
• Other Uses (Power/Motors/Heating): Maximum 5% drop. (e.g., 11.5V on a standard 230V single-phase supply).

The Voltage Drop Formula

The standard equation utilised by this engine—and detailed in Appendix 4 of the IET Wiring Regulations—is:

Vd = (mV/A/m × Ib × L) / 1000

Where:

• Vd: Total Voltage Drop in Volts (V).
• mV/A/m: The millivolt drop per amp per metre for the specific cable size and type (from BS 7671 Tables).
• Ib: The Design Current in Amps. Note: We use the actual design load (Ib), not the rating of the protective device (In).
• L: The route length of the cable in metres.

What if my circuit fails the calculation?

If your calculated voltage drop exceeds the statutory limits (displaying a red FAIL on our engine), the primary solution is to increase the cross-sectional area (CSA) of the cable. A thicker cable has a lower resistance (and thus a lower mV/A/m value), which will reduce the voltage drop over long distance runs.