Worked Example: Voltage Drop for a 200 m Office Lighting Circuit per BS 7671:2018

A new open-plan office floor in a commercial building requires a dedicated lighting circuit. The distribution board is located in a riser cupboard at one end of the floor, and the final luminaire on the circuit is 200 metres away at the far end of the building. The architect has specified a 2.5 mm² twin-and-earth cable, which is standard for lighting circuits in the UK. However, with 200 metres of cable run, voltage drop is the primary concern. This example demonstrates a common scenario where voltage drop — not current capacity — governs cable selection, and shows how to verify compliance with BS 7671:2018+A2, Table 4Ab.

From BS 7671:2018, Table 4D2B (voltage drop for flat twin-and-earth PVC copper cables, Reference Method C), extract the resistive and reactive components of the cable impedance per metre:

For 2.5 mm² cable:

r = 18.0 mV/A/m   (resistive component, from Table 4D2B column 3)
x = 0.29 mV/A/m   (reactive component, from Table 4D2B column 4)

For 4 mm² cable (we will need this later):

r = 11.0 mV/A/m   (resistive component)
x = 0.29 mV/A/m   (reactive component)

Note: The reactive component is essentially the same for both sizes because it depends primarily on conductor spacing (which is similar for twin-and-earth cables), not conductor area. The resistive component is inversely proportional to conductor area and dominates for small cables.

The voltage drop formula incorporating power factor from BS 7671 Appendix 4, Section 6:

dV = I_b x L x (r x cos(phi) + x x sin(phi)) / 1000     -- (Eq. 1)

where:
  I_b      = 20 A (design current)
  L        = 200 m (cable route length)
  r        = 18.0 mV/A/m
  x        = 0.29 mV/A/m
  cos(phi) = 0.95
  sin(phi) = 0.3122  (from sin = sqrt(1 - 0.95^2))

dV = 20 x 200 x (18.0 x 0.95 + 0.29 x 0.3122) / 1000

dV = 4,000 x (17.10 + 0.0905) / 1000

dV = 4,000 x 17.19 / 1000

dV = 68.77 V

Wait — that result looks impossibly high. Let’s check: Table 4D2B gives values in mV/A/m as the total voltage drop for the full circuit (go and return), so the formula is correct. But 68.77 V on a 230 V circuit is:

dV% = 68.77 / 230 x 100 = 29.9%                          -- (Eq. 2)

29.9% voltage drop — this is catastrophically high. The 3% limit is 6.9 V. The 2.5 mm² cable fails by a factor of 10. This is why the architect’s initial specification of 2.5 mm² for a 200 m lighting run is completely unworkable.

Instead of trial-and-error, we can calculate the maximum allowable mV/A/m value and select accordingly:

Maximum allowable dV = 3% x 230 V = 6.9 V                -- (Eq. 3)

Required (mV/A/m)_max = dV_max x 1000 / (I_b x L)

(mV/A/m)_max = 6,900 / (20 x 200)

(mV/A/m)_max = 6,900 / 4,000

(mV/A/m)_max = 1.725 mV/A/m

Now compare this to the tabulated values from Table 4D2B:

Remarkably, to achieve 3% voltage drop over 200 m at 20 A, we need a 35 mm² cable — 14 times the original 2.5 mm² specification. This is cost-prohibitive and physically impractical for a lighting circuit.

The practical solution is not to run a single 200 m circuit. Instead, we split the floor into sub-circuits fed from local distribution boards, or use a sub-main to bring the supply closer to the load centre.

Revised design: Install a local distribution board at the midpoint of the floor (100 m from the riser), fed by a sub-main. Each lighting circuit then runs a maximum of 50 m from the local DB.

For a 4 mm² cable on a 50 m circuit with 20 A design current:

dV = I_b x L x (r x cos(phi) + x x sin(phi)) / 1000     -- (Eq. 4)

dV = 20 x 50 x (11.0 x 0.95 + 0.29 x 0.3122) / 1000

dV = 1,000 x (10.45 + 0.0905) / 1000

dV = 1,000 x 10.54 / 1000

dV = 10.54 V

Still too high at 4.6%. Let’s check 4 mm² at a more typical 30 m lighting circuit:

dV = 20 x 30 x 10.54 / 1000 = 6.32 V                     -- (Eq. 5)

dV% = 6.32 / 230 x 100 = 2.75%

2.75% — within the 3% limit. A 4 mm² cable on a 30 m circuit is compliant.

However, the original question asks us to verify the 200 m run. If the design truly requires a single 200 m circuit (which is poor practice but sometimes unavoidable in existing buildings), the answer is 35 mm².

For the recommended design of splitting the 200 m run into circuits of manageable length, here is the full verification for a 4 mm² cable at 30 m:

Cable: 4 mm^2 Cu twin-and-earth, PVC 70C
Route: 30 m, clipped direct (Method C)
Load: 20 A at 0.95 PF

Voltage drop:
  dV = 20 x 30 x (11.0 x 0.95 + 0.29 x 0.3122) / 1000
  dV = 600 x 10.54 / 1000
  dV = 6.32 V
  dV% = 6.32 / 230 x 100 = 2.75%                         -- (Eq. 6)

Limit: 3% (6.9 V) per BS 7671 Table 4Ab
Result: 2.75% < 3%   PASS

Current capacity check (Reference Method C, Table 4D2A):
  I_z(4mm2) = 32 A
  I_n = 20 A
  32 A > 20 A   PASS

Both current capacity and voltage drop are satisfied with a 4 mm² cable on a 30 m circuit.

Recommended solution: Do not attempt to run a single 200 m lighting circuit. Instead, install a local distribution board and limit individual lighting circuits to 30 m using 4 mm² Cu twin-and-earth cable. The voltage drop is 2.75%, compliant with the 3% limit per BS 7671 Table 4Ab.

This example demonstrates the fundamental principle that voltage drop, not current capacity, governs cable selection on long circuits. A 2.5 mm² cable has ample current capacity for 20 A, but it cannot deliver that current over 200 m without losing nearly a third of the supply voltage.

• BS 7671:2018+A2, Table 4Ab — Maximum voltage drop limits (3% lighting, 5% power)
• BS 7671:2018+A2, Table 4D2A — Current-carrying capacity, twin-and-earth cables
• BS 7671:2018+A2, Table 4D2B — Voltage drop (mV/A/m), twin-and-earth cables
• BS 7671:2018+A2, Appendix 4, Section 6 — Voltage drop calculation method
• IET Guidance Note 1, Section 6.6 — Voltage drop worked examples

Use the ECalPro Voltage Drop Calculator to verify this calculation or explore how different cable sizes, route lengths, and power factors affect voltage drop compliance. The calculator supports BS 7671, AS/NZS 3008, IEC 60364, and NEC standards.