BS7671 Voltage-Drop & Current Calculator

Refs (BS 7671:2018)
• Table 4F2B: Voltage Drop at 90°C.
• Table 4Ab: Max permitted voltage-drop: lighting ≤ 3%, other final circuits ≤ 5%.

Supply Vs (V): Solver: Passes:

*Quadratic only active if exactly one constant-power load.

Cable Selection & Resistance

CSA (mm²): Ω/m @ 20 °C Temp (°C): auto-calc-temp Length (m):
Override R_cable Ω V-drop/m/A: – mV/A/m

R_cable = Ω/m × length × factor × Ct; clamped ≥ 1 μΩ. Or override manually.

#	Name	Type	R (Ω)	P (W)	I (A)	V-drop (V)	Action
0			–	–	–	–	–

Resistive vs. Constant-Power Loads

Resistive Loads:
• Current ∝ Voltage (I = V/R).
• Examples: heaters, toasters, incandescent lamps.
• If voltage droops, they draw less current and power falls (P = V²/R).

Constant-Power Loads:
• Internal regulators (switch-mode PSUs, LED drivers, modern electronics).
• They try to keep P = V·I ≈ constant, so if V falls, I rises (until limits).
• Examples: PCs, TVs, phone chargers, LED lighting drivers.

Note: Motors & ballasts have their own nonlinear V–I behaviour and may need special treatment.

Why Iterative Bisection?

• Our circuit is Vs → Rc → (parallel loads).
• Each load i has Iᵢ = Vload/Rᵢ (resistive) or Iᵢ = Pᵢ/Vload (constant-power), with
Vload = Vs – Rc·I_total.
• Summing gives h(I) = Σ Iᵢ(Vs – Rc·I) – I = 0.

• When there’s exactly one constant-power load (or all loads lumped into one P), h(I) becomes a quadratic:
a·I² + b·I + c = 0 → closed-form solution.
• With mixed or multiple loads, h(I) is higher-order or rational—no single quadratic applies.

• Bisection brackets the root I in [0, Vs/Rc] and halves the interval each pass,
guaranteeing convergence whenever h(0)>0 and h(Imax)<0.
• Each pass doubles the precision: ~20–30 passes → ~10⁻⁶ accuracy.

Equations & Examples

1) Resistive loads (easiest)
• Cable: R_c = ρ·L·Ct (or Ω/m × length × factor × Ct).
• Parallel loads: 1/R_par = Σ (1/R_i).
• Total current: I = V_s / (R_c + R_par).
• Load voltage: V_load = V_s − I·R_c.
Example: V_s=240 V, R_c=0.5 Ω, R_par=10 Ω → I=240/10.5=22.86 A, V_load=228.6 V.

2) Constant-power loads (bisection)
• Each load: I_i = P_i / V_load.
• With mixed loads: h(I) = Σ I_i(V_s − R_c·I) − I = 0.
• Iterate: bracket I in [0, V_s/R_c], then bisect until h(I) ≈ 0.
Example: V_s=240 V, R_c=0.5 Ω, P=1000 W → solve h(I)=0 → I≈4.33 A, V_load≈237.8 V.

3) Single constant-power load (quadratic)
• One P only: R_c·I² − V_s·I + P = 0.
• Solve: I = [V_s − √(V_s² − 4·R_c·P)] / (2·R_c).
Example: V_s=240 V, R_c=0.5 Ω, P=1000 W → I≈4.33 A.

Why 25 mm² Service Cables?

In ideal conditions, a 45 m run of 16 mm² Cu at 90 °C carrying 25 kW (≈107 A)
drops only about 7.11 V (2.96 % of 240 V), which meets the voltage-drop limit.

But BS 7671 compliance requires more than just voltage-drop checks:

Current-carrying & Derating:
• 16 mm² base rating ~87–100 A in free air, but in conduit/bundles it derates to ~50–60 A.
• 25 mm² remains lawful after grouping & ambient-temperature corrections.
Protective Coordination & Fault Withstand:
• Larger CSA handles higher fault currents (I²t) and resists damage until the breaker clears.
Future Loads & DNO Rules:
• Meter tails/main feeds are often mandated at 25 mm² to allow EV chargers, heat pumps, etc.
Temperature & Longevity:
• Oversizing keeps conductors cooler, reducing insulation ageing.

Summary: Although 16 mm² can pass a single voltage-drop test at 107 A, real installations must also satisfy current rating, derating, fault-clearance, future proofing and DNO rules—hence the widespread use of 25 mm².