Worked example: shower with factors35 degrees and grouped with two others. What cable now?

8.5 kW shower · domestic · flat T&E · reference method 102 · 20 m run
BS EN 60898 circuit breaker
NEW: ambient temperature 35 °C · grouped with 2 OTHER circuits (3 in total)

1) Ib = P ÷ V = 8500 ÷ 230 = 36.96 A
2) In = 40 A Type B        (Ib ≤ In ✓)

3) Correction factors
   Ca (70 °C thermoplastic, 35 °C)        = 0.94
   Cg (3 circuits, bunched/enclosed)       = 0.70

4) It needed = In ÷ (Ca × Cg) = 40 ÷ (0.94 × 0.70) = 60.79 A

5–6) T&E, method 102, need It ≥ 60.79 A
   10 mm² → 47 A  ✗
   16 mm² → 63 A  ✓  → 16 mm² twin and earth

7) Voltage drop: 16 mm² → 2.8 mV/A/m (max 5% = 11.5 V)
   VD = (2.8 × 36.96 × 20) ÷ 1000 = 2.07 V ✓

Answer: 16 mm² T&E on a 40 A Type B.
(Without the factors it was 10 mm².)

Remember Ib ≤ In ≤ Iz, where Iz = It × the factors. You don't make the protective device bigger to fix it. You upsize the cable, because the device must still operate before the cable overheats.

BS 7671 · Appendix 4 · Table 4D5 70 °C flat T&E (It and mV/A/m) · p.484 On-Site Guide · Table F6 (same values, handier) · p.193 Electrician's Guide to the Building Regs · T&E current ratings table by reference method · p.41

Voltage drop limits in BS 7671

BS 7671 · Appendix 4 · Table 4Ab voltage drop: 3% lighting, 5% other uses · p.458 The On-Site Guide sets out the VD formula as ÷1000 (radial) and ÷4000 (ring). Learn it for the 18th Edition exam.

max mV/A/m = (max VD × 1000) ÷ (Ib × L)
           = (11.5 × 1000) ÷ (36.96 × 20) = 15.56 mV/A/m
→ any cable at or below 15.56 mV/A/m passes volt drop on this run.
You won't be asked to do this in the exam. You'll pick the value from the table.