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Checking Zs against the table, and where 2.19 comes fromWhy multiply the table by 0.8?

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The worked example

System      = TN-S        Max Ze for TN-S = 0.8 Ω
Ze measured  = 0.42 Ω   -> comfortably inside
R1 + R2     = 0.36 Ω

Zs = 0.42 + 0.36 = 0.78 Ω

Now check it. BS 7671 · Table 41.3 · p.75 gives the maximum tabulated Zs. For a 20 A Type B circuit breaker to BS EN 60898 that is 2.19 Ω.

Where 2.19 Ω actually comes from

Zs = (Uo × Cmin) ÷ Ia

Uo    = 230 V
Cmin  = 0.95   (the minimum voltage factor, for voltage variations)
Ia    = 5 × In for Type B = 5 × 20 = 100 A

Zs = (230 × 0.95) ÷ 100
   = 218.5 ÷ 100
   = 2.185  ->  2.19 Ω
The formula is printed in the regulations as Zs × Ia ≤ U0 × Cmin: BS 7671 · Reg 411.4.4 · p.73, and again at Reg 411.5.4 under Table 41.5 on p.77. You are not expected to derive it in the exam — but knowing where the number comes from stops you reaching for the wrong row.

Then apply the 80% rule to get a maximum measured value:

The 80% check

80% of 2.19 Ω = 1.75 Ω

Our calculated Zs = 0.78 Ω  ->  well inside. Pass.
✋Why 80%? The table assumes the conductors are at operating temperature. Your test circuit is cold. Once the circuit is in use the current warms the cable and the resistance rises, so the measured figure has to sit below the tabulated one.
The 80% figures are printed too. Check against On-Site Guide · Table B6.
TN-S · 20 A Type B circuit breaker Zₑ measured 0.42 max for TN-S is 0.8 Ω — comfortably inside R₁ + R₂ 0.36 Zₛ = 0.78 Ω 0.78 maximum tabulated Zₛ = 2.19 Ω Table 41.3 PASS — well under
2.19 Ω is not arbitrary: Zₛ = (Uₒ × Cmin) ÷ Iₐ = (230 × 0.95) ÷ (5 × 20) for a 20 A Type B.

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