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Calculating prospective fault current, and what makes it fall laterFour spotlights became eight. What changed?

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Using the Zs from the volt drop example

Lowest fault current — for checking disconnection:
I   = (Uo × Cmin) ÷ Zs
    = (230 × 0.95) ÷ 0.98
    = 218.5 ÷ 0.98
    = 222.96 A  =  0.223 kA
For breaking capacity, measure the maximum. A device sees its highest fault current at its own terminals, where the impedance is lowest. Guidance Note 3 · 2.6.16 · pp.88–89 has you measure the maximum prospective fault current, record the greater of PSCC and PEFC, and check it is below the device's breaking capacity. The calculation above gives the minimum, not the maximum.
Record it to three decimal places unless told otherwise — that is what the instrument gives you.
Where Cmin comes from. It is the minimum voltage factor, to take account of voltage variations. BS 7671 · Reg 411.4.4 · p.73 defines it, and its Note 1 gives 0.95 for low-voltage supplies under the ESQCR. It is not tied to conductor size.

Now the question worth understanding: what makes the fault current lower later on?

Work it from Ohm's law

I = U ÷ R  —  and R can be swapped for Z

So for the current to fall, the impedance must rise.
What raises Zs after you leave
TemperatureThe cable is in use now and the copper is warm
LengthThe client extends the circuit — more sockets, a spur, an extension
Dry ground, on a TT systemThe electrode reading drifts up
The example that makes it land. Four spotlights in a kitchen when you certify it. You are called back five years later for a periodic and there are eight. Nobody added a circuit — they just added cable. More length, more resistance, less fault current than you recorded.
Uo is not the variable. 230 V is nominal and only moves within the supply tolerances — the +10% / −6% table at the end of BS 7671 · Appendix 2 · p.436.

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