The symbols you need coldIb, In, Iz, It. Which is which?

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I₂ and the 1.45 factor

A second formula worth knowing, from Reg 433.1.1(c) (p.103), and it explains a number people meet without understanding.

I₂ ≤ 1.45 × I_z
I₂ = the current that ensures effective operation of the device · 1.45 is a factor, not a current

For a circuit breaker, I₂ is 1.45 × I_n: the current at which it is sure to operate within the conventional time. On the Appendix 3 curve that sits in the slow thermal (overload) region, not the instantaneous part. So once I_n ≤ I_z, a circuit breaker meets I₂ ≤ 1.45 × I_z automatically.

A 6 A type B lighting breaker

Effective operation:  1.45 × 6  = 8.7 A
  → sure to trip, within the conventional time

Instantaneous trip:   5 × 6     = 30 A
  → operates within 0.1 s

A lighting circuit normally draws 2–3 A,
so 8.7 A is a long way above normal use.

!For a BS 3036 semi-enclosed rewireable fuse the rating factor is 0.725 — exactly half of 1.45, because a rewireable fuse needs about twice its rating to operate in the conventional time. Different device, very different behaviour.

Overload versus fault, on one graph. The long sweeping curve is the overload region — a modest excess can take a very long time to trip. The straight vertical part at the bottom is the fault region, where a big current trips it almost instantly.

433.1.1(c) · p.103 · I₂ ≤ 1.45 × I_z Appendix 3 · Figure 3A4

At its rated 6 A the breaker carries current all day. At 8.7 A — 1.45 times the rating — the breaker is sure to operate, but only within the conventional time, not instantly; the further above it you go, the quicker it operates. Only at 30 A, five times the rating, does the magnetic trip act instantly. People meet 1.45 as if it were an instant trip current; it is the factor that gives I₂, the current ensuring effective operation.

The symbols you need cold

Make sure you know these, and highlight them in the abbreviations list towards the end of Part 2.

Symbol	Means	Where it comes from
I_b	Design current	Calculated from the load: P ÷ V
I_n	Nominal rating of the protective device	Chosen so I_b ≤ I_n
I_z	Current-carrying capacity of the cable in its installed conditions	I_t × the correction factors
I_t	Tabulated current-carrying capacity	Read from the Appendix 4 tables
I₂	Current ensuring effective operation of the device	1.45 × I_n for a circuit breaker · must not exceed 1.45 × I_z
I_a	Current causing automatic operation in time	5 × I_n for a type B

≤The chain that ties them together: I_b ≤ I_n ≤ I_z, where I_z = I_t × the factors — so you need I_t ≥ I_n ÷ the factors. Get that order straight and most cable-design questions unlock.

Part 2 · symbols (pp.52–53) and abbreviated terms (pp.54–55), at the end of the part

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