It is not necessary to know Iz if method use is5.1 and 5.2

Question

Basic cable calculation Checking Iz From BS7671 Method is used in items 5.1 and 5.2 Ib or In / Correction factors 
 
 It should be noted that the value of (It) appearing against the chosen cross-sectional area is not (Iz). 
 It is not necessary to know Iz where the size of conductor is chosen by this method. But it is possible to get incorrect coordination with cable calculations that seem acceptable? 
 
 Ib 28A 
 In 32A 
 L 40m 
 Ca 0.94 
 Cg 0.7 
 
 Cable T&E 
 Clipped direct 
 
 Calculation 
 
 It &ge; Ib &divide; correction factors so: It &ge; 28 &divide; 0.94 x 0.7 = 42.5A 
 
 Table 4D2A Method C 6mm2 cable - 46A It = 46 
 
 Volt drop 8.17V 
 
 So 6mm cabe looks OK 
 
 But if I do check Iz 
 
 Iz = It x Correction factors 46 x 0.94 x 0.7 = 30.26A 
 
 Ib &le; In &le; Iz 28 &le; 32 &le; 30.26

AJJewsbury · Accepted Answer

I think there are two different ideas getting mixed up here and making things look more complicated than they need be. 
 Firstly there's the idea of choosing "how much current you need the cable to carry" - if it only needs to carry what the load normally draws, then you can use Ib, if on the other hand it may need to carry more than that, up to the limit of the protective device (i.e. overload protection is needed) then you use the rating of the protective device - In. 
 Quite separately from that, there's the relationship between the cable's current carrying capacity in "standard" conditions (i.e. as tabulated, before applying correction factors) and that it will actually have under the conditions it's installed in (Iz). 
 The "obvious" way to calculate things might be to take the tabulated value, multiply by all the correction factors and compare that with "how much current you need the cable to carry". The trouble with that approach (especially in the days when electrician's didn't have a computer to hand) is that you end up having to do the same calculation umpteen times over to check a variety of cable sizes. 
 So they came up with an alternative approach of doing the calculation the "other way around" - i.e instead of calculating how much each size of cable could carry under the actual conditions, you'd instead calculate what the current carrying capacity of a just-the-right size cable would be if it was installed under standard conditions instead of the actual conditions - because you can then calculate one value and look it up in the tables once. 
 That might sound a bit odd, but mathematically it's quite simple - instead of multiplying two numbers to give a final answer, you take the final answer and divide by one number to give you the other. 
 So instead of calculating Iz for each cable size as It * C (and compare the result with "how much current you need the cable to carry"),you'd take "how much current you need the cable to carry" and divide by the correction factors to give you what would be the "tabulated" value of your ideal cable - you just then look up a size that's no smaller than that. 
 does that make things any clearer? 
 - Andy.

It is not necessary to know Iz if method use is5.1 and 5.2

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