gkenyon said:

I_z=I_t×C_g×C_a×C_s×C_d×C_i×C_c(×C_f) ... see equation (2.4.0) in EIDG (page 19 in the 5th Ed) with the addition of see equation (2.4.3) in EIDG (page 21 in 5th Ed).

Am I missing something? Surely that is not true if using a rewireable fuse. The current carrying capacity, Iz, is not affected by the type of fuse. 

The current carrying capacity, Iz, is not affected by the type of fuse. 

Sort of - BS 3036 fuses are slower to respond to overloads so where overload protection is required, you may need a larger cable to give the same margin of safety as you'd have with other devices (which normally have a fusing factor of 1.45 or less) - so yes it's strictly true that Iz shouldn't alter, it is a convenient way of preserving the margin when matching a cable to the fuse rating. Like a while back when we had separate tables for 'course' and 'close' protection - with lower tabulated values for course (i.e. rewireable fuses).

   - Andy.

Thanks for all your help with this.

Don't want to keep bothering, but I think I am getting there..

In the 3rd drawing I have 3 scenario's

Using the 3 formula mentioned in EIDG 

Calculating Iz 
Calculating It  using In
Calculating It using Ib

lyledunn said:

Surely that is not true if using a rewireable fuse. The current carrying capacity, Iz, is not affected by the type of fuse.

YEs ... that is what Cf is for .. and why it's in brackets because it can be ignored for cases where rewireable fuses are not used.

I_z=I_t×C_g×C_a×C_s×C_d×C_i×C_c(×C_f)

Not quite sure what is bothering you. Isn't the upper half of your image saying exactly what I did?

In practical terms, you look up the tables and select cable size accordingly. It might be that on the north side of a house up north you could reduce the ambient temperature, but why bother?

Even here on a hot day on the south coast, ambient temperature is not 30 ºC, though it could be higher if the sun is shining on the electrical equipment.

Perhaps a bit of engineering judgement is required?

Using the 3 formula mentioned in EIDG 

Calculating Iz 
Calculating It  using In
Calculating It using Ib

At first glance, it all looks reasonable to me

   - Andy.

Chris Pearson said:

Not quite sure what is bothering you

I don't think that Iz is very clearly explained, in all of the books or articles I have read.

I see lots of articles that seem to misuse the terms
The fact that the calculated value, has no name just It≥  bugs me.  Because this can easily be mis thought of as Iz.

I have seen no publications that give worked examples of  Iz  (2.4.0 formula  EIDG)

A basic misunderstanding is that this is down rating the CCC of the cable, Not reducing the current, so a smaller cable could be used.


The guidance notes are good but frustrating, because I find they tell you but don't teach you. I want deeper explanations.

The actual basic cable calculation is really straight forward.

But Iz is confusing until you get it, which I think was picked up here

AJJewsbury said:

- and it's often not explained at well either.

I can imagine that my basic questions, and misunderstanding can be a bit frustrating, but I like to ask on here because you guys know your stuff and are generous with your time to explain, without any condescension. Which I really appreciate.

gkenyon said:

I_z=I_t×C_g×C_a×C_s×C_d×C_i×C_c(×C_f

 With Cf at the end, I deem the above to be technically incorrect. It may well assist in facilitating  how we arrive at;

but something technically incorrect is not a good starting point for an explanation. If it were correct, then we would need to use Cf for circuits that did not need overload protection but were protected by a 3036 fuse requiring only fault current protection.

Correctly, the EIDG does not include Cf as shown in the top equation as the type of overcurrent protective device does not affect Iz.

Whilst we ultimately end up at the same place, the strict application of regulation 433.1.202  gives the maximum rating of fuse to be used when the current carrying capacity (Iz) of the smallest conductor of the circuit is multiplied by 0.725 for a 3036 device with a fusing factor of two.

Since that would be kind of impractical, Cf is applied as a divisor to In to give the minimum value of Iz (or It if referring to the ccc in a table)

Suppose we had a 2,5mm2 single-phase cable clipped direct as Table 4D2A column 6. No correction factors apply other than we are using a 20A 3036 fuse for overload protection. It=27A and by dint Iz = 27A

(a) what is the value of Iz?

Since Cg, Ca, Ci etc are all 1 then Iz =It * 1 = 27A  (Cf is ignored as it does not alter the actual value of Iz). Iz is definitely not 27 * 0.725 as would  be the reasoning of the topmost equation.

(b) what is the maximum rating of 3036 fuse that could be employed?

In must be less than or equal to;  0.725 *27 = 19.575A . 

That is hardly a practical method when we are looking for the minimum current carrying capacity to be used, although In could be dropped to 15A depending on the value of Ib.

(c) What is the minimum value of Iz that can be used; Iz minimum =  In/0.725 = 20/0.725 = 27.6A. Since the 2.5 can only carry 27A, strict application would indicate that when using a 20A 3036 fuse for the circuit, the arrangement would not comply.

In that regard, a 4mm2 cable would need to be used with an It of 36A.

Now of course, it would have been infinitely easier just to apply the second formula from the top to find the minimum value of It;

It  must be greater than or equal to In / 1*0.725  = 27.6A   

(d) Minimum size of cable if only fault protection is required and Ib =20A;

Ib must be less than or equal to In

and Ib must be less than or equal to Iz

It must be greater than or equal to Ib/applicable correction factors. 

In this case Cf is not applied so; 20/1 =20. So the minimum value of Iz is 20 and since 2.5mm2 cable is used, It is 27A. Thus, the conditions are satisfied (although further calculations for thermal protection will be required).

JohnE said:

I see lots of articles that seem to misuse the terms
The fact that the calculated value, has no name just It≥  bugs me.

I don't disagree with that statement.

JohnE said:

Because this can easily be mis thought of as Iz.

At the end of the day, the normative requirements in Part 4 of BS 7671, state the requirements in terms of Iz.

Appendix 4 explains quite clearly, that you are looking up a value It with a view to it meeting or exceeding the minimum acceptable value Iz for the installation conditions in question.

JohnE said:

The guidance notes are good but frustrating, because I find they tell you but don't teach you.

That's interesting ... In my opinion, the guidance notes in general are not, in themselves supposed to be teaching materials, but provide information as either a 'refresher', 'cpd', or to inform those who are being taught using appropriate materials.

I can provide this feedback, and perhaps some appropriate publications could be devised to fill the gap, if there is one?

lyledunn said:

but something technically incorrect is not a good starting point for an explanation. If it were correct, then we would need to use Cf for circuits that did not need overload protection but were protected by a 3036 fuse requiring only fault current protection.

But you can use C_f for all circuits?

433.1.202 gives you the value of Cf for a BS 3036 fuse different to that in general for 433.1.1 (iii) where Cf for other OCPDs is 1, if you take the view that Cf=1 applies to 433.1.1 (iii) an this is clarified for BS 3036 fuses by application of Cf = 0.725 ?

This is the view that is portrayed in 5.1.1 (a)  of Appendix 4 to BS 7671?