# It and Iz

Im getting a bit confused with It and Iz, and which is out final value when calculating cable size

It  - the value of current tabulated in this appendix for the type of cable and installation method concerned, for a single circuit in the ambient temperature stated in the current-carrying capacity tables

Iz -  the current-carrying capacity of a cable for continuous service, under the particular installation conditions concerned.

In =32  Ca 0.87

so    It ≥  32 / 0.87

It ≥ 36.78A

Look up tables in BS 7671  clipped direct (c) one  2 core cable

4mm=36A

6mm= 46A

So we go with 46A

So our It (Tabulated value we look up) is 46A and we will use a 6mm2 cable

So what is Iz?

Iz ≥ In ≥ Ib  understand the relationship

Now Iz = It x correction factors.
Iz = 46 x 0.87 =  40.02A  which is greater than In (32A) so thats OK

But we got a value of 36.78 A when applying correction factors to In

We also have this formula...

It ≥ Iz / correction factors

40.02 / 0.87 = 46

We end up with the same number.

We have a calculated value of 36.78A  in the book the cable choice is 6mm which is rated at 46A

I'm struggling to see what we would call our final cable choice? Its not Iz , but the description (continuous service) makes it sound like it is

What is Iz

Hope that make some sense

Its based around this IET article

Parents
• Thank you both for two really helpful replies.

I have tried to follow the process you mentioned as in the EIDG

This is a really simple example I have chosen. probably too simple

We have the information we know.
Then we have the desired  Iz

Is this right?

• That doesn't look quite right to me. Fundamentally we need the cable as installed to carry the required current (in this case Ib = In) so I would have thought that Iz ≥ In would do. The correction factors convert It to Iz where the installation conditions differ from what's assumed by the tables.

I'd probably say your 26.3A is the minimum required It (rather than Iz).

Yes, 20.67A is the Iz of the cable as installed - which meets your required CCC  of 20A.

- Andy.

• Cf only applies for BS 3036 semi-encloses rewireable fuses.

Regards Ca, Your example doesn't say what the ambient temperature is ... but Reference Method C includes for that if the ambient is 30 degrees.

Similarly, what is your assumption for Ci ?

• Ci 50mm 0.88 Ca 40oC

• 2.4.3 in EiDG

• 2.4.3 in EiDG

Humm, I won't have the latest version to check against, but I wonder if you're spotted a typo there. To my mind it should be It ≥ In/C (or Iz  ≥ In).

- Andy.