the maximum length of cable permissible

[Banburyman]

[FONT=&quot]I found this question in "practical giude to inspection, testing and certification" book

A lighting circuit is to be wired in l mm2 twin and earth thermoplastic [/FONT][FONT=&quot]cable, the circuit is protected by a 5 A BS 3036 fuse. What would be the [/FONT][FONT=&quot]maximum length of cable permissible to comply with the earth fault loop impedance requirements (Zs)? (Ze, is 0.45.)[/FONT]

 

[acat]

Untitled

Hi if you follow the link that should give you the answer you need


Chris

 

[Banburyman]

it is how I did it
Zsmax for 5A 3036 is 9.58 [FONT=&quot]Ω
rule of thumb 9.58 * 0.75 = 7,17[/FONT] [FONT=&quot]Ω
[/FONT][FONT=&quot]Zsmax - [/FONT][FONT=&quot]Ze[/FONT][FONT=&quot] =
7,17[/FONT] [FONT=&quot]Ω[/FONT] - 0.45 [FONT=&quot]Ω[/FONT] = 6.72 [FONT=&quot]Ω[/FONT]
[FONT=&quot]R1+R2 = 36,20 m[/FONT] [FONT=&quot]Ω/m

Length * (R1+R2)/1000 = 6,72[/FONT] [FONT=&quot]Ω[/FONT][FONT=&quot]
lenght = 185 meter

I try to prepare myself to C&G2391 exam but it is very difficult as I have to use a book with so many errors.
According to Practical Guide to Inspection Testing and Certification the answer is 498meters (this is one of many errors in this book)
thanks a lot I just needed somebody to confirm my calculation
[/FONT]

 

[old dog]

the rule of thumb is now 0.8 in the 17th edition

 

[grandad]

looks like they used the max zs value of 100% of 17.7 ohms for the 5 second disconnection time .
remember the .8 rule is only to give us a safety margin, not the max zs allowed.