A bit of help needed here!

[Goody]

I am filling in an EIC for the electrical installation with a light & a socket circuit for the newly built “Out-house” (garage conversion) and I have used the existing link from the main DB.

At the main DB -> Ze = 0.10 Ω , PFC = 4.0 KA and R1+R2 of the 2.5 mm link cable from the main DB to the Out-house DB = 0.32 Ω

Should I now assume that Ze at the Out-house DB is 0.10+0.32 = 0.42 Ω?, if not, how do I calculate Zs for the Out-house lighting circuit whose R1+R2 = 0.42 Ω

Again on the EIC which “DB’s” details should I enter for “Main Switch or circuit-Breaker” and “DISTRIBUTION BOARD DETAILS” boxes? Please refer to the pictures. I think main DB's details are not relevant here.

Many thanks!

 

[spark 68]

Hi Goody,

As you say the Ze or Zdb for your second DB is 0.42, your lighting circuit R1+R2 is also 0.42 (on this board ?) so I would say the Zs for this circuit is 0.84

Ze + (Rx1+Rx2) + (Ry1+Ry2) = Zs where x is the submain and y is the Final circuit of the second DB, if that makes sense