R=pl/a and R=R1+R2

[Bear]

Hey all, i honestly cant work this out, i think its the way they ask the questions and am not sure whats going where,
am failing on my maths like normal, so i would love it if somebody can show me how to break this question down and how i answer it correctly. (ie. show my working)

Given that the resistance per meter of 2.5MM² cable is 7.41mohms/m at 20ºC ,
What would the approximate R1+R2 reading be at the extreame end
of a 25m long radial socket circuit?

Now the answer is 0.37ohms, and it refernces to R=pl/a and R=R1+R2,

but am not sure where am going wrong or how am calculating it. am i right in thinking that my R=R1+R2.

R=7.41*25/1000, 0.185

Or am i completely wrong here?

Thank you to all that help me out, i know most of my help has come from here and i appreciate it all.

 

[telectrix]

your 0.185 is R1 (or R2). double it to get R1+R2..... 0.37.

 

[westward10]

You need to consider the line and cpc, the 7.41 they are giving is for one conductor. If you have two 2.5 conductors that would be two lots of 7.41 for the line and cpc, in other words 14.82. Try it again with 14.82.

 

[Bear]

OMG!!!! honestly i feel daft right now, simple answer thanks guys,

so in a question like that its asking for the line and cpc to be calculated together,

thanks

 

[Strima]

As above, double it.

Edit: Assuming the CPC is also 2.5mm.

 

[westward10]

OMG!!!! honestly i feel daft right now, simple answer thanks guys,

so in a question like that its asking for the line and cpc to be calculated together,

thanks

Easy done mate.

 

[telectrix]

OMG!!!! honestly i feel daft right now, simple answer thanks guys,

so in a question like that its asking for the line and cpc to be calculated together,

thanks

we all have a brain fart now and then. even us old coffin dodgers make mistakes.