Little help please just working my way through some practice exams for me I&T

[Greggy]

come across this one been stuck on it for ages please help also please tell me how you came to the answer as I can guess the answer but I always like to know how. Here is the question thanks in advance. A periodic inspection on an existing single phase installation which has 12 circuits protected by BS3036 fuses shows its Ze measured at 0.09Ω. This is potentially dangerous because a. it shows no earthing arrangements b. the prospective fault current may be to high for the circuit protection c. the BS3036 fuses will not meet the disconnection times d. the prospective fault current maybe to low.

 

[Spoon]

Hi Greggy. Best bet is to post what you think the answer is and your reasons why, then you usually will get feedback.
The people here dont like doing homework for people unless they try first...

 

[Greggy]

I have been looking for sometime I have table 7.4 up in my onsite guide but I am truly stuck on this one

 

[Silly Sausage]

For starters, try looking up the breaking (or short-circuit) capacity of BS3036 fuses.

 

[Greggy]

where would I find this in my OSG or 17th edition because I cant and yes I have looked in the index

 

[Zebra Stripe]

Hey Greggy, relax, this is how you learn.
Look through both books, read, get side tracked by another subject within the book, read take it in and eventually you find the answer...try starting in section 7 OSG.

 

[DPG]

At least tell us which one answer is an obvious wrong one (and why). Daz

 

[Greggy]

Anyone please......please help

 

[DPG]

At least 2 people have already helped - did you check Archy's or Zebra's suggestions? What's your thoughts on answer a? Daz

 

[Zebra Stripe]

Greggy, part of the answer lies in the OSG section 7. Think about what would increase if the impedance of a circuit is low and would a certain thing be adequate to withstand this...shirley, Ive given it away again?

 

[Richard Burns]

If you have a resistance (Ze) and your nominal voltage of 230V.

Use Ohm's law to calculate the current, which will be approximately the prospective fault current.
I = V/R
With this value have a look at the answers (and the posts above) and see if you can see what the result will be.