help with questions on 2391

[amateursparky]

for q26 i dont get how i can work out the answer without r1 and rn end to end values

 

[SPECIAL LOCATION]

You are given the length of the ring...

and from tables in the OSG you can work out the resistances per meter is for most common CSA's...

:Salute  

 

[amateursparky]

sorry can you someon spoon feed me a little. were in the osg am i looking? i feel out of my depth here.

 

[Doc Hudson]

sorry can you someon spoon feed me a little. were in the osg am i looking? i feel out of my depth here.

I do hope you are not out of your depth with something so basic?  The contents start on page 3 and go thro to page 5 of the OSG, if you cannot work out which section(s) you may or may not need to calculate the expected resistance of a cable with a given length and CSA, then really you should not be attempting any exams yet. May be good to contact your tutor if they have missed this out of your course content?

Doc H.

 

[amateursparky]

sorry. found it now.

Panic is setting in. I have failed my practical because i rushed it.

I knew how to do all the things iI missed and my tuor knews it but she had to fail me because be unfair on others.

She will also be doin a diservice to the industry if she passed me.

 I will be trying to find someone i can train under before setting out on my own. It's not easy being 33.

The goverment favours lads who are more interested in getting laid then laying cables(pun intended)/getting smashed at the bar than laying cables.

In 30 years time perhaps i will be having this conversion with someone else.

 

[kerching]

So,what did you get for your answer?

 

[TonyMitchell]

As the OP seems to have disappeared from this thread, I'll give it a go, as am new to RFCs & still learning (I do temporary power, where it's all radials)...

OSG Table I1 shows (R1+R2) for 4mm2+1.5mm2 CPC as 16.71mΩ/metre

(58*16.71)/1000 = 0.97Ω

97/4 = 0.24Ω

I'd say the multiple choice answer is (b) 0.27Ω - approximately 1/4 of the (R1+R2) plus cpc loop resistances.

This is well within the maximum of 1.1Ω shown in GN3 Table A4, for a 32A Type B protective device.

 

[Sidney]

^^that'll do...^^

 

[Sensai]

OSG Table I1 shows (R1+R2) for 4mm2+1.5mm2 CPC as 16.71mΩ/metre but what you actually need to do is use the 1.5mm2 line conductor column to get 12.10 as your multiplier. Just because the column says line or protective conductor doesn't matter it's the cable size you need to base it on. That's the trick to this question from City and Guilds. So in effect the sum is:

(58*12.10)/1000 = 0.70Ω

So the multiple choice answer is (c) 0.70Ω

Which is well within the maximum of 1.1Ω shown in GN3 Table A4, for the 32A Type B protective device.