A Question from the 2392/52

[Bear]

Morning all,

Am studying for my multiple choice questions next month, and my brain does not seem to get these type of questions? Want to now if am doing it right and understanding the theory correctly. The question below and then the way i got my answer.

A radial socket outlet circuit is supplied by 2.5mm2 cable and is 20m long. Given that the Ze of the installation is 0.43Ω and that the resistivity of the cable is 7.41mΩ/m at 20°C, what is the calculated Zs?

Answers
A. 0.9Ω B. 0.6Ω C. 0.58Ω D. 1.2Ω

So how i worked it out. Zs=Ze+(R1+R2) Zs= 0.43+(7.41x20/1000) =0.5782

Now i rounded it off to 0.58Ω and the answer was right, But should i be rounding off? was the calculation correct? and what if i dont know the length of the cable?

Any help is appreciated guys and girls, am pulling my hair out lol

 

[telectrix]

spot on from my seat.

 

[Cid]

The answer is correct. That is the sort of figure you'd expect in a domestic situation.

In the real world you don't need to know the length of the cable. You work out the resistance by doing continuity test.

 

[Zerax]

I don't know whether you plugged the numbers into a formula that you'd remembered then used a calculator or not... but just be cautious that you don't lose sight of why you work it out that way...

For me... my memory is very bad (and no, it's not an age thing, it always has been !) so I find it better to know the logic, rather than a rote learned formula.

So my logic is... if you have two resistances in series, you add them... so I need to add-on the resistance of the cable run... the longer the cable, the higher the resistance... if it's 7.41 milliΩ per metre and 20 metres long... it's double 7.41, then add an nought... so that's just less than 150 on top of the 430 we already have... making 580mΩ... or 0.58Ω

 

[telectrix]

I don't know whether you plugged the numbers into a formula that you'd remembered then used a calculator or not... but just be cautious that you don't lose sight of why you work it out that way...

For me... my memory is very bad (and no, it's not an age thing, it always has been !) so I find it better to know the logic, rather than a rote learned formula.

So my logic is... if you have two resistances in series, you add them... so I need to add-on the resistance of the cable run... the longer the cable, the higher the resistance... if it's 7.41 milliΩ per metre and 20 metres long... it's double 7.41, then add an nought... so that's just less than 150 on top of the 430 we already have... making 580mΩ... or 0.58Ω

that's my thinking also. i never divide by 1000, (same with volt drop). i get answer in mV or m ohhms. then menatally div. by the 1000....do most in my head.

 

[Bear]

Thanks all.

Like i say am just studying like mad right now i haven't done any electrical stuff in ten years, due to job change i need to relearn it all.

Think the problem is overthinking it all sometimes. thanks

 

[newfutile]

I don’t think any of the answers are right ! tho that’s most likely the answer there looking for.
20 m of cable R1+R2 is 40 m of conductor ,however the answer then is 0.7264

 

[R-fur]

Re rounding it off, most calculations are done to two decimal places so making 0.5782 in to 0.58 is perfectly acceptable.

 

[FatAlan]

I don’t think any of the answers are right ! tho that’s most likely the answer there looking for.
20 m of cable R1+R2 is 40 m of conductor ,however the answer then is 0.7264

My thoughts as well FWIW. The danger of multi choice questions which if badly worded can be quite ambiguous.