2391 written exam last nite

[bobby101]

did the exam last nite i thought the exam was not to bad still quite tough and the dreaded VD question came up by VD i mean voltage drop i think i managed to work it all out but i made the stupid stupid stupid mistake of saying it was within six percent instead of five dont know how heavily i will get penilised for that but was quite impressed with myself for being one of the very few to work it out dont get me wrong i did make a few more schoolboy errors on the exam too but fingers crossed got my practical on mon at 8am aarrgghh had a few beers last nite though how did the rest of you find it and what questions did you think were nuts cause i think there were a few

 

[GStueyXR]

the last 2 part of the question told you that the circuit didnt comply with Vdrop requirements

 

[T&Earth]

Hi All fellow 2391

I also did my 2391 Thursday night, just reading these post has brought all back to me!! I found it difficuilt not realy knowing what the question was asking.

Lets hope we have all passed

Steve

 

[ezzzekiel]

the long wait is the worst, and ive just heard the guy marking is on sick leave so could take upto 6 months





just a joke before you start biting at the bit - sure you all did fine, and if not you have better understanding next time round.

 

[bobby101]

hi guys i just tried a formula calculator online regarding the voltage drop and it came out at 3.52 volts 10mm cable 40 amp load and 20 meters in length ???????
also suprised nobody picked up the fact the calculation i posted had a deliberate mistake on it honest ...... can you spot it !!!!

 

[scotsparky]

The circuit was 80 metres long....

But they did not give you the Voltage drop figures, just the resistive load of the cable....

The answer is 13.5 volts.

i read 20metres . you use the resistive load to work out the volt drop and you have to know the maximums.

230v power 5% =11.5v
230v lights 3% = 6.8v
400V power 5% =20v
400v lights 3% =12v

 

[Baal]

It was 80 Metres

 

[bobby101]

why would you double the lenth if there is no voltage present in the neutral conductor !!! !!!

 

[GStueyXR]

Can anyone say, was it 20 or 80 meters ???

 

[bobby101]

well like i said i did the calc online via a preprogrammed calculation and it did not double the length of the cable for the readin or result

 

[GStueyXR]

why would you double the lenth if there is no voltage present in the neutral conductor !!! !!!

Because there is a voltage in the neutral conductor.... think of it as 3 resistors in series..... Line conductor-load-neutral.

The neutral conductor dosen't know its a neutral conductor.... If there was no potential difference, then current could not flow. And we know that current flowing around a circuit is equal at every point barring paralel circuits.

 

[Baal]

I think it should be doubled, the reason it's not normally in this calc is that the volt drop tables in BS7671 already allow for 2-conductors whereas the resistance values given in the exam were for single conductors.

 

[GStueyXR]

----------------------R1 ( 5 Ω )--------- Rload ( 10 Ω )--------- RN ( 5 Ω )

-----------------|=======|---------|========|-------|=======|------------

R1=Line Conductor
Rload = Load
RN = Neutral conductor

If we put 100V through this circuit, then 5A would flow through it 100/(5+10+5).... So we know therefore that V=IXR.

VDrop at R1 = 25V, Rload = 50V....

Leaving us with a PD from the point of connection with the load-neutral connection, all the way back to the centre point of the transformer Of 25V .........




Volt drop in any circuit is 100% therefore, the Vdrop must be 25V across neutral in this circuit



This in my opinion, is underpinning knowledge....That's why they tested it.

 

[Toonlad]

I worked it out as single conductor.
So if im wrong,i should get 1/2 points for half the workings imo, so not too much to worry about.

Good luck to those waiting for there results,its all done now, just need to wait the 40 odd working days for the results.

 

[brucelee]

hi all
I am sure it was 80 meters in length, 45A load, a 10mm conductor and if the mohm/M was 1.83
I did it like this and think i trashed it
Mohm/M X L divide by1000
1.83 X 2 X 80 divide by 1000
= 292.8 divide by 1000
=0.2928
=0.3ohm
then this is where i got lost
0.3 x 45 = 13.5v
all the time i was thinking its something to do with V=IXR but was confused and thinking mv/A/M x IB x l divide by 1000

was the resistance of the cable 1.83 or as in a earlier post 1.87

 

[GStueyXR]

it was 1.87

 

[brucelee]

cheers mate i thought it was but in Gn3 its 1.83 but it doesnt really make a difference because of rounding up when working resistance out you still get same answer
should have not panacked as now thinking rationally its pretty easy to do just never been shown that way
all the time iwas thinking its something to do with V=IXR
i would like to see the paper again to check where i messed up

 

[SafetyFirst]

i really didnt under stand the first question.... lol... think i put accsesoys comply to bs or equilavent standard... installation complys to bs7671 and not defected as to cause danger,,, or something along those line... any one else remember the question... or what they put???

I spent about 5 mins trying to interpret that first question and i still couldn't work out what they were asking me. I started writing about check for damage ie overload and crossed it out wondering if it was supply characteristics so skipped it and left a mark on question book thinking i'll come back to it later on. Left another one or two out in the same way thinking i'll come back to them after answering section 2. I still managed to spend over an hour on section A even though i had time to spare in any mock papers i've done.

I ran out of time just finished anwering qu 25 so i'm at least 20+ marks down already, hopefully the ones i did answer i got top marks otherwise a resit awaits.

I actually spent more time reading the questions than answering them, this exam is crazily worded, for those who pass this i have every respect, but i think soemtimes they want rocket scientists only to pass this course. I've revised like mad but nothing could prepare me for the purposefully silly way they ask questions.

I remember some questions that were asked but remember the wording in most of the questions was not wording that ought to be used for an exam in any conditions, unless there's an exam out there for 'interprating really awkwardly worded questions'!
What makes it even sillier is that City and Guilds have the cheek to constantly say in each of there examiners reports that students fail to grasp and understand what is asked of them in the question!

Well bugger me change the wording into practical easy to understand ENGLISH!!!!

 

[brucelee]

could not have put it better my mate when we get results we should all send them a email telling them exactly what we thik of there wording this was how i felt after?

 

[SafetyFirst]

could not have put it better my mate when we get results we should all send them a email telling them exactly what we thik of there wording this was how i felt after?

I'm of the same feeling. In fact i want to send them a letter right now, i wonder if any folks on here have before? It is rediculous. However i daren't send any views now incase it has any negative bearing on my result. Can you remember many of the questions to reproduce them on here anyone? I will post a few on here that i remember another time after i've drowned me sorrows over that GREEN moment!

 

[scotsparky]

----------------------R1 ( 5 Ω )--------- Rload ( 10 Ω )--------- RN ( 5 Ω )

-----------------|=======|---------|========|-------|=======|------------

R1=Line Conductor
Rload = Load
RN = Neutral conductor

If we put 100V through this circuit, then 5A would flow through it 100/(5+10+5).... So we know therefore that V=IXR.

VDrop at R1 = 25V, Rload = 50V....

Leaving us with a PD from the point of connection with the load-neutral connection, all the way back to the centre point of the transformer Of 25V .......




Volt drop in any circuit is 100% therefore, the Vdrop must be 25V across neutral in this circuit



This in my opinion, is underpinning knowledge....That's why they tested it.

the volt drop if for the end of line of the circuit. wher you would read a ZS not thfor the neutral as no load is drawn off the neutral side.

you also said there is always a voltage in the nutral interesting comment what about ballanced three phase supplyes where no neutral is used?

 

[scotsparky]

I think it should be doubled, the reason it's not normally in this calc is that the volt drop tables in BS7671 already allow for 2-conductors whereas the resistance values given in the exam were for single conductors.

what volt drop tables? never seen them can you point them out. or are you thinking of the tables in appendix 4 which give a resistive value in mA?

 

[brucelee]

think he does mean the mv/A/m tables that are with the current carrying capacity tables and it does say voltage drop in appendix 4

 

[GStueyXR]

you also said there is always a voltage in the nutral interesting comment what about ballanced three phase supplyes where no neutral is used?

When a circuit is energised, there is a potential in neutral..... Without it, no current can flow.

You know neutral has a resistance (remember little Rn) So lets say neutral = 1 ohm but V=0 do the calcs yourself mate.........

NO CURRENT FLOWS.... But we know equal current flows in all parts of circuit.... think of the water analogy !!!

And 3 phase dosent need a neutral, you're right. Because it uses the potential difference between line conductors. No offence meant here, but this is basic stuff... Guess what though, single phase dosent need a neutral either..... it'll quite happily use your butt as a path to earth.

 

[scotsparky]

i have done the calcs using the information. i passed 2391-10 in 2008 first time and 2391-20 first time and an playing the devils advocate a bit.

I mentioned the 3 phase to try and prove a point. you use the tables in appendix 4 and the lenghty of the cable. you have no need to double the lenght of run for voltage drop.

 

[GStueyXR]

i have done the calcs using the information. i passed 2391-10 in 2008 first time and 2391-20 first time and an playing the devils advocate a bit.

I mentioned the 3 phase to try and prove a point. you use the tables in appendix 4 and the lenghty of the cable. you have no need to double the lenght of run for voltage drop.

I will not argue that... My argument is the neutral puts up a resistance... So it must therefore be built into the calculations in the tables. There is no need to seperate them, after all L and N conductors must be equal size.

BUT, we were not given tabulated values of Vdrop.