Today's 2395

[Newbyspark]

Just finished written and online 2395. Passed online with 90%, found the written pretty hard. Anyone else done it today.

 

[Spark_shark]

Just done the written test, and think it went ok, but will have to wait and see! How did u get on?

 

[long512]

just done the written as a re sit. i had to put in loads of revision so hopefully went well? had a couple of "brain farts" but after checking my notes i'm ok. thought i new enough to get away with just a little revision last time, that cost me £120!!
best bit of advice to anyone else doing it, download and read the examiners report its available to all and gives solid advice.

 

[Newbyspark]

Couple of questions I struggled with. Have wait and see.

 

[Newbyspark]

just done the written as a re sit. i had to put in loads of revision so hopefully went well? had a couple of "brain farts" but after checking my notes i'm ok. thought i new enough to get away with just a little revision last time, that cost me £120!!
best bit of advice to anyone else doing it, download and read the examiners report its available to all and gives solid advice.

How did you work out volt drop question ?

 

[andyk1969]

sat the written tonight also think it was easier than the 2394 i took in december

volt drop question should has been 9.2 volts 0.23 ohms*40amps= 9.2 volts (ohms law)

 

[Spark_shark]

Think it was 0.23X40= 9.2 volts. As this was >11.5 (5% of 239) it's compiles. How about u?

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Beat me to it andy! Lol

 

[Spark_shark]

Less than 11.5 should I say!!

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And 230 not 239*

 

[long512]

How did you work out volt drop question ?

Wrong by them replies! lol
got the correct answer but then looks like i have used an unrequired diversity factor!!
may pick up some marks for the correct equation written down and the correct percentage allowed??

Doubt it though!

 

[wuzz]

I did the exam last night and on that equation I used the correction factor of 1.2 in the equation!!

 

[cmac70]

Done mine last night went a bit brain dead on last question. So hopefully have passed.

 

[tony mc]

Whats a 1.2 correction factor when calculating volt drop ?

Can anyone post the question up ?

I take it from some of the repies that you have a value for either Rn or R1 doubled it and multiplied by the protetive device to get your answers!

 

[Silly Sausage]

Whats a 1.2 correction factor when calculating volt drop ?

Can anyone post the question up ?

I take it from some of the repies that you have a value for either Rn or R1 doubled it and multiplied by the protetive device to get your answers!

I did my 2391 over 10 years ago, this never came up once!
Was mentioned on a recent thread, think you're getting confused with resistance testing.

 

[wuzz]

I can't remember the question word for word but it was something like this, you have combined line resistance of 0.23 and a load of 40amps calculate the voltdrop? I went for 0.23x1.2x40=11.04Vd . 1.2 for the correction factor for the operating temperature at 70·

 

[Silly Sausage]

The vd tables are for conductors running @ 70[SUP]o[/SUP] (etc).
You don't need to compensate, they are the worst case figures.
I'm more than happy to be corrected though!

 

[Spark_shark]

I can't remember whether the qu mentioned if 0.23 was the measured live conductors resistance (ie R1+RN). I myself, didn't use any correction factors, but can see why people have. I thought the that 1.2 correction factor was for correcting the R1+R2 part of the Zs=Ze+(R1+R2) equation, as its carried out as a dead test, and therefore the conductors aren't at operating temperature? But thinking about it... The same logic could be applied to an R1+RN test?

 

[tony mc]

I very much doubt that the question would be asking for correction factors in relation to a VD question!

IMO you would be using them to calculate the ZS of a cct making sure it would comply but you would need lots of info in the question along with tables from the OSG or regs book along with the ambient temperture of the room.

The rule of thumb method is often asked for when checking to see if a cct will comply but that has nothing to do with VD !

If you were given a value of either Rn or R1 and the size of MCB it would be a case of a quick calculation Rn x 2 x MCB (size) = ?? and as you know you VD % for the circuit ( 3 or 5% ) and you know the Uo say (230 V ) the answer you get will either comply or not!

 

[Spark_shark]

I agree tony, but not sure about the mcb bit? It gave the load current of the circuit and resistance values, the mcb does come into the question...

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Doesn't come into it*

 

[long512]

I very much doubt that the question would be asking for correction factors in relation to a VD question!

IMO you would be using them to calculate the ZS of a cct making sure it would comply but you would need lots of info in the question along with tables from the OSG or regs book along with the ambient temperture of the room.

The rule of thumb method is often asked for when checking to see if a cct will comply but that has nothing to do with VD !

If you were given a value of either Rn or R1 and the size of MCB it would be a case of a quick calculation Rn x 2 x MCB (size) = ?? and as you know you VD % for the circuit ( 3 or 5% ) and you know the Uo say (230 V ) the answer you get will either comply or not!

yes mate i agree, just had a self inflicted "Brain fart"! put me self under a bit of pressure due to already failing once!
My reason for failing the first time, i believe, was an over confident tutor. Kept telling us we were doing well when we in fact were rubish.
showed when we sat the exam, 3 out of 12 passed.

 

[mick1985]

Sat the 2391 around 4-5 years ago failed. It's a tough exam that

 

[Amp David]

I can't remember whether the qu mentioned if 0.23 was the measured live conductors resistance (ie R1+RN). I myself, didn't use any correction factors, but can see why people have. I thought the that 1.2 correction factor was for correcting the R1+R2 part of the Zs=Ze+(R1+R2) equation, as its carried out as a dead test, and therefore the conductors aren't at operating temperature? But thinking about it... The same logic could be applied to an R1+RN test?

It mentioned that it was the total measured resistance of R1 and Rn and that the testing was done at an ambient temp of 20 deg C.

 

[Spark_shark]

Looks like the correction factor is needed! This was from the previous chief examiners report on c and g website...

"When asked to calculate the voltage drop for a given scenario most candidates failed to include a temperature compensation factor in their calculation. As testing was carried out at 20 oC and voltage drop is calculated with the conductors at operating temperature this compensation needs to be included in the calculation."

damn it! I didn't use any correction factor, so don't know if I'll have picked up any points for that, or the question that followed on from it asking if it complied or not? I but it complied, which it would've with or without the correction factor, as both were <11.5 volts

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I put*

 

[long512]

well, spark_shark you had me thinking i had done the right thing there. Until i remember'd that for reasons unknown to me my correction factor was something like:-

40x0.23=9.2v correction factor of 1.2 required, but i thimk i have done 1 / 1.2, wich then lowers the bloody voltage to 7.6v!!!!!!!!!!!!!!!!
when it should have gone up to 11.04v!!!!!!!!!!
were i got 1 divided by 1.2 from?????????????? put it down to resit nerves!

 

[Amp David]

Done the online tonight and managed a 98%:wink_smile: