C&G 2391-101 July 2007

Well the 19 July 2007 finally arrived which for me meant at 18:30 I began the City and Guild's 2391-101 examinination.

The room was full of candidates sitting the exam, of those that took the course with me only one was missing.

Having completed the passed papers with relative ease, I did not feel under pressure, that was until I actually turned the paper over.

The first 20 questions took me one hour to complete. The second 6 questions took 90 minutes to complete. So the two and a half hour exam took that long for me and all the others but for a small minority.

Although the questions were in a similar vane to those I had practised with, the actual wording meant that for me at least, two or more possible answers could have been given.

The scenario for Part B was a Public house extension. Two questions involved an air curtain. (I have since found out that an air curtain is the name for the heating device fitted over many shop doorways.) This led to a little bit of a stall, since I had to try and think what one was.

Question 20 and 26 seemed to both ask the same question and I had to check back through the question paper to verify I had not made a mistake.

Why is PFC prospective fault current important in selection of a protective device? Okay the answer is easy. The protective device needs to be of a sufficient rating to break this current without substaining damage. The other part was easy too, which would need to be recorded the PSCC or PEFC. The answer is the highest since it is this current that determines rating required.

The DB was 20 Way with 14 ways used, the individual circuits tested at >200Mohms. What would the overall test be.

so 1/Rt = 14/R ; 1/Rt= 0.07Mohms

Thus we have 14.286 Mohm.

Perhaps the longest question to answer was on the ring final circuit test showing calculations.

You were given the loop length of I believe 40 Metres. A table gave resistance per metre length of all the csas used in the new installation.
So for 2.5mm this was 7.41 and for 1.5mm 12.1.

So for stage 1 you would have 40*7.41/1000 for phase and neutral conductors. The CPC would be 40*12.1/1000.
So PHr and Nr would both be 0.2964 while the cpc would be 0.484 ohms.

Stage 2 would be (Phr+Nr)/4 ; (0.2964+0.2964)/4 = 0.1482 ohms

while Stage 3 would give (PHr + CPCr)/4 ; (0.2964 + 0.484)/4 =0.1951 ohms

In addition to the calculation you also had to go through the procedure to conduct the 3 stages.