tonights 2391 exam....(10th march 2011) | Page 15 | on ElectriciansForums

Discuss tonights 2391 exam....(10th march 2011) in the Electrician Courses : Electrical Quals area at ElectriciansForums.net

L

lynj

one thing to say and that is did c&g forget how to be staightforward? :rolleyes: felt like some of the questions were half thought......

anyone else take it?
 
Cheers guys, the only thing I'm concerned about is testing the 3 phase commando plug that is on one of the rigs. I know its on there because I've seen it and I have a "photographic" memory ;);).

Testing is not that difficult but when you've got an examiner sitting on your shoulder makes you think if you've missed a step. As you've said, take a methodical approach and everything will be fine.

Practical tomorrow, cscs exam Saturday and starting a new job Monday, happy days. One week trial/depending on work load but its better than sat at home waiting for the phone to ring, and its a good firm.
 
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the three phase command is easy mate i had never tested one before
just link L1 and cpc at DB then find the correct terminals at the socket by pushing probes in the end dont need gs 38 4mm tips for dead tests of socket or the rear connections of the socket easy mate then record the highest as thats the worst case scenario
eg
L1 - cpc = 0.2ohm
L2 - cpc = 0.3
L3 - cpc = 0.2

L2 - cpc is the R1 + R2 you need to record on your schedule of test results
on the relevent line
 
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for the 3 phase commando it covers 3 lines of the schedule complete middle line and use brackets for all three lines
when you test the zs for the distribution circuit from the 3 phase db this is also your Ze or ZDB for the single phase board and calculate Zs all circuits except sockets mate even calc the zs for 3 phase commando and then right calc ZS in the remarks column at side of the relevent circuit

on radial circuit with fused spur check R1+RN also to prove continuity
on two way switches operate individually whilst conducting further tests
for continuity proves polarity when measuring R1 + R2 and for IR it ensures all circuit tested
do
all dead tests dont forget to do the main protective bonding and disconnect 1 end to remove parallel paths start on 3 phase dead tests put things back as you go then single ccu then polarity live then Ze pfc on three phase put covers on switch on make sure everything reconnected first do Zs and then rcd tests and PRESS the test button then rest of functional tests finish cert etc
before you start complete as much of cert info as you can like client details etc records of previous certs etc last test date and so on so so you have as much of the main pir done before you start the actual testing
DONT FORGET SAFE ISOLATION AND RETAIN THE KEY IN POCKET
 
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PAUL dont worry mate the examiner keeps his distance just watches from afar apart from at key moments especially the safe isolation then he basically leaves you to it and if you ask him he will tell you when you have 30 mins left out the 90mins ask his permission to isolate and lock off check meter ok leads and functioning
 
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Ok I've managed to get a quick look at the paper. And question 26 asks about the rating of the RCD needed to provide fault protection on the newly installed lighting circuit.

Tab 41.5 on p50 of BS7671 gives max Zs values for ratings of RCDs used for fault protection on TT installations. And where a 500mA RCD is used max Zs is stated as 100 ohms.

The Scenario states that the earth electrode resistance is 100 ohms so therefore the Zs for the lighting circuit is going to be 100 ohms plus the R1 + R2. So it would be a a 300mA RCD not 500ma as most replies on this thread have stated.

But then again you might get it marked correct as I'm doubtful that C & G have thought it through properly.
 
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RA doesnt include R1 i dont think mate its just R2 thats included
and if C&G just gave 100ohms as the Ra then i agree with 500mA protection
its just easier to measure R1 + R2 then add this to Ze but really the R2 should be measured with method 2 R2 with use of long wander lead and then added to the Ze
but Ra as far as i am aware doesnt include R1 just got me looking at the regs now lol
 
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Read note 1 under the table. it will tell you how to arrive at the Zs. Ra x IdN same or below 50v.

So 50/100 = 0.5 or 500mA.

Ra is the sum of the earth electrode and the earth conductor (Reg 411.5.3). as the earth conductor will be negligible possiblly 0.05ohms then your only concern in the question is the 100ohm of the electrode
 
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yes mate it does say where ra not known it may be replaced by Zs note 2
and used where the device is a overcurrent protective device the following shall be satisfied
Zs X Ia less than or equal to Uo
for TT system where RCd is used for earth fault protection
Ra X I delta n is less than or equal to 50v

be good to see this paper after all comments about did no one find a copy in there bag lol
 
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Ra is the sum of resistances of the earth electrode and the protective conductor connecting it to the exposed conductive part

Yes this is true. But believe me, I have checked the wording and it does NOT say RA is 100 ohms. What it does say is "TT system with earth electrode resistance of 100 ohms."

Q26 then asks you to determine the the max permissible operating current for the BSEN 61008-1 RCD protecting the lighting circuit. This has to include the R1 + R2 of the lighting circuit??? Think about it... if you get an earth fault at the extremity of the circuit then the fault current is determined by the value in ohms at that point.

Hope I'm not confusing things just shows the confussion, deliberately or not, caused by the wording in C & G exams.
 
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