C&G 2391-101 July 2007 Exams

I am going to put the questions in my own way but the answers will be the same as those for the exam.

1. What are the three main area to be verified on a new installation?

These would be equipment and material complies with relevant British Standard - all parts are correctly selected and erected - no parts are visibily damaged or defective

2. Ignoring compliance with BS 7671, earthing arrangement and load capacity what else must the designer must ensure when the work is not a new installation?

3. Give three locations that could have to comply to other statutory regulations other than the EAWR.

Approved Document Part P is a statutory document and thus dwellings
Locations that have explosive atmospheres other than Mines and Quarries
Cinemas

4. Give an example of each of the following
Limitation of discharge of energy Electric fence used on farms
Placing out of reach Pickups on electric cranes or overhead wires
Undervoltage Protective Devices Motor controllers

5. (i) What statutory document places a duty on person carrying out inspection and testing Electricity At Work Regulations 1989.
(ii) Which two groups of people does the the inspector have a duty to ensure safety of? Although worded confusingly I think himself and others.

6. What instrument should be used to carry out a continuity test of a circuit protective conductor? Low resistance ohmmeter typically with a digital instrument having a 0.01 ohm resolution
What would be unit of measurement of the above instrument? Ohms
Why would the above instrument and scale be selected? The instrument would be need to be able to measure the low resistance encountered with this kind of test

7. On a TT system which would be the first three live tests to be carried out?
Earth Electrode Resistance
Earth fault loop impedance Ze
Prospective fault current

8. Give correct names for three protective conductors that maybe connected to the MET
Main earthing conductor
Main equipotential bonding conductors
Circuit Protective Conductors

9. A cable have a resistance of 0.183 ohms is 25m in length. What be the resistance if
you had 50m of the same cable 0.366
25m in length but twice the csa 0.0915
25m in length but with half the csa 0.366

10. What are the three reasons for carrying out the ring final circuit test

Continuity of the phase, neutral, protective conductor around the ring. That no interconnections exist. Correct wiring of the ring.

11. Assuming a circuit is safely isolated give three checks to be made prior to carrying out an insulation test.

12. Excluding saunas, bathrooms, and swimming pools, name three other special locations
Construction sites.
Caravan Parks
Agricultral and horticultral sites

13. International Protection rating consists of two numbers, what do the numbers represent
The first letter
The second letter
If an X replaces a number, what would that mean
The first letter signifys protection against ingress of solids eg IP4X 1mm
The second letter signifys protection against ingree of liquids.
Where the X replaces a number it means that the protection as not been checked or that it is irrelevant.

14. When polarity testing as part of initial inspection and testing is being carried out, what are the two stages that shold be carried out. Secondly which instrument would be used?

15. External to an electrical installation what are the component parts of the earth fault impedance for a a TT system.
Even though the question does not specifically request a drawing it does help with the explanation. At the consumer Phase conductor -Main earthing conductor -Earth electrode to General Mass of Earth. At the Distributor, General Mass of earth - Earth Electrode - Main earthing conductor - Neutral on Transformer Secondary.


16. Before an installation is powered up, Zs needs to be obtained. What test value needs to be used. What else is required to be known. Finally write out the formula
Test value is (R1+R), this is CPC + phase conductor resistance)
External earth loop impedance Ze would need to be known
Zs = Ze + (R1+R2)

17. What two methods could be used to ensure the measured value of Zs complies with BS 7671.
What is the maximum accepted value for tabulated values in BS 7671 of 0.35 and 14.1
Where the ambient temperature is known then a compensation factor can be used to determine the maximum value of Zs.
The Rule of thumb is used and is the worst case scenario and reduces values in BS7671 by 75%.
Using rule of thumb
0.35 would be 0.35 * 0.75 = 0.2625 ohms
14.1 would be 14.1 *0.75 = 10.575 ohms

18. A 30mA is required to have three instrument test applied. What are they and what currents would need to be applied.
At 50% of 30mA, 15mA the RCD should not trip
At 100% of 30mA, or 30mA the RCD whould trip in under 200mS
At 500% of 30mA or 150mA the RCD should trip in under 40mS


19. What are the two types of fault current measured at source, which needs to be recorded.
PEFC and PSCC
The largest of the two needs to be recorded

20. Why is prospective fault current important in selection of protective devices?
PFC is important since the protective device would need to be able to break the fault current under short circuit conditions

For questions 21 to 26 I am going to assume you have already have access to the exam paper.

21a Electrical Installation Certificate which would include a schedule of inspection and schedule of test results.
21b The designer
21c Usage and external influences
21d BS 7671, Guidance Note 3, GS38 are all non statutory, Electricity at work regulations is a statutory document
21e Confirm that the installation adheres to the requirements of BS 7671 and that there is no visible damage to the installation
21f Earthing Arrangement - Number and type of live conductors - Nature of Supply parameters - Supply protective device characteristics

22a Although the question is perhaps a little vague I beleive the answer is that since the buildings are attached, the new services would need to be taken to the main earthing terminal

22b

• Circuit verified dead using method outlined in GS38
• CPC disconnected at main DB and shorted to the phase conductor
• At the new DB a low resistance ohmmeter is connected to the phase and disconnected CPC.
• The value obtained is (R1+R2)

22c We have 30 metres of 25mm csa cable, given that the milliohm/ metre is 0.727 then (R1+R2) will be 2*25*0.727/1000 which is 0.03635 ohms. Zs is Ze+(R1+R2) from information Ze is 0.1 ohm therefore
Zs= 0.1 + 0.3635 ohms
Zs = 0.4635 ohms

22d The SWA would be glanded at both ends and this would mean that the resistance of the armourings would be in parallel with the CPC resulting a lower value of Zs.

23a You would need the three drawings given in GN3 or similar to explain the three stages of the test since this is a ring final circuit.

The phase and neutral conductor loop values are 40*7.41/1000 using the equation used above, thus giving 0.2964 ohms

The CPC is 40*12.1/1000 or 0.484 ohms

So taking the three loop tests we would PR1 to PR2 (PR) and NR1 to NR2 (NR) as 0.2964 and CPCR1 to CPCR2 (CPCR) being 0.484 ohms

Stage two requires that PR1 is connected to NR2 and PR2 is connected to NR1.

At each socket the value expected is (PR + NR)/4

(0.2964+0.2964)/4 = 0.1482 ohms

Stage three consists of PR1 being connected to CPC2 and PR2 being connected to CPC1

At each socket the value expected this time is (PR + CPCR)/4

(0.2964+0.484)/4 = 0.1951 ohms.

24a Precautions could be removing loads from the circuits such as lamps, being aware that the test voltage could cause damage to some electronic equipment and taking action like linking the phase and neutral conductors together.

24b Factors are the conditions of the individual circuits being tested, since the circuits are in parallel the value of the test will be less than that of the lowest test that would be obtained on individual circuits.

24c The test voltage needs to be 500 Volts. A test result lower than 2 Megohms would require further investigation although the minimum accepted result is 0.5 Megohms.

24d 1/Rt = 1/R1 +1/R2 etc

Since there are 14 circuits with values of minimum 200 Megohms. 1/Rt = 14/200 therfore

Rt = 200/14 Mohms.

= 14.2857 Mohms.

25a and b Circuit 10 is 80 metres using 1.5 mm T&E so CPC will have a csa of 1 mm. The maximum Zs is 4.8 ohms.

The Phase conductor (R1) resistance will be 80*12.1/1000 = 0.968 ohms using values in figure 2.
The Neutral conductor (R2) resistance will be 80*18.1/1000 = 1.448 ohms

Zs = Zdb + (R1+R2) where Zdb is earth loop impedance of new distribution board which from question 22 is 0.4635 ohms .

Zs = 0.4635 + (0.968+1.448)

Zs = 2.8795 ohms

Using the 3/4 rule of thumb maximun Zs value becomes 4.8 *3 /4 = 3.6 ohms

The value of Zs would be acceptable.

(Hard to believe that only 6 marks have been awarded for all that work and why did it require reference to a previous result)

25c A drawing here will assist to explain the procedure

At the furthest point of the circuit with all switches closed connect the earth loop impedance tester to the CPC and then the connect the other lead to the phase conductor, remembering that this is a live test and the risk needs to assessed. The result obtained will be Zs for the circuit. If the circuit was a two or more way arrangement, each way would need to be tested and the highest value would be recorded as the Zs.

26a. The basic drawing can be found in the On-Site Guide. TN-C-S would mean that the neutral and earth would be seperated in the distributors metering at the consumers end. The circuit would also include the main protection device, the protection device for the new DB and also the 20A MCB feeding the air curtain (Just what is an air curtain, not exactly common and therefore likely to create a problem to those taking the exam, IMHO).

26b. PEFC is given by Uo/Ze for the source thus PEFC = 230/.1 = 2.3kA

26c. Since Zdb is higher than Ze it can be assumed that the PEFC will be less than at the source.

26d. For this question we could say refer to question 20 since for me the answer is the same. This time the mark is not 1 but 2 however.

The PFC is important in selection of protective devices since this is the current that the device will need to break in case of a fault of negligible resistance.

These are the answers I have since worked out or found since taking the exam and although many seem to be similar I cannot be certain

I you disagree with the answers please add a comment

What is PART P?

The Building Act 1984 in section(1)1 made provisions that regulations maybe made for various purposes. This meant that rather than going to parliment each time additions were needed, the Secretary of State can approve documents. Approved Document P: Electrical safety - 2006 edition is one of these documents, and is commonly referred to as Part P.

Prior to the introduction of Part P, anyone could install electrical equipment into a dwelling. The IEE did have regulations in place in the form of BS 7671, but these were not statutory and were often ignored by some installers (including DIYers). Part P does not limit who can install electrical equipment but does require that certain work will require prior notification to a building control body. There is an exception and that is where the installer is registered with an electrical self-certification scheme.

IT earthing arrangement, where used.

Coal mining is one of the places that use an IT earthing arrangement.
Transformers used underground are of a Delta/Star arrangement, namely DY11. The transformer secondary star point is not connected directly to earth but through a current limiting impedance device.
The secondary side goes through a breaker that from memory contains a protection device that would trip if the resistance to earth drops below 20000ohms. Since the transformed supply could be used to supply a number of different parts of the mine called sections. Each section needs to be protected to afford discrimination and this means that insted of 20 k ohms the value is raised to 40 k ohms. Where final circuits are involved then a value of 60K would create a trip of the protective devices. Notice I said all the protective devices.

From memory three resistors are used to create a false neutral which in turn is connected to earth via another resistor. Should an earth fault on any of the circuits fed from the transformer occur then every one of the protective devices would have a voltage drop across the resistor connecting earth to the false star points. This voltage drop is sensed by electronics and causes the protective device to operate.

Prior to the arrangement above a core balance transformer was used. Each phase would go through the core or ring and thus each in turn would set up a magnetic field in the core. Since the current flowing in the three phases would under conditions be balanced, each magnetic field would cancel out the others thus giving no overall net magnetic field.
Under a fault condition there would be an overall magnetic field due to the imbalance and a this would cause a current to flow in a coil wrapped around the core. This current is sensed and would cause the protective device to operate.

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.

C&G 2391 Inspection and Testing Lighting?

Many questions seem to include either a PIR or else discharge lighting. From this I have come to the conclusion that the person setting the question is wanting an insulation test to indicate that the phase and neutral need connecting together to prevent damage.
Discharge lighting generally have a choke to limit the current flow, this creates a reactive component which is reduced by fitting of a capacitor. If you check the rating of capacitors fitted they are normally rated at 250Volts. Testing with a 500Volt insulation resistance meter could damage the capacitor.
PIRs in a circuit contain electronics so should not be tested phase to neutral.

So how would you do an insulation test on a lighting circuit?

• Disconnect circuit from supply and verify circuit is dead using approved method outlined in GS38
• Check that the testing will not create danger to persons, livestock or property
• Link together the phase and neutral
• Check your instrument or it's leads are not damaged.
• Select 500 Volt.
• Securely clip one lead to the earth bar
• Prove connection by testing to another earth terminal, this should show 0.0 megohms or short circuit
• Clip to the phase/neutral connection and press test, wait for a steady reading and record.
• Allow the instrument to remain connected to discharge any stored charge created by the test.
• If test is greater than 2 Meg ohms, disconnect leads, reconnect phase and neutrals back into respective places, restore power.
• Tests of under 2 Megohms would need further investigation but is not a fail as this is below 0.5 Mohms.

C&G 2391 Inspection and Testing EIC, PIR, MEIWC

How I have been revising to pass the exam.

EIC will need to be written as Electrical Installation Certificate
PIR will need to be written as Periodic Inspection Report
MEIWC will need to be written as Minor Electrical Installation Works Certificate

Note that both these need attached to the first two but not the MEIWC.

1. A Schedule of inspection
2. A Schedule of test results

You will lose marks if you simply say Schedule of Tests.

To answer which is required, remember the clue is in the question.

A new circuit as been installed? Installed equates to installation, thus could be one of two, but since circuit added, is not a minor undertaking. This leaves us with Electrical Installation Certificate.

A report on the condition is required? There is only one report and this is the Periodic Inspection Report.

Initial Verification, or making sure in the first place, relates to the Electrical Installation Certificate.

Statutory documents are those passed by an Act of Parliment and could be The Health and Safety at Work Act 1974 or more likely the Electricity At Work Regulations 1989.

None statutory documents are likely to be BS 7671, Guidance Note 3, On site guide.

3/4 Rule of thumb Inspection and Testing. 2391 Exam

So what is this rule of thumb?

When carrying out testing of an electrical installation the value of the phase and circuit protective conductors resistance is measured together to give (R1+R2).
Ze which could have been either measured, calculated or obtained by enquiry is then added to (R1+R2) to obtain Zs, but you knew that.
For a particular protective device BS 7671 gives tables as to the maximum value of Zs that can achieve disconnection within a specified time.
The problem is that when the value of (R1+R2) is measured the temperature of the conductors are likely to be at ambient temperature. Trouble is that the tables in BS 7671 are not relevant to Zs values measured at ambient temperature.
The resistance of copper is proportional to the temperature, so if the temperature drops the resistance drops.
Rather than having to measure the ambient temperature the 'rule of thumb' method is allowed.

To compensate for the lower temperature the maximum value of the Zs allowed needs to be lowered. 3/4 of the value of Zs is the worst case and thus acceptable to use.


A Zs in the table of say 4 ohms. would result in a corrected value of 4 times 3/4 which is now 3 ohms.

From a C&G 2391 viewpoint you will be given a Zs maximum value for a circuit and a set of Zs values that were calculated or else given (R1+R2) to which you need to add Ze to.

Show the formula you are using so that even if your maths are wrong you will still gets marks awarded.
Corrected Value For Zs = ZsMax X 3/4.

Provided that the calculated value is less than the corrected value then the results are acceptable.

C&G 2391 Inspection and Testing Exam Date Getting Closer

The C&G 2391 Inspection and Testing Exam is this coming Thursday. I have been going through the past papers and timing myself and believe I must be missing something with my answers.
Part A which should take a hour to complete is taking me 20 to 25 minutes tops!!
Part B which should take a hour and half is only taking 30 to 45 minutes!!!

Some of the model answers to the past papers are almost the same answers to what I get. The keywords are there but perhaps in a different part of the sentence to those of the model answers. what could I be missing?

I can assume that the C&G are allowing the remaining time for thinking since I would not call myself a super fast writer, especially with a pen and paper.

I have been looking through the adverts for testing instruments to carry out inspection and testing of electrical installations. I wondered about whether is would be wise to get seperate instruments but soon dispelled this notion.

• A multifunction instrument is initially cheaper to purchase.
  
• The calibration cost is not that much dearer for a multifunction unit than the cost for a single funtion instrument.
  
• If I lose a function or else a single instrument packs in, I cannot carry out the inspection either way so no advantage either way.
  
• Some manufacturers claim that it is easier to use seperate instruments since there are not so many dials. I think that if someone is competent to carry out inspection and testing then a few dials are not going to be too taxing.
  
• Having worked on some large industrial complexes the last thing I want is 6 seperate boxes to have to carry around, plus my tools.
  
• One set of batteries to replace or carry spares for rather than 3 just in case 50% of the single testers decide to get flat batteries half way through a job.
  

My conclusion would be a multifunction unit for the reasons given above. Now do I need the extras?

Phase rotation is a must for me having an industrial background, since it can cause damage when large numbers of motors all go in reverse!!!!

I have never had to carry out an Earth Electrode Test but can forsee the day coming when I need to check on an 11kV/400 Volt transformer earth connection.

You ever worked on a construction site or one like i,t with water sludge dirt oil and grease all over the place? If so, then like me you can see the benefit of being able to record the results to download later. Ever dropped your notebook in dirty water and know all your results are lost?

I want an IP67 rated multifunction tester that can be dropped from 3 metres, if only one was made. So will have to settle for a shower resistant one.

C&G 2391 Inspection and Testing Day Questions for Roger 3

A few more questions for Roger in respect of C&G 2391 Inspection and Testing

1. GS38 lists three steps when using an approved test lamp to check that a circuit is dead, list the three steps.
2. What is the recommended period between calibrations of test instruments.
3. Who determines the extent and limitations on a PIR?
4. Name one dead test that could cause injury through electric shock and one live test that could cause injury through electric shock to persons other than the inspector, explain the reasons why?
5. A system having a TN-S earthing arrangement contains a fixed electric heater that develops an earth fault, sketch the fault path for this arrangement.
6. What test is required for site applied insulation?
7. The following are recorded on the schedule of tests, what instruments would have been used to obtain the results? (40mS, 6.23kA, >100M ohms, 0.03 ohms)
8. Why does the main earthing conductor need to be removed when carrying out a Ze (Zdb) calculation?
9. What precaution could be employed when carrying out an insulation test between live and earth to prevent damage to sensitive components?
10. A circuit has a Zs of 500 ohms, show your calculation as to the the RCD to be choosen.
11. What is the main difference between a type B and type D MCB?
12. Why when carrying out a polarity check on the incoming supply of a new installation would this not reveal an earth neutral fault?
13. What disconnection time would be required for a 110 Volt system?
14. Under what circumstances would a PIR not be required for an industrial installation.
15. What is the requirement for obstacles used to prevent direct contact.
16. Give two examples where height is used to protect against direct contact.
17. What is the difference between conductive metal parts and extraneous metal parts?
18. Give two examples of enviromental influences that may require an inspector to shorten the period of inspection?
19. Give five reasons why a PIR may be required on domestic premises.
20. Where there is likely to be parallel paths present during a PIR, apart from tests what else must be done to check the cpc?
21. What does the abbreviations PEN, PME, MET mean with respect to BS 7671.
22. Give three systems that are not covered by BS 7671 but need to be considered during a periodic inspection?

C&G 2391 Inspection and Testing Day Questions for Roger 2

A few more questions for Roger in respect of C&G 2391 Inspection and Testing

1. What does the letters IP stand for. What is the minimum IP rating for the top of an electrical enclosure? Finally what size particle does the IP rating relate to.
2. What document assigns an inspector the legal status of Duty Holder?
3. Where a single person is not responsible for a new electrical installation what title roles would need to sign the certificate?
4. What notice would not need to be present on a 230V DB but would be required on a 400Volt DB?
5. An insurance company requires a report on an electrical installation, which report would need to be produced and what additional pages would be required?
6. What is meant by the terms SELV and PELV.
7. What two methods can be employed for measuring the earth electrode on a TT system.
8. What is meant by the term TN-C-S?
9. Explain what is meant by direct contact and indirect contact. Name three methods of protection against direct contact.
10. When testing equipotential bonding what is the minimum resistance allowed? What is meant by equipotential?
11. Name three precautions to be taken when carrying out insulation tests.
12. What voltage setting needs to be selected when testing SELV and what voltage when testing a 230 Volt circuit.
13. Between what voltages must a low resistance ohm meter produce?
14. When using 2 30mA RCDs in series how can discrimination be achieved?
15. The mechanical operation of an RCD should be tested at what interval?
16. A domestic installation contains no CPCs on the lighting circuit, how old could you assume the electrical installation to be?
17. Explain what is meant by the 3/4 rule of thumb in relation to Zs.
18. Where an instrument is not capable of testing above 230volts, what calculation can be used to work out the phase to phase prospective fault current in a three phase circuit.
19. Explain what is meant by the 'touch voltage' and what is the maximum allowed?
20. Why is important to obtain the prospective fault current?
21. When carrying out a stage three test on ring final circuit the resistance rises around the ring and then begins to drop towards the end what is the likely cause.
22. Stage three of a ring final circuit obtains (R1+R2) but what else does it check.
23. On a ring final circuit the first and last socket gives 0.4 ohms but further round the ring the value drops to 0.2 ohms, what could be the cause?
24. In a standard twin and earth PVC cable, which conductor(s) will have the lowest resistance and which will have the highest?
25. Given an insulation resistance of 1.8 Meg ohms obtained from an overall test carried out on a 230 volt six way consumer unit. If 5 of the circuits have insulation resistances of 50, 50, 30, 50, 40 what is the insulation resistance of the sixth circuit and would this meet the minimum required for the installation?

More to come later?

C&G 2391 Inspection and Testing Day Questions for Roger

Roger and I took the course together and I said I would ask him some questions that I will make up. So you might wish to check your knowledge ready for the exam.

1. There are three reports that an inspector can issue, what are these? Secondly give an example where each would be used.
2. What is the period between inspections for electrical installations in commercial premises? What reasons could an inspector reduce the period of time between inspections for a domestic premise?
3. What is the none statutory publication relating to test instruments? What size fuse and type is recommended to be fitted in an instrument's test leads.
4. Which earthing arrangement would have a maximum of 21 ohms if obtained by enquiry or the maximum allowed under BS 7671?
5. Give the order in which the first 5 dead tests would be carried out?
6. List the six tests that could be carried out on an 30mA RCD and what results would be expected for each test?
7. Explain what is meant by the terms Zdb, Ze, Zs, (R1+R2)? Show the formula for calculation of the earth fault impedance.
8. EEBADS is an abbreviation for what? What disconnection time is required for sockets when protected by a BS 88 fuse and also the time for other applications.
9. Give four examples of services in a building that would require bonding? What is the wording on the label required where an earth connection connects to services.
10. What is the difference between SELV and PELV? What voltage would an insulation resistance tester be set to test either SELV or PELV? What is the minimum resistance allowed?
11. When carrying out an insulation resistance test on a 6 Way consumer unit the following tests where recorded 5, 3, 20, 10, 20, 50 meg ohms. What is the overall resistance for the system and is this acceptable. Give three possible reasons for the lower tests.
12. What precautions would need to be taken prior to carrying out an insulation test.
13. The csa of the conductors in a cable are doubled, what effect would this have on the resistance. If copper is heated will the resistance increase or decrease?
14. A prospective phase/neutral fault current of 2.3kA is obtained along with a phase/earth prospective fault current of 3.4kA, which value would need to be recorded and why?
15. Ring final circuits can be tested using three stages. Briefly explain the stages? When testing at the sockets the resistance begins to increase after the third socket, what could be a possible reason?
16. Give three examples of where the polarity needs to be checked.

Not a full 20 but still enough to have a go at testing your knowledge.

C&G 2391-1 Exam and forms

My exam is set for 18:30 on the 19 July 2007, so keep your fingers crossed for me.

Looking at passed papers and trying to predict what questions are likely to be asked may work but in my experience of externally set exams it is better to concentrate on testing your knowledge of the subject. Sat reading through your notes on it's own is not likely to help much.

For me it is best to choose a particular subject matter and write all you can remember on the subject then correct any errors and then add omissions from your notes. In this way you are amplifying your knowledge and re-enforcing your memory links such that you will have instant recall not only for the exam but later on in life.

You will have to believe me that I have none of the books or notes to hand as I write this. My subject matter is on the forms that are required by the BS7671 related to electrical testing and inspection.

A new installation requires the completion of an Electrical Installation Certificate that can be one of two formats. The first format contains spaces for two designers to sign, the installer and finally the actual inspector. The second is a shorter format and is where the designer, installer and inspector are the same person. Both forms require inclusion of two other documents, these being a schedule of inspection and a schedule of tests. It is also required that all documents are numbered generally in the format 1 of 5, thus giving the total number of pages, in this case 5 and also the actual page number, again in the case page 1. It is this certificate that gives initial verification that the new circuit or system complies with BS 7671.

At this point it is worth pointing out that BS 7671 is not a statuotory document but in almost all cases will compliance with BS7671 ensure compliance with the stauotory document, the Electricity At Work Act 1989. Under the EAWR the Inspector is called the Duty Holder and this title must not be taken likely since the maximum prison sentence could be 10 years. Testing instruments come under the EAWR and so their use or miss use could land the inspector in a court of law. To aid those involved in electrical testing the Health and Safety Exceutive has published a document called GS38 which again is not statutory but compliance with it will almost in all cases ensure compliance with the EAWR 1989.

The Schedule of Inspection is a list of items that need to be considered when inspecting an installation. All parts of the document should be filled in with a tick for compliance a cross for none compliance, N/A where the particular item of inspection is not relevant, LIM where that part is noted in the Limits and Exceptions.

The Schedule of tests contain results of tests carried out on the installation.

Test 1 Earthing conductor
Test 2 Equipotential Bonding
Test 3 Circuit Protective conductors (R1+R2)
Test 4 Ring Final Circuits
Test 5 Insulation Testing
Test 6 Earth Fault impedance (Would require verification of Ze)
Test 7 Prospective fault current

Time has won, more later, cya soon

C&G 2391 Inspection and Testing Day 1

Day 1 of the course began with the normal fire exit procedures, and personal introductions. The tutor is a 50 y/o guy called Dave. Waiting for me was two books Inspection and Testing Guidance Note 3 known as GS3 for short and the ON-SITE GUIDE BS 7671: 2001 (2004).
The day was filled with brief snippets that would we were reassured several times come together over the remaining two days.
Will had more tonight as to content of Day one and Day two of the course.

I will say so far that there is nothing that seems to challenge either me or the other delegates.

C&G 2391, cable csa easy marks

On seeing some of the past papers it appears that some easy marks can be gained by remembering basic electrical theory in relation to cables.

The conductors within a cable have a given resistance for a given cross sectional area (csa) depending upon the metal used. Copper is not the only metal since aluminium is also used.

The questions tend to ask if the length if increased or decreased and also if the csa is increased or decreased what value of resistance would be obtained.

Altering the length of the cable is equivalent to resistors in series whereas altering the csa is equivalent to resistors in parallel.

Let us assume I have a cable of a length and csa that gives me 1.2 ohm. (long cable). If I double it's length then I have 2*1.2 ohm or 2.4 ohms. If I half it's length then I have 1.2/2 or 0.6 ohms.

The same cable above is now doubled in csa so this will lower the resistance since we have effectively two resistors in parallel thus for same length the resistance will be 1.2/2 or 0.6 ohms.

So if we double it's length and also double the csa we get (2*1.2)/2 or 1.2 ohm. Notice that the resistance is the same, namely 1.2 ohm. If we half it's length and half it's csa then we have (1.2/2)*2 or once again 1.2 ohm.

If you imagine that the cable is an actual resistor it may be easier to remember when doing the exam.

Why is Zs important in electrical installations?

If we take some fuse-wire and pass through it different currents we could create a time current characteristic graph. The graph would show that the relationship between current flow and time for the fuse-wire to burn out is not linear. As the current increase the time for the fuse-wire to burn out decreases much faster giving a curved line.

Since current flow is proportional to resistance for a given voltage then a fault having a zero resistance would create the greatest current hence the fastest trip time. There is however additional resistance to consider and this is the resistance or impedance of the supply (Ze) the phase conductor and circuit protective conductor CPC or (R1+R2), these combined make up Zs.

The higher Zs is, then the longer it takes for the fuse to blow since current flow will be a less. It is possible that the value of Zs is so high that even when a short is placed at the furthest point there is insufficient current flow to cause the fuse to blow.

BS 7671 lays down the times that over-current devices must operate and this means that for a particular device there needs to be a maximum value of Zs that will allow the device to operate within this time.

C&G 2391, 3/4 rule in relation (R1+R2)

Most metals have a positive temperature coefficient of resistance. This means that the higher the temperature the higher the resistance. You did know that I hope?
Measure the resistance of a 100 Watt lamp when cold and then calculate the current flow when connected to a 230 Volt supply. Bet you did not get under half an amp?

So what does that have to do with the price of eggs or indeed calculation of R1+R2 with respect to Zs?

When using a low value ohmmeter the wiring, the supply will be dead. This means that there will be no heating of the conductors, since no current is flowing, so the resistance obtained is actual less than that which would be measured.

There is a calculation that can be carried out providing that temperature measurements can be taken but it is accepted practice to apply the 3/4 approximation.

Hang on though why is Zs all important when it comes to electrical testing anyway? More in the next posting.

C&G 2391, insulation resistance

It would appear that some taking the 2391 exam are forgetting basic principles when it comes to insulation resistance.

Resistance in series, the formula is Rt = R1+R2+R3+R.... (Resistance value will be higher)

Resistance in parallel the formula is Rt = 1/(1/R1+1/R2+1/R3+1/R....) (Resistance value will be lower).

Where multiple circuits are to be tested from the same distribution board or consumer unit then the overall insulation resistance can be calculated using the parallel formula.

So for example, tests of 5, 20, 10, 20, 2 are obtained from individual circuits insulation tests on a consumer unit. What is the overall insulation resistance for the installation?

So we have Rt = 1/(1/5 + 1/20 + 1/10 + 1/20 +1/2) (Mohms)

thus Rt = 1/(0.2+0.o5+0.1+0.05+0.5) (Mohms)

Rt= 1/0.9 or 1.11111 (Mohms)

Notice that I have shown all the working out here. This is so that if I had made a simple error in the calculation I may still get marks for using the correct formula!!!!!