Re-take - Useful Information for 2394 :

[amberleaf]

Yes this Question has been asked on 2394:
PS . there are still in the Regulations BS-3036 fuse

S1A devices are 1KA. in sizes 5 ,15 , 20 , 30 , 45 , & 60A
S2A devices are 2kA. in sizes 5 ,15 , 20 , 30 , 45 , 60 & 100A
S4A devices are 4kA. in sizes 30, 45, 60 & 100A

Q ) Which of the following protective device is never suitable for determining prospective earth fault current at the origin of an installation
□ BS-3036 fuse
□ BS-88-3 fuse
□ BS-88-2 fuse
□ BS-EN-60898-1 circuit breaker .. 6kA / 10kA .

The maximum breaking capacity of a BS-3036 fuse is 4kA , so it is not a suitable protective device where the fault current is 10kA .

• Note : that the devices are available in three types relating to short-circuit capacity, i.e. 1, 2 & 4kA.

Out but not gone . ?? ... BS-3036 fuse

O.S.G. P/72 .
Regulation 533.1.1.3 of BS-7671:2008 refer
states that a fuse shall preferably be of the cartridge type but this does not preclude the use and installation of semi-enclosed fuses. Where a semi- enclosed fuse is selected, the carrier shall be fitted with an element in accordance with the manufacturer’s instructions,

533.1.1.3 permits the fitting of a single element of tinned copper wire of the appropriate diameter specified in Table 53.1 P/140

Terminology
Those fuses referred to as " rewirable fuses " are correctly termed semi enclosed fuses as they are partially enclosed within the fuse-carrier.

semi-enclosed fuse ... BS-3036 fuse
a fuse in which the fuse element is neither in free air ( other than the air in any external containing case not forming part of the fuse ) nor totally enclosed

Rated breaking capacity
BS-3036 defines breaking capacity as: breaking-capacity rating a prospective current stated by the manufacturer to be the greatest

BS 3036 are suitable in certain applications only.
BS 3036 semi-enclosed fuses cannot be relied upon to operate within 4 hours at 1.45 times the nominal current of the fuse element.
Correct protection can be obtained by modifying the normal condition ( In ≤ Iz ) such that the fuse rating does not exceed 1.45 ÷ 2 = 0.725 times the rating of the circuit conductor

For this reason, larger cables may need to be selected where overload protection is provided by semi-enclosed fuses than when it is provided by a cartridge fuse or fuses or circuit-breaker ( Appendix 4 of BS 7671 and Guidance Note 6 give further guidance ). 2008

Fault Current Protection . from my old 16th Edition notes .

The purpose of fault current protection is to disconnect the supply speedily and restrict damage and danger as far as possible .
All parts of an installation must be protected against the highest ( pƒc ) that can be anticipated at any particular point in the system The ( pƒc ) is at is highest at the intake position and will decrease ( or attenuate ) through the installation as the résistance of cables is added to the fault part .

Short Circuit Rating : Under short-circuit conditions there will be a considerable current surge . This is the ( pƒc ) .. for a standard 100A supply the ( PFC ) often quoted by DNO is 16 000A ( kA ) The value is never as high as that .

much depends upon the arrangement of the supply network and the distance from the substation

Rated short-circuit capacity ( Icn ) of the circuit breaker conforming to BS-EN-60898-1 is the value of the ultimate short circuit breaking capacity assigned by the manufacture .

circuit breaker also has a corresponding " in-service " short circuit breaking capacity ( Ics ) which is either equal to , or lower than , the rated short circuit capacity ( Icn )

Conveniently , ( Icn & Ics ) are the same value for circuit breakers conforming to BS-EN-60898 having an ultimate short-circuit capacity of ( 6kA ) or Less .

 

[amberleaf]

Useful junk .

Overcurrent protective device . BS-EN-60898-1
it is only the rated short circuit capacity ( Icn ) that is marked on the circuit breaker 6000A (kA)

Younger members . ( Maximum )
Regulations : specify various maximum disconnection times for different types of circuit(s) for TN - TT - systems

41.3. ( Maximum ) disconnection times
Final circuits not exceeding 32A .. TN - 0.4s .. TT - 0.2s
Final circuits exceeding 32A & for distribution circuits .. TN - 5s .. TT - 1s

from table 41.1. the above values are applicable for a nominal voltage 120V < Uo ≤ 230V

Circuit breaker BS-EN-60898-1 has the characteristics such that it will disconnect the supply in less than 0.1s providing that the maximum earth fault loop impedance permitted by BS-7671: for the circuit being protected is not exceeded .

( PFC ) of an earth fault depends on the maximum earth fault loop impedance at the point of a fault .
if that impedance is lower than the maximum permitted value in Table 41.3. Chapter 41 ... ( Protection against Electric Shock ) disconnection of the supply will occur in less than 0.1s , circuit breaker conforms to BS-EN-60898-1

 

[amberleaf]

2394:
Putting your answers on paper . Due to the time constraints of the exam , do not waste time by copying out the question .
Facts . The markings of your answers does not include any penalties or additional marks for spelling or grammar .

Useful junk .

Overcurrent protective device .
circuit breakers conforming BS-EN-60898-1 is deemed to comply . but there is a need to take account of the increase in temperature and résistance of circuit conductors as a result of overcurrets .

Regulation reminds us . Circuit loop impedance . 612.9. Earth fault loop impedance

Circuit loop impedance given in regulations cover this condition and should not be exceeded where the conductor's are at their normal operating temperature . Regulation reminds us ; if the conductors are at a different temperature when tested , which is usually around ( 20ºC ) the reading should be adjusted accordingly .

This requirement often presents a problem for the person carrying out tests . who is unlikely to be certain of the normal operating temperature . Where this is the case , it is safe to assume the maximum permissible operating temperature of the cable which , in the case of ( thermoplastic PVC ) insulated cables , is ( 70ºC )

2394 : derived for typical test condition's .. View for thought's

Most of your answers are in BS-7671:2011: & GN-3 .

 

[amberleaf]

.. 2014 .

100mA BS-EN-61008-1 RCD is installed in a TT installation to provide fault protection , Which of the following identifies the maximum test current to be applied and the maximum disconnection time at that test current when testing the RCD
□ 100mA & 200mA
□ 100mA & 300mA
□ 500mA & 200mA
□ 500mA & 300mA

The RCD is provided for fault protection and not additional protection and the ( 1 x I∆n ) is the maximum test current that needs to be applied , As this is a BS-EN-61008-1 device the maximum disconnection time is 300mS .

A maximum disconnection time of 200mS applies to some older RCDs manufactured to a British Standard , but not to BS-EN-61008-1 device(s)

Older RCDs .. BS - only .. 200mS
BS-EN-61008-1 .. 300mS

A 30mA BS-EN-61009-1 RCBO is installed on a socket-outlet circuit to provide additional protection , Which of the following identifies the maximum test current to be applied and the maximum disconnection time at that test current .
□ 30mA & 300mA
□ 30mA & 200mA
□ 150mA & 300mA
□ 150mA & 40mS

You're Q . 5 x 30mA = 150mA

The RCBO is provided for additional protection and the ( 5 x I∆n ) is the maximum fault current that needs to be applied . As this device is providing additional protection , the maximum disconnection time is ( 40mS ) as required by BS-7671: Regulation 414.1.1.

 

[amberleaf]

Younger members . in collage
please tell you classmates about this forum . I have did everything but sit your Exam 2394. Amberleaf

 

[amberleaf]

Which of the following is not a suitable method of confirming phase sequence , Read the Question carefully under Exam conditions

□ Measurement using a rotating disc type instrument
□ Measurement using an indicator type instrument
□ Measurement using an approved voltage indicator
□ Checking polarity and connections throughout the installation

the method of determining phase sequence are given in GN-3 , An approved voltage indicator cannot be used to determine phase sequence .

GN-3 . P/57 . PS what can be used . Phase sequence testing . take your Q from above ▲ ▲ ▲
• rotating disc type . Yeah
• indicator lamp type . Yeah

 

[amberleaf]

let's stretch it out a bit .

Q) The purpose of a functional test on an RCD carried out quarterly , using the integral test button , is to confirm

□ The correct functioning of the RCD in the event of a fault
□ That the operation of the RCD complies with BS-7671:
□ That the RCD operates in the event of a short circuit
□ The correct mechanical operation of the RCD mechanism

Operating the RCD test button will only confirm the mechanical operation of the RCD

GN-3 . P/59 . Integral test device .
612.13.1. An Integral test device is incorporated in each RCD , This device enables the functioning of the mechanical parts of the RCD to be verified by pressing the button marked ( T ) or Test

GN-3 . points out some down falls . to watch for . As an Inspector . !! ( Have you Verify )
• The continuity of the earthing conductor or the associated circuit protective conductors Ps RCDs Earth fault . etc
• Any earth electrode or other means of earthing
• Any other part of the associated installation earthing
• The sensitivity of the device .

Point to Remember , Yes we all use Multifunction testers nowadays .

GN-3 P/58 . Operation of residual current devices . 411.4.5.
for each of the tests , readings should be taken on both positive and negative half-cycles and the longer operating time recorded

Point to note : The RCD test button will only operate the RCD if it is energized

 

[amberleaf]

A test is carried out on the interlock for switching an alternative power supply , Which of the following is the type of test being undertaken
□ Functional
□ Continuity
□ Load
□ Installation

This test confirms the operation of the interlocking device and is therefore a , Your Q . Functional test .

 

[amberleaf]

Q) Following a test of earth fault loop impedanc4e ( Zs ) the following results are compared with the values given in BS-7671: Which of the following describes the purpose of the comparison

□ To confirm correct test method are used
□ To confirm conductor csa is suitable for the circuit
□ To determine whether disconnection times will be achieved under earth fault conditions
□ To determine whether the protective devices will operate under short circuit conditions

The operation of a protective device under earth fault conditions relies upon a sufficiently large current flowing , from Ohms Law the supply voltage and impedance of the system will determine the fault current . The ( Zs ) value is the variable and can therefore be used to determine whether disconnection will be achieved within the required time .

 

[amberleaf]

Q) Checking that the measured test results meet the required values will enable the inspector to confirm that the electrical installation is :
□ Correctly installed
□ Safe to be in service
□ Compliant with BS-5266
□ Never going to be dangerous

Providing the test result obtained during the inspection & testing meet the requirements of BS-7671: and the design requirements , the installation will be suitable for use .

 

[amberleaf]

( 1 ) Inspection & Testing 2394 & 2395

Section B - All questions carry equal marks , Answer all three questions . show all calculations

Questions 4 to 6 all refer to the enclosed scenario , see source document . Ensure you read this scenario before attempting these questions

4 a) The loop length for the office ring final circuit is 60m and all the socket-outlets are connected directly into the ring , Determine , showing all calculations

i) the expected ( R[SUP]1[/SUP] + R[SUP]2[/SUP] ) value ... 7 marks

ii) the expected measured ( Zs ) value ... 4 marks

b ) When measuring ( Zs ) for the toilet lighting circuit the circuit breaker operates .
State why the circuit breaker tripped during the earth fault loop impedance test ... 4 marks

5) The circuit for the tree phase saw bench is installed to the local isolator for the saw.
a) A test of earth fault loop impedance is to be carried out on the saw circuit .

i) State the test instrument to be used . ... 2 marks
ii) State the document that specifies the requirements for the test leads ... 2 marks
iii) Describe , in detail , how the test would be carried out . .. 6 marks

b) i) Determine , using the information in Figures 1 and 2 , the expected ( Zs ) value for the circuit . Show all calculations .. 3 marks

ii) Explain , in detail , why the measured value for this circuit is lower than the value calculated in b) i) above .. 2 marks

6 ) Describe , with the aid of a fully labelled diagram , the earth fault loop path for the outside lighting circuit . .. 15 marks

 

[amberleaf]

( 2 )

Section B -
Remember that the answers to the following questions must relate to the scenario contained in the Source Document .

4 ) a ) The loop length for the office ring final circuit is 60m and all the socket-outlets are connected directly into the ring , Determine ,
showing all calculations

i) the expected ( R[SUP]1[/SUP] + R[SUP]2[/SUP] ) test value .. 7 marks

Answer : r[SUP]1[/SUP] + r[SUP]2 [/SUP] = 60 x ( 7.41 + 12.10 ) 1000 = 1.17Ω .. ( sum in a sum ; 7.41 + 12.10 = 19.51 ; 60 x 19.51 ÷ 1000 = 1.17Ω )
R[SUP]1[/SUP] + R[SUP]2 [/SUP]= 1.17 ÷ 4 = 0.293Ω

There is more than one way to determine this value but any correct method would be given the marks . The use of a temperature correction factor in this calculation is not appropriate because the question asks for an expected " test " value and the Source Document states that testing is to be carried out at ( 20ºC ) which is the same temperature as that which applies to the mΩ/m values in Fig 2

The calculation could have also been laid out as show below ;

r[SUP]1 [/SUP]= 60 x 7.41 / 1000 = 0.445Ω
r[SUP]2 [/SUP] = 60 x 12.10 / 1000 = 0.726Ω
R[SUP]1[/SUP] + R[SUP]2 [/SUP]= 0.445 + 0.726 / 4 = 0.293Ω .... ( sum in a sum ; 0.445 + 0.726 = 1.171 ÷ 4 = 0.29275 . ( 0.293Ω )

4 a) ii) the expected measured ( Zs ) value .. 4 marks
Answer ; Zs = Ze + R[SUP]1[/SUP] + R[SUP]2 [/SUP]= 0.11 + 0.29 = 0.40Ω

C .
The answer to this question is dependent upon the answer given in, a) i) above . it is not normal practice to penalise a candidate twice for the same error. With this in mind , marks would be awarded for the correct formula , correct total , including units .

4 b) When measuring ( Zs ) for the toilet lighting circuit the circuit breaker operates , State why the circuit breaker tripped during the earth fault loop impedance test .. 4 marks

Answer :
The test current may be greater than the instantaneous tripping current of the circuit breaker causing it to operate .

C .
The examiner is looking for any statement that indicates an understanding of what has happened , Reference to an RCD tripping would not gain any marks as the circuit is not protected by an RCD , This information is shown in Figure 1 . in the Source Document .

5) The circuit for the three phase saw bench is installed to the local isolator for the saw .
a) A test of earth fault loop impedance is to be carried out on the saw circuit .
i) State the test instrument to be used ... 2 marks

5 a) ii) State the document that specifies the requirements for the test leads .. 2 marks

Answer : Leads to GS-38 .

5 a ) iii) Describe , in detail , how the test would be carried out .. 6 marks

Answer :
Supply on
Local isolator
• Access live terminals in the local isolator
• At the incoming terminals of the local isolator test
º L1 to isolator earthing terminal
º L2 to isolator earthing terminal
º L3 to isolator earthing terminal
• Close the local isolator cover
• Record highest result

 

[amberleaf]

( 3 )

C .
The structure of the question is intended to help the candidate identify all relevant information .

A list has been used to answer a) iii) because it is easy to write down , easy to check and time efficient . Each statement must include sufficient information to make it clear how the test would be carried out . it is not necessary to include the instrument title and lead requirement as part of this answer as they have already been identified in i) and ii) above

The test is carried out between each line conductor and the isolator earthing terminal . if a candidate described the isolator earthing terminal as " the earthing terminal " earth " or " the cpc " this would also gain the marks.

if the wrong test was described in the answer , such as R[SUP]1[/SUP] + R[SUP]2 [/SUP]test and the results is then added to ( Zs ) then no marks would be awarded . The question specifically asks for an earth fault loop impedance test and not any test method that could be used to determine ( Zs )

Answers that include dangerous procedures would score zero marks .

5 b ) i) Determine , using the information in Figure 1 & 2 , the expected ( Zs ) value for the circuit . Show all calculations ... 3 marks

Answer : Zs = Ze + R[SUP]1[/SUP] + R[SUP]2 [/SUP] = 0.11 + 2 x 7.41 x 15 / 1000 = 0.11 + 0.22 = 0.33Ω

C .
Marks are awarded for each stage of the calculation . Setting out the calculation and showing each step makes it easy for the examiner to follow the process . it also helps the examiner award marks when the answer is not fully correct . Answers that are not clear , do not contain all steps and contain errors are likely to score fewer marks than an answer containing errors but is clearly laid out . This is because in the latter case the examiner can easily identify the correct parts of the calculation.

5 b ) ii) Explain , in detail , why the measured value for this circuit is lower than the value calculated in b) i) above .. 2 marks

Answer : Résistance ( R[SUP]2[/SUP] ) of is reduced due to steel conduit being in parallel with the cpc .

C .
Although the answer is only worth 2 marks , the question does state " in detail " The examiner is looking for a complete answer , Answers such as " parallel paths " will not score 2 marks as this is not in detail .

6 ) Describe , with the aid of a fully labelled diagram , the earth fault loop path for the outside lighting circuit 15 marks .

Answer : Drawing . TN-C-S System Earth Fault .

The fault path is :
from the point of fault
Along the cpc to the MET
from the MET via the earthing conductor to the supplier's PEN terminal
Along the PEN conductor to the supply transformer winding
Through the transformer winding , along the line conductor to the fault

C .
This answer contains both a fully labelled diagram and a description of the path , Providing the diagram is very clear and the path is shown on the diagram , as is the case above , then full marks would be awarded , The description of the path gives additional information for the examiner to consider when awarding marks . it is in the interest of the candidates to give as much information as possible . if an incorrect system is described / drawn then no marks would be awarded . This is also true if the diagram is incomplete and the type of system cannot be determined .

 

[amberleaf]

...................................................................................................................................................................................................................................................

Source Document . ◄ ▼▼▼
Scenario ( Section B - Questions 4 to 6 )

An existing building has been converted into a builders store on the ground floor with the offices above The electrical installation and supply to the building forms part of a 400 / 230V TN-C-S system . ( Ze is 0.11Ω ) and the prospective fault current ( Ipƒ ) is 4.1 kA

All circuits within the store area are wired in ( 70ºC ) thermoplastic single-core cables , with copper conductors , in surface mounted , galvanised steel trunking and conduit .

All circuits in the office area are wired in ( 70ºC ) flat thermoplastic multicore and cpc cable installed in pvc trunking and the ceiling voids .

Inspection and testing is to be carried out in an ambient temperature of ( 20ºC )

detail of conductor résistance in mΩ/m at ( 20ºC ) are shown in ( Figure 2 )

( Figure 2 )
Conductor size mm[SUP]2 [/SUP]1.5 ............ Résistance in mΩ/m at ( 20ºC ) 12.10
Conductor size mm[SUP]2 [/SUP]2.5 ............ Résistance in mΩ/m at ( 20ºC ) 7.41
Conductor size mm[SUP]2 [/SUP]10 ............ Résistance in mΩ/m at ( 20ºC ) 1.83

One drawing to come . ► ( Figure 1 ) worst case scenario I'll type it out

 

[amberleaf]

Question 2394 :

What is faster that Usain Bolt , My fingers typing . Gentlemen please hit the thank you button at your own leisure

 

[Richard Burns]

Source document for Section B scenario, the answers to the above questions must relate to this scenario.

 

[amberleaf]

New :

(*1 ) 2394 : Q/As On Paper you're Inspecting & Testing . To prove yourself . As deemed by BS-7671: & GN-3

Answers , & Comments and advice where appropriate .

Section A
1 a) i) State one statutory document to which the inspector may refer whilst carrying out an inspection and test of an electrical installation .. (1/M)

Ans ; The Electricity at Work Regulation 1989 ◄ The Wording The correct meaning . end off

C;
because the question has asked of one answer , only the first answer given by the candidate would be considered , This is because candidates sometimes write down everything they can remember in the hope that one of the answers may be correct .
This is not showing ( KNOWLEDGE or UNDERSTANDING ) ........................ Make a statement

" ► " EaWR " or " EWR " are acceptable , ►►►►► Electricity at Work " ACT " ►►►►► is incorrect

( PS what does Point's make Prizes ) Your Ticket .

1 a) ii) State tow non-statutory publications to which the inspector may refer whilst carrying out an inspection and test of an electrical installation .. ( 2/M )

Ans ; BS-7671:2008(2011) & IET Guidance Note 3 .. ◄ Your Q . The Wording The correct meaning . end off

C;
Because the question has asked for two answers , only the first two answers given by the candidate would be considered , Other Guidance Notes would be acceptable but the MOST relevant is Guidance Note 3 .

The On-Site Guide is a perfectly acceptable alternative answer . ( Note not ◄► O.S.G. & GN-3 ) Full wording at all times .

 

[amberleaf]

(*2 ) 2394 :

1 a) iii) State the title given in ( LAW ) to the inspector whilst carrying out the inspection of an electrical installation ... 1/M

Ans ; Dutyholder

C;
This question refers to the requirements of The Electricity at Work Regulation's and the term " Dutyholder " refers to anyone having a duty of care to other

1 a) iv) State the legal status of the inspector .. 1/M

Ans ; Competent

C;
Regulation 16 of The Electricity at Work Regulation's requires persons to be competent when carrying out any activity on an electrical system where the work may give rise to danger .

1 b) State the scope of the Minor Electrical Installation Works Certificate ... 2/M

Ans ; Alteration or addition to an existing circuit .

C;
A key part of this answer is the reference to " Existing Circuit " An alteration or addition to an existing installation is NOT within the SCOPE of a Minor Electrical Installation Works Certificate ............ 2394 : Take heed

Any description covering the essential points " alteration " or " addition and " existing circuit " would get the marks .

 

[amberleaf]

(*3 ) 2394 : you will get this one . Yeah

2 a ) i) List five items to be inspected on PVC conduct system .. 5/M

Ans ;
• Conduit system overall
• Fixings
• Bends
• Draw-in points
• Joints

2 a) ii) Explain what is being inspected for each of the items identified in a ) i ) above .. 5/M

Ans ;
• Conduit system overall - conduit system is complete
• Fixings - Sufficient saddles fixed correctly
• Bends - Bending radius not too tight and bend not damaged
• Draw-in points - Sufficient number so that cables can be draw in without damage
• Joints - are secure and fit correctly between lengths and at boxes

C;
The question is in ( Two parts ) i) asks for what is to be inspected and ii) requires the candidate to identify precisely the nature of the ( INSPECTION ) Sometimes these ( Two parts ) are combined into one question

it is important for the candidate to state clearly what is being ( INSPECTED ) and what in particular they would be looking for

" Conduit system not damaged " " conduit complies with appropriate British Standard " , " expansion joints fitted where necessary " and " bushes are tight " are all acceptable answers

Any answer relating to cables , cable terminations and conduit capacity would not be acceptable , although the capacity of the conduit could have been addressed by reference to the conduit being the correct size specified by the designer

 

[amberleaf]

(*4 ) 2394 :

3 a ) Explain , in detail , why an earth fault loop impedance test would need to be carried out on ( Existing circuits ) after the changing of a consumer unit within a domestic installation .. 5/M

You're Q here . ▼▼▼ Have you altered any characteristics ; On Paper you're Inspecting & Testing . To prove yourself . As deemed by BS-7671: & GN-3 . PS we are all sing out the same hymn sheet here .

Ans ;

Existing circuit protective devices have been removed and new protective devices have been fitted , The characteristics of the new devices are likely to be different for previous devices , Therefore it must be confirmed that the earth fault loop impedance is low enough so that the required disconnection times will be met .

C;
The question ask for an explanation " in detail " so the description needs to be comprehensive if full marks are to be awarded , Answers that referred to disconnection and re-connection of conductors and the reliability of those connections would score marks

3 b) i) Explain , why the earthing conductor in an installation must be disconnected from the MET when measuring ( Ze ) ... 2/M

Ans ; To remove parallel earth paths so that the intended fault path can be confirmed to be reliable

C;
if the parallel earth paths are not removed during the test then the reliability of the test result is in DOUBT .