Re-take - Useful Information for 2394 :

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Extracts BS-7671: 2001 (2004) ◄

Unwanted tripping

Unwanted tripping of RCDs can occur when a protective-conductor-current or leakage-current causes unnecessary operation of the RCD.

An RCD must be so selected and the electrical circuits so subdivided that any protective conductor current that may be expected to occur during normal operation of the connected load(s) will be unlikely to cause unnecessary tripping of the device (Regulation 531-02-04 refers). Such tripping can occur on heating elements, cooking appliances etc., which may have elements that absorb a small amount of moisture through imperfect elementend seals when cold. When energised, this moisture provides a conductive path for increased leakage and could operate the RCD. The moisture dries out as the element heats up. Although not precluded in BS 7671, it is not a requirement to use an RCD on such circuits if other satisfactory means of protection are available. Providing an RCD with a higher rated residual operating current may solve the problem but the requirements of the Regulations would still have to be met.

 

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Extracts BS-7671:2008(2013) :icon_bs:

Unwanted operation :svengo:
Unwanted operation of RCDs can occur when a :aureola: protective-conductor-current causes the RCD to operate under non-fault conditions, i.e. the accumulative of protective conductor currents developed by the switch-mode power supplies of computers, e.g. too many computers on one circuit. An RCD must be so selected and the electrical circuits so subdivided that any protective conductor current that may be expected to occur during normal operation of the connected load(s) will be unlikely to cause unnecessary operation of the device (see Regulation 531.2.4). Such operation can occur on circuits with heating elements of cooking appliances etc., where elements can absorb a small amount of moisture through imperfect seals when cold. When energised, this moisture provides a conductive path for current to flow and could operate the RCD. The moisture dries out as the element heats up. Although not precluded in BS 7671, it is not a requirement to use an RCD on such circuits but the requirements of the Regulations would still have to be met, i.e. cables in walls, Regulation 522.6.101.

 

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Ring-final-circuit .
Learning curve .

This also has the benefit that successful test results in Step 2 & Step 3 will confirm that the socket has been correctly wired and the polarity is correct.
End-to-end
r[SUP]1[/SUP] = 0.6Ω
r[SUP]N [/SUP] = 0.6Ω
r[SUP]2 [/SUP]= 1.0Ω

R[SUP]1[/SUP] and R[SUP]N[/SUP] should be the same as the ( CSA ) areas of neutral & line-conductor(s) should be the same in a single-phase-circuit
( T&E ) 2.5mm[SUP]2[/SUP] & circuit-protective-conductor 1.5mm[SUP]2 [/SUP] ... ( CSA ) of the CPC is indeed smaller than the ( CSA ) Line & neutral

The reading expected in Step 2 .
i) ( R[SUP]1[/SUP] + R[SUP]N[/SUP] ) / 4
ii)
iii) ( 0.6 + 0.6 = 1.2 ) ÷ 4
iv) = 0.3Ω

The reading expected in Step 3 .
i) ( R[SUP]1[/SUP] + R[SUP]N[/SUP] ) / 4
ii) ( 0.6 + 1.0 = 1.6 ) ÷ 4
iii) = 0.4Ω

The actual readings obtained in Step 2 & 3 have been recorded .
Socket-outlet (1) Step 2 0.31Ω , Step 3 0.41Ω
Socket-outlet (2) Step 2 0.3Ω , Step 3 0.4Ω
Socket-outlet (3) Step 2 0.32Ω , Step 3 0.41Ω

like any test result , very close to the expected values .

 

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2394
Q/As

A circuit breaker to BS-3871 is marked “ M6 ″ explain what this marking means and describe how the same information would be marked on a circuit breaker to BS-EN-60898-1.

M6 means the circuit breaker has a breaking capacity of 6kA. On a BS-EN-60898-1 circuit breaker it would be marked as the figure 6000 inside a rectangular box.

 

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2394
Q/As
Earth fault loop impedance can be calculated from measured values of Ze, ( R[SUP]1[/SUP] + R[SUP]2[/SUP] )
a) state the equation to be used
b) explain why a measured value of earth fault loop impedance may be less than a value calculated as above.

a) Zs = Ze + ( R[SUP]1[/SUP] + R[SUP]2[/SUP] )
b) Ze is measured with the earthing-conductor disconnected from the main earthing terminal. Zs is measured with all bonding connected and this may provide parallel paths which reduce the apparent value of Zs.

 

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2394
Q/As

An insulation resistance test is to be carried out on a lighting circuit. State the action to be taken in respect of :
a) lamps
b) light switches
c) two-way light switches

a) remove lamps or switch off locally if removal is not possible
b) put light switches ‘on‘ unless the switch is being used to isolate lamps which cannot be removed
c) operate two-way switches alternately while applying the test voltage to ensure that both strappers and the switch wire are tested


​

 

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2394
Q/As
An insulation resistance test is to be carried out on a lighting circuit. State the action to be taken in respect of vulnerable equipment including smoke detectors and self-contained emergency lighting luminaries.

Isolate the vulnerable equipment or disconnect it. If this is not possible then only test between all (( live conductors )) connected together and earth.


​

 

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2394 / 2395
Q/As

Explain briefly why it is desirable to minimise the amount of dismantling during a periodic inspection and test.

Dismantling introduces the risk of:
i) damage
ii) incorrect reassembly
iii) forgetting to reassemble

 

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Q/As

State the maximum rating of a RCD used for:-
a) protecting a 63A socket outlet on a construction site
b) fire protection in an area with a high risk of fire due to the materials
c) protecting a socket outlet supplying a caravan.

a) 500mA
b) 300mA
c) 30mA​

 

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Q/As Yeah they do ask this one .

Describe briefly how to measure the prospective fault current at a three-phase distribution board.

Measure the prospective short-circuit current between each line and neutral. Take the highest reading and double it to get the prospective fault current.

 

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2394
Q/As

1) List two statutory documents applying to the conduct of inspection and testing and one statutory document requiring installations to meet the standards set out in BS 7671.

2) List two main things to be agreed with the client prior to inspecting and testing an existing installation and state where these should be recorded.
3) List five things to check before using a test instrument and its leads.
4) State the instruments which would have the following ranges:-
0-2Ω
0-200 MΩ
0-20 kA

1) Two statutory documents relating to the conduct of inspection and testing:
Health and Safety at Work etc. Act 1974
Electricity at Work Regulations 1989

Electricity Safety, Quality and Continuity Regulations 2002 requires installations to meet requirements of BS 7671.

2) Two things to be agreed with client:
Extent
Limitations
Record these on the Electrical Installation Condition Report form.

3) Five points should be given. See GS-38.
Instrument :​
- in good condition
- calibrated
- battery ok
- suitable for voltage it is to be used on
- suitable ranges for the measurement to be taken.

Leads:
in good condition
adequately insulated for the voltage they are to be used on
fused and/or fitted with resistors to limit current
probes fitted with finger guards
probe tips with bare minimum of exposed metal (2mm or less) or
else covered by retractable spring loaded shrouds
adequate length
coloured to aid identification

4) a) low resistance ohmmeter,
b) insulation resistance ohmmeter,
c) prospective fault current meter

 

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List two factors affecting the measured value of earth fault loop impedance of a given circuit.

a) State a simple rule for taking account of the factors in a) above when comparing measured values of
b) Zs with the maximum values given in BS 7671.

a) circuit loading and ambient temperature
b) Measured values of Zs should not exceed 80% of the maximum given in the tables in Chapter 41 of BS 7671. (Appendix 14 of BS-7671)​

 

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a) State the practical advantage that measuring Rl + R2 has over the alternative method of measuring c.p.c. continuity.
b) Say when it is necessary to use the alternative method.
c) State the percentage rise in the resistance of a copper conductor between 20°C and 70°C.

a) Measuring R1 + R2 requires meter leads of standard length whereas the alternative long lead method, measuring R2, requires a lead the length of the circuit.
b) The alternative ‘long lead’ method must be used for checking the continuity of bonding conductors.
c) 20%
​

 

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2394
Q/As

Outline the dangers which could arise during the following tests:
a) insulation resistance test
b) earth fault loop impedance test
c) RCD test

a) The use of high voltages creates a risk of shock and the risk of damaging vulnerable equipment.
b) If the c.p.c. is broken then every exposed conductive part downstream from the break will be live during the test and present a shock risk.
c) Danger is the same as for b)

A SELV circuit in zone 0 of a bathroom is being inspected. State:-
a) the BS number of the safety isolating transformer
b) the maximum voltages allowed ( both ac and dc )
c) where the transformer should be located

a) the BS-EN number of the safety isolating transformer
b) 12Vacor30Vdc
c) Outside zones 0, 1 and 2

The continuity of the conductors of a ring final circuit is checked using the method given in IET Guidance Note 3. List four things which are confirmed by satisfactory readings. Assume that testing has been done through the front of the sockets using a test lead with a plug.

1) the ring is continuous
2) the value of ( R[SUP]1[/SUP] + R[SUP]2[/SUP] )
3) there are no interconnections in the ring
4) socket outlets are correctly connected

 

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2395 Practical assessment

You will do an insulation résistance test on a single-phase consumer unit , Here you would have to identify 2 faults which would either be a short-circuit , low-résistance , or and earth-fault .

This is very straightforward and should not pose a problem if you know , how to work with insulation résistance tester .

Points to remember . (( Examiner ))
This might differ from examiner to examiner . if they tell you , :yes: do it in the sequence they tell you , you will be better off doing it their way . if you do not then they will fail you , and challenging them on it will probably not get you very far . :35:

 

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2395 Practical assessment

There are also 2 faults to identify out of a possible 7 that the assessor can put on the rig. This could be something like a low insulation résistance, to an open ring-final-circuit. There are not difficult faults trying to trick you, but rather faults you should pick up if you do the tests and interpret the test results correctly.

 

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You are pressed for time in the practical part and have quite a few things to remember. Sometimes a small thing like forgetting to prove dead in the correct manner could cost you the whole test. If you do not isolate, lock off and prove dead in accordance with GS-38 then you will fail.

 

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A very large number of candidates were unable to demonstrate an understanding of voltage drop and its determination during a periodic inspection and test. This is a common and recurring situation across the 2395-302 series of examinations.

A large number of candidates appeared to be aware of the determination of voltage drop during the design of an electrical installation and then attempt to apply that to a periodic inspection. The main incorrect responses included:

• Being unable to explain why voltage drop cannot be determined by direct measurements at the origin and furthest point of the circuit.
• Stating voltage drop can only be determined using the design calculation mV/A/m x Ib x L ÷ 1000 even when this information is not available for an existing installation.
• No correction for conductor operating temperature where resistance is measured at 20°C

Where the method of measurement of conductors was correctly identified a number of candidates lost marks by incorrectly stating measuring R[SUP]1[/SUP] + R[SUP]2 [/SUP]and not ( R[SUP]1[/SUP] & R[SUP]N[/SUP] ) as required for voltage drop. ◄◄◄◄ :6:

( Vd ) Volt drop using test readings .

“ learning curve only ”

Always try to answer the questions in full using the correct terminology

Using ( R[SUP]1[/SUP] + R[SUP]2 [/SUP])
Often the R[SUP]2[/SUP] ( CPC ) will be a conductor with a smaller CSA than the live conductor(s), this of course will result in the calculation showing a higher voltage drop than there would be in reality ( because the current flow in R[SUP]1[/SUP] & R[SUP]N[/SUP] ) ... live conductor(s) L / N

Providing the R[SUP]1[/SUP] + R[SUP]2 [/SUP]calculation give a result of less than the permitted value of 3% ( 6.9V ) or 5% ( 11.5V ) depending on the type of circuit, then all is good . Remember this is only a check , But where the R[SUP]2 [/SUP]calculation for a smaller CPC gives a higher than permitted volt drop it will be worth tying again using R[SUP]N [/SUP]instead .

As an example let’s take a circuit is wired using a 2.5mm[SUP]2[/SUP] / 1.5mm[SUP]2[/SUP] T&E , it has an R[SUP]1[/SUP] + R[SUP]2 [/SUP]value of ( 0.6Ω ) and the circuit is protected by a 20A circuit-breaker

Using the R[SUP]1[/SUP] + R[SUP]2 [/SUP]value of ( 0.6Ω ) you can now calculate the voltage drop for the circuit . (( 0.6 x 20 x 1.20 = 14.4V )) This is to high, if you use R[SUP]1[/SUP] + R[SUP]N [/SUP]you may end up with an acceptable result

How to calculate R[SUP]N [/SUP]
The problem is you do not have a reading for ( R[SUP]N [/SUP]) but a simple calculation will give you all of the information you need
CSA line x ( R[SUP]1[/SUP] + R[SUP]2 [/SUP]) = R[SUP]2 [/SUP]
CSA line + CSA cpc

To put figures to this :
2.5 = 0.625 x 0.6 = 0.375Ω
2.5 + 1.5mm[SUP]2 [/SUP]

0.375Ω is the résistance of R[SUP]2 [/SUP], therefore if you subtract this value from the R[SUP]1[/SUP] + R[SUP]2 [/SUP] value you will have the value of R[SUP]2 [/SUP]0.6Ω – 0.375Ω = 0.225Ω

The résistance of the 2.5mm[SUP]2[/SUP] line-conductor is 0.225Ω
Therefore the (( 2.5mm[SUP]2[/SUP] Neutral-conductor will be the same )) if you now double this value you will have (( R[SUP]1[/SUP] + R[SUP]N [/SUP] )) 0.225 + 0.225 = 0.45Ω

Now you can carry out the voltage drop calculation using 0.45Ω as the résistance value . 0.45 x 20 1.20 = 10.8V

3% ( 6.9V ) or 5% ( 11.5V )

 

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value . 0.45 x 20 x 1.20 = 10.8V

 

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Voltage drop in conductors

To check the suitability of the current carrying capacity it is simply a matter of looking at the installation method , and then checking on the current carrying capacity tables for the cable in Appendix 4 of BS-7671

To ensure that the cable meets the voltage drop requirements is slightly more complex , A simple method is to measure the voltage at the origin of the circuit , and then measure the voltage at the end of the circuit with the load connected and switched on , The difference between the two measurements will be the voltage drop .

if the first method is impractical, then a résistance test should be carried out between the Line and Neutral of the circuit. This test is carried out using the same method as the R[SUP]1[/SUP] + R[SUP]2 [/SUP]test although of the test being between Line and CPC (( it is between the Line and Neutral for the circuit ))
Once the résistance of the R[SUP]1[/SUP] + R[SUP]N [/SUP]circuit has been measured it should be multiplied by the current that will flow in the circuit , This will give you the voltage drop for the circuit .

Example . A circuit is wired in 2.5mm[SUP]2[/SUP] & is 25 metres in length . The current in the circuit is 18 amps
Measured value of résistance is 0.37Ω

Voltage drop = I x R = V .. 18 x 0.37 = 6.66 V