***Cont../ Useful Information for Electricians & Apprentices***

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O.S.G.. The use of other methods’ of determining Maximum Demand is Not Precludedwhere Specified by the Installation Designer

FirstlyI make no Apologies for the Way am Writing on any Matters . it can be a first day Apprentice or some one Needing aJog of Memory .
Sowe are all in the Same Boat . “ To Learn “

For the Apprentices . The Day we stop learning is the Day we hang Upour Tool-Bag

CookerDesign Current Calculations

Thefirst thing you have to do is get Your Head around the Calculations !!

(From a Design point of View ) 2392-10

DomesticInstallation Oven(s) & Hob(s) are to be Calculated upon their MAXIMUM LOADING
Startwith a simple Calculation ( An Oven has a rating of 2kW ) 2000

(I = P/V ) Formula … I = 2000 ÷ 230V = 8.70A …. Weare Using the Unit Amps


2392-10/ Domestic Installation Oven(s)

Ovenhas 4 Rings ( 2 x 1kW ) & ( 2 x 1.5kW ) & Grill ( 2kW ) & Oven (3kW )

-Controlled via a CookerSwitch with a Socket outlet .

Asa Designer . we’ll have to Apply Diversity ??

Important )- Diversity allowance to be Applied to the FULL LOAD CURRENT for CookingAppliances .

TheO.S.G. is telling us . Purpose of the Final Circuit fed from theConductors )
O.S.G.Table 1B p/97 – column (3) Cooking Appliances → At the Top of the Page Note : Type ofPremises ( 2392-10 → Household Installations ) Domestic Installation(s)

DomesticInstallation(s) Only O.S.G. - 10A + 30% f.l – Full Load ) of connected Cooking Appliances in the Excess of 10A+ 5A if a socket-outlet is incorporated in the Control Unit . ( C.C.U. ) – 45A + 13A Socket Switched with Neon .

Fromyour point of View ( The First 10A ofthe rated current plus 30% of the reminder ( Plus) 5A if the Control Unit incorporates s Socket.

Calculations)- You bank “ Hold OFF“ the first 10 Amps of the Maximum Load Current )
The10A will be used at the End of the Calculations’

-So your Work out the Total Power Rating & then calculate the Full Load Current

Calculations)- Power = ( 2 x 1 ) + ( 2 x 1.5 ) + ( 2+ 3 ) = 10kW

I= 10000 ÷ 230V = 43.48A … round it up to the first four numbers43.47826087 ( 48 ) 43.48A

UsingDiversity allowance stated ↑↑ ( 43.48A sub 10A = 33.48A )

I= 33.48 x 30 ÷ 100 = 10.04A

Youradding the ( 5A ) for Socket outlet . I = 10A + 10.04 + 5A = 25.04A )- Asa Designer this is your Expected Current Demand .

Remember )- Supply Cables Rated to suit DesignCurrent ( Iz )

 

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I got this one on my Exams . ( PIR ) :crazy:

Extent & Limitations of the inspection . Note 5 . ( Limitations / Regulation – 634.2 . :13:

 

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Ihave NOT got . ( First Amendedment ) Green / am working off . Red Edition

2392-10. -&-s are looking for .??? Example . BS-7671: 2008 .333 / 337

ElectricalInstallation Certificate : ( EIC )

ThisReport is only Valid if a Schedule of Inspection(s) & a Scheduleof Test Result(s) . 2 x :leaving:


PeriodicInspection Report : ( PIR ) - To be appended to the Report .

(PIR ) Firstly )- Reporting on the Condition of an Existing Electrical Installation . -&-s :13:

ThisReport is only Valid if a Schedule of Inspection(s) & a Scheduleof Test Result(s) . are appended .

PS.Am not working to NICEIC. Test Sheets

 

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WanderLead ( R2 )

Megger- 1552 . Megger / Wander Lead 50m ( 1.17Ω ) Not Nulledyet . :leaving:

Megger- 1552 . “ Auto – Null “ . 0.00Ω

Apprentices. it is Important to Null out any Lead before you startany work .

TestMethod 1 . Will yield : ( R1 + R2 ) GN-3. p/34
TestMethod 2 . Will yield : ( R2 ) GN-3 .p/35

TestMethod ( 1 & 2) are alternative ways of Testing the Continuity of ProtectiveConductor

Scheduleof Test Results ( R2 )
Scheduleof Test Results ( R1 + R2 )

Two methods may be adopted forthis Test but t is Important to Note : that when Testing the Continuity of ( CPCs ) with respect to ( SupplementaryProtective Bondingconductors . WhereRequired ◄◄ ) & MainProtective Bonding Conductors that in order to avoidParallel Paths .

Parallel Paths / it isnecessary to disconnect one end of the conductor in each Case to Obtain anAccurate value .

Parallel Paths ( Parallel - Circuit(s) has more than One Resistor ( Used as a Figure ofSpeech ) . ( Anything that UsesElectricity to do Work ) Learning Curve.
getsits name from having MultipleParallel Paths . to move along .

TheWhy’s / MultipleParallel Paths - This is done so that Lights don’t stopWorking . Yeah

• Fromthis Equation is that the more Circuits you add to a ( ParallelCircuit ) – The more thing you Use : Division of Installation / 314 . Dividedinto Circuits . Etc
TheLower the Total Résistance becomes . Rememberthat as the Total Résistance decreases . The Total Current increases . So ; themore things you Use . the more Current has to flow through the Cables in the PlasteredWalls .

TheInverse of the Total Résistance of the Circuit . ( Effective Résistance ) isequal to the sum of the Inverses of the Individual Résistance .

1/Rt = 1/R[SUP]1[/SUP] + 1/R[SUP]2 [/SUP]+ R[SUP]3[/SUP]+ …. ( This one come around on 2392-10 Exam )

UsefulJunk .
OhmsLaw . may be used in a ( Parallel Circuit ) aslong as you remember that you can use the formula with EitherPartial values or with Total Values . don’t mixparts

V – Total = I total : R total .
V –Part = I part : R part .

- Thepotential drops of each circuit equals the potential rise of the source .

V[SUP]T[/SUP]= V[SUP]1[/SUP]= V[SUP]2 [/SUP]= V[SUP]3[/SUP]= …..

- The total current is Equal to the sum of thecurrent in the circuits .

I[SUP]T[/SUP]= I[SUP]1[/SUP]+ I[SUP]2[/SUP]+ I[SUP]3[/SUP]+ …..

 

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Informationfor Inspector . 2391-10 .

RequirementInformation . Regulation . 610.2 . :38:

BS-7671:2008. requires that the followinginformation shall be made available to the person or persons carrying out theInspection & Testing .

 

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2392-10: Handy in yourExams . :19:

RingFinal Circuit(s)

Statethe effect on Conductor Résistance when cablesare Connected .

Whencables are connected in Parallel the overall Résistancewill be Less than the Résistance of the conductor of the lowest value .

Theformula for Calculating Résistance of Conductor(s) in Parallelis ??

1/R[SUP]T [/SUP] =1/R[SUP]1[/SUP] + 1/R[SUP]2 [/SUP] + 1/R[SUP]3[/SUP][SUP] [/SUP]+ ….. Ω

Where( RT ) is the overall Résistance .

R[SUP]1 [/SUP], R[SUP]2[/SUP][SUP] [/SUP], R[SUP]3[/SUP][SUP] [/SUP], Etc. are the Individual Résistancevalue of each conductor for as many conductors that are Connected in Parallel

Example:

Threeconductors of Résistance ( 1Ω , 2Ω ,& 4Ω ) are connected in Parallel . Calculating the total overall Résistance .

1/R[SUP]T [/SUP] = 1/R[SUP]1[/SUP] + 1/R[SUP]2 [/SUP] + 1/R[SUP]3 [/SUP][SUP] [/SUP]

1/R[SUP]T[/SUP][SUP] [/SUP] = 1/1 + 1/2 + 1/4 .

1/R[SUP]T [/SUP] = 1 + 0.5 + 0.25

1/R[SUP]T [/SUP] = 1.75Ω ( 1 ÷ 1.75Ω = 0.57Ω )

1/R[SUP]T[/SUP][SUP] [/SUP] = 0.57Ω

 

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2392-10: Handy in yourExams . ( TT System ) ◄◄◄ :19:

Measuringusing Earth Electrode Test Instrument )

ThreeMeasurements are taken : 79Ω + 85Ω + 80Ω

Addedal Together ( 244Ω ) …………… 244Ω ÷ 3 = 81.33Ω

Why( 3 ) 3 / Rods . Yeah :chillpill:

 

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2391-10: Ring Final Circuit . :26:

• Statethe relationship between Conductor length & Conductor Résistance .

Résistance. in Ohms , of a Conductor or Cable Increases with Length .


• InsulationRésistance . :yesnod:

Variationin Cable Length .

Asa Cable increases in length . unlike the conductor résistance which increaseswith length . Insulation Résistance decreases with length .

Theformula for calculating ( New Value ) of Insulation Résistance is as follows .

InsulationRésistance = Original length x InitialInsulation Résistance . / New Length

Example:

ACable of length ( 50m ) has an Insulation Résistancevalue of ( 200MΩ ) .
Ifits length is increased to ( 100m ) . Calculate its Insulation Résistance

InsulationRésistance = Original length x InitialInsulation Résistance . / New Length

=50 x 200 ---- 100

=10.000 ----- 100

= 100MΩ

 

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Somethingto think about . Yeah . :yesnod:

Insulation Résistance

Statethe precautions to be . taken before testing Insulation Résistance in Terms of.

i)Isolation of circuit(s) & equipment .
ii)Voltage sensitive equipment .
iii) Electronic components .

a)Isolate of circuit(s) & Equipment .

1)Select a ( GS-38 ) ApprovedTest Lamp or Voltage Indicator .
2)Check ( Test /Lamp / Voltage ) indicatoris functioning correctly by testing on a ( Known Supplyor Proving Unit )
3)Determine whether Isolation is for ( Whole Installation or ( Specific Circuit(s) .
4)Locate means of Isolation ensuring there is no ( Standby Supply Arrangement )
5)Switch ( OFF & Lock OFF Isolation device(s)e.g. Fuses Switch . Isolator . MCB .
6)Verify . ( Using Test Lamp / Voltage Indicator. Circuit(s) are Dead between Line(s) Neutral & Earth.
7)Erect ( Warning Notice(s). in appropriate position(s) . 2392-10 . atConsumer Unit .
8)Re-check ( Test Lamps / VoltageIndicator ) is still functioning correctly .
9)Retain Key(s) . for Locking OFF Devices /

b) Before carrying out InsulationRésistance Testing it must beestablished that there is ( No Voltage Sensitive Equipment Connected ) thatmay be damaged or Malfunction because of the test voltage applied .

Examples:
Maybe Computers . Programmable Logic Controllers ( PLCs ) Programmers . etc. .

TheEquipment would either have to be disconnected during test . ( Testing / 612.3.3 ) ……….. PLUS – Note . Wherethe circuit includes Electronic Devices .

OrInsulation Test with Line(s) & Neutral connected together & Test between theses& the Protective Conductor .

c)Before carrying out Insulation Résistance Testing it must also be established that there is ( No Electronic Components ) that may be damaged ormalfunction because of the Test Voltage Applied .

Example - Dimmer Switch . This would have to beRemoved during test & the switch wires connected together .

- Describe Methods of Testing Insulation Résistance :

Beforecarrying out Insulation Résistance Testing ensure that . ??

1)All lamps . capacitors & loads are disconnected .
2)Voltage sensitive electronic equipment such ad Dimmer Switches . Touch Switches. Delay Timers . Power Controllers . Fluorescent lamps starter . Emergency Lighting . RCDs are disconnected .
3)There is No Electrical Connections between Line(s) . Neutral &Earth .

 

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Useful Junk . ( If )

2392-10 :
RCD . the contacts will Automatically be Opened . Thereby providingagainst Electric Shock . The Term . “ RESIDUAL “ in Residual Current Device implies that ( There) is a residual current flowing in the ( Circuit )Over & above that required to provide ( Power to aLoad ) :yesnod:

 

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2391-10 : ( This maybe helpful in Exams ) WORDING . Yeah :6:

612.2.1. Continuity ofProtective Conductors :

Protective Conductors .

Describethe need for . & Methods of . verifying the ( Continuity) of Protective Conductors & the “ Interpretation of results. ◄◄ :willy_nilly:

“ Answer “

TheContinuity of the ProtectiveConductor is very important since it ( Forms Part) of the Earth Fault Loop Path .
Inthe event of a Fault to Earth the Faultcurrent flows through the Protective Conductors& the Earth Fault Loop Path as a whole .
Ifthe impedance of the Path was too high the Fault current may not be largeenough to operate the Protective Device thusresulting in ( Exposed Metalwork becoming Live )due to the circulating Fault current .

 

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2392-10: :6:

Question.

InsulationRésistance . ( IR) Table 61 . :willy_nilly:

SELV & PELV (- upto ( 50V ) or ( 120V d.c. )

SELV ( Howdo I know / p/31 – Regulations . ) Extra – Low .


Describe the Test to Verify Separation between ( SELV ) circuits & other circuits ( -&-s )

SeparatedExtra Low Voltage ( SELV ) circuits operated at voltages up ( 50V ) or ( 120V d.c. ) This voltage can be obtainedin a number of ways such as a ( Double / Wound Transformer ) having a supply primary coil connected tothe 230V a.c. . Supply & an outputsecondary coil at the required Extra Low Voltage .
Thetwo coils are not & should not be connected in a ANY way 7 a test must be carried out to prove this .

Byusing an Insulation Resistance Tester . set to ( 250Vd.c. ) test voltage on the ( MΩ ) rangecarry out the following test procedure . With the transformer disconnected fromthe supply connected one test lead to the Primary coil & the Other testlead to the secondary coil .

Pressthe test button & note the reading in ( MΩ ) .

Tocomply with BS-7671:2008 . This reading should be no less than ( 0.5MΩ )

 

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Apprentices:

Inspection & Testing . :coolgleamA:

Whatis the definition of Inspection .
Inspectionis a general term that can have a wide variety of meaning . Will usually bemore specific / VisualInspection or Non-destructive Testing .

GN-3p/63 . ( PIR )
621.2.
Theformal Inspections should be carried out in accordance with Chapter 62 ofBS-7671 . This requires an Inspection comprising careful scrutiny of the Installation. carried out without Dismantling or Partial Dismantling as required .together with theAppropriate Tests of Chapter 61 .

EarthFault Loop Impedance .
TheImpedance of the Earth Fault Loop . Line - to - Earth Loop . Starting& Ending at the point of Earth Fault .

RingFinal Circuit . Final Circuit arranged in the form of a Ring & connectedto a Single Point of Supply .

ResidualCurrent Device . ( RCD )

( Residual / Dual . meaning two . Line& Neutral – Two liveconductors .

ItUses the Idea that the Current in the Line &Neutral must be Exactly the Same . Yeah ( Balanced )

TheRCD does is compare the Current in the Line& Neural . If they’re not Identicalit Trips Out .

 

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2392-10:

Regulationsinitially call for Visual Inspection . but someItems such as Correct Connection of Conductors .
Canbe done during the actual Testing . :coolgleamA:

 

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2392-10.

O.S.G.p/86
612.7. EarthElectrode Résistance . ( Ra ) - 200Ω

612.7. : Electrode underTest is being used in conjunction with an ( RCD)

A Loop Impedance Tester ( Zs )

TTSystem(s) . The value of the Earth Electrode Résistance ( Ra ) in Ohms. Multiplied bythe Operating Current in Amperes of the Protective Device . ( I∆n ) should NOTexceed ( 50V ) ( Ra ) = 200Ω . Then the maximum RCD operating Current should NOTexceed ( 250mA ) :coolgleamA:

 

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Apreferred sequence of Tests is recommended . Where Relevant . 2392-10 :willy_nilly:

TN-S / TN-C-S .

612.2.1. Continuity of protective conductors .
612.2.2. Continuity of Ring Final conductors .
612.3.Insulation Résistance . ( IR )
612.6. Polarity .

612.7. Earth Fault LoopImpedance .
612.10. Additional Protection .
612.11. Protective Fault Current . ( PFC )
612.13. Functional Testing .

 

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2392-10.

Lets make the Books work for Us . O.S.G. p/166 . Table 9A .

612.2.1. Continuity of ProtectiveConductors .

AllProtective Conductors . including Main Protective Bonding Conductors &Supplementary Protective Bonding Conductors . ( Where Required )

ConductorSize ( mm[SUP]2[/SUP] ) 2.5 : Résistance ( mΩ/m ) 7.41 .
ConductorSize ( mm[SUP]2[/SUP] ) 4.0 : Résistance ( mΩ/m ) 4.61 .
ConductorSize ( mm[SUP]2[/SUP] ) 6.0 : Résistance ( mΩ/m ) 3.08 .

ConductorSize ( mm[SUP]2[/SUP] ) 10.0 : Résistance ( mΩ/m ) 1.83 .
ConductorSize ( mm[SUP]2[/SUP] ) 16.0 : Résistance ( mΩ/m ) 1.15 .


I WorkingLong Hours . Sunday is the Only DayOff .

I do Apologize if I make any Small Mistakes . Thank youfor Understanding . :willy_nilly:

 

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Topquark. You a Star Sir . Thank You . Amber . I do Apologize

612.13. Functional Testing .

 

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ASimple Résistance test between the Ends of the Line . Neutral or Circuit Protective conductors willonly Indicate that a Circuit Exits .

Whetherthere are Interconnections or Not .

Same value whatever diameter is measured .

(L1 / N2 ) ↔ ( N1 / L2) Circuit formed by Cross- connection .

InitialVerification :
Aftercross connection of Line & CPC you’d check between Line & CPC at every Socket & Spurs would becomeApparent . ( Spur ) will give a Higher Reading .Ω


Apprentices: ( Firstly – You Reading confirms thatthere is No Open Circuit on the Ring Final Circuit) Yeah .

RingFinal Circuit . ( Dead Test )

Instrument: Set on Ohm’s .

Under(- Schedule ofTest Results .
CircuitLoop Impedance Ω
RingFinal Circuit / Only measured – ( End to End )

Line/ Line - Little ( r1 ) = 0.43Ω . Measured – ( Endto End ) Bigger Copper / Lower Résistance .
Neutral/ Neutral - Little ( r/n ) = 0.42Ω . Measured – ( Endto End ) Bigger Copper / Lower Résistance .

CPC/ CPC - Little ( r2 ) = 0.73Ω . Measured – ( End to End ) SmallerCSA / High Résistance


Topquark. I can’t Editany Post . on my Section(s) Thankyou . :coolgleamA:

 

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OhmicValues (- 70m Ring Final Circuit . –wired in 2.5mm[SUP]2[/SUP] / 1.5mm[SUP]2[/SUP] Twin . T&E ) :willy_nilly:

Readings at Each Socket . Examples Only.

Measurements. L1 / L2 . 0.26Ω - ( 0.015Ω )
Measurements. N1 / N2 . 0.26Ω - ( 0.015Ω )

Measurements. CPC -1 / CPC - 2 . 0.32Ω & 0.34Ω . - ( 0.02Ω )

ForSpurs . each metre in length will add the flowing résistance to the above values.

 

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Polarityis towards the end of the recommended test sequence . 612.6 :svengo:

Polarityon the Ring Final Circuit conductors is Achievedsimply by conducting the Ring Circuit Test .

Asfor Radial socket outlets . (- Continuity of the CPC will have already been Proved by linking Line& CPC & Measuring between the same terminalsat each socket .
OnlyLine / CPC reversalsneed to be Checked .

612.2.Continuity of Conductors . We are conducting this Test at the same time as thatfor Continuity of CPCs . ( In Theory ) :bulb2:

Wedo this for Ring Final Circuit . CPCs End – End
Wedo this for Radial . Line / CPC

( In Theory ) Radial Circuit . Line / CPC . we are using Wander Lead Return .

Linkingtogether Line & Neutralat the Origin & Testing between the same Terminals at “ Each Socket “ Line / CPC reversalwill result in NO Reading at the Socket inQuestion .

Lighting Polarity (- Link at Consumer Unit ) Line & CPC – Pendent Set . ( Polarity / proved by Operating Switch ) Switch wire / CPC . Switchis Opened & to confirm then Closed Polarity

 

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2392-10: :aureola:

Statethree different types of Protective Conductorthat are commonly connected to the Main Earth Terminal ( MET ) of an Electrical Installation .

i)Earthing Conductor .
ii)Circuit Protective Conductor .
iii)Main Protective Bonding Conductor .

 

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Self-employed Electricians going to working in Industry . :crazy:

The Pop Up machines now come under the ( PAV ) Category ( Push Around Verticals )

Which also includes the ( 1a ) Category .This course can be done solely for these .

But be more advantageous for you to dothe ( 3a ) Category ( Scissor Lifts ) whichwould also cover you for using the ( PAVs ) and any in the ( 1a ) Category . proving that you have had “Familiarisation “ on them .

What your needing . ( PAL ) Powered Access Licence .
Type ( 3a / 3b ) ◄◄

IPAF : ( 1 person ) One Day Course .Training £ 228.00 – Vat paid

Lasts : 5 Years .

This will show you the Machines to Use . WWW.ipaf.org

 

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FourDocument that the ( Inspector ) may need to consult during the Inspection& Testing procedures . :vanish:

BS-7671:2008.
IET– Guidance Note – 3
IET– On – Site – Guide .
TheElectricity Safety Quality & Continuity Regulations 2002 .

 

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Inspection& Testing of Electrical Installations . :vanish:

Usingdiagrams to support your answer explain the three steps required to perform aContinuity of Ring Final Circuit Conductors Test.

Explainany precautions & checks you would make before starting the Test .

Using (- A Low résistance ohm-meter continuity tester will beused for this Test .

WithTest Leads & Probes conforming to HSE . GN-38
Beforeany measurements are taken the résistance value of the Test Leads .
&Probes are noted & this value is subtracted from any Test Results . Alternatively if the Test Instrument has anAuto null facility this should be applied before any measurements are recorded.
Withthe electrical installation safety Isolated& Proved Dead.

Indentify& disconnect the Line . Neutral . & CPCconductors of Ring Final Circuit at the Distribution Board .

Measure& Record the End to End résistance of the following.

Therésistance of the Phase to Line Loop = (r1 )
Therésistance of the Neutral to Neutral loop = ( r/n )
Therésistance of the CPC to CPC loop = ( r2 )

Thevalues of ( r1 + r/n ) & r2 willindicate whether the conductors are continuousor not .

A FiniteSmall reading will indicate the conductors are continuous .

A high Résistance value will indicatethe conductors are Open Circuit . or Faulty . orConductor terminations are loose .

TheLine & Neutral conductors both being typically 2.5mm[SUP]2 [/SUP] should have equal values of résistance:- Measured within 0.05Ω of each other .within 0.05Ω
Wouldalso apply if the Ring Final Conductors were larger ( 4.0mm[SUP]2 [/SUP] ) CPC being 1.5mm[SUP]2 [/SUP] . for
2.5mm[SUP]2 [/SUP] / 1.5mm[SUP]2 [/SUP] T&E - has a smaller Cross-sectional area than theLine 2.5mm[SUP]2 [/SUP] conductors . As suchthe CPC résistance will slightly higherthan the Line conductors’ .

2.5mm[SUP]2 [/SUP] / 1.5mm[SUP]2 [/SUP] T&E - The résistance of the CPC should beapproximately ( 1.67 times ) greater than the résistance of the Line conductors.
Performa quick calculation to check the Expected value of the CPC is approximately ( 1.67times ) larger than the Line Conductors .

 

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Whattype of Inspection & Tests . Will be required for the ElectricalInstallation within this building . ??

Forthe ( NEW ) Electrical Installation . an InitialInspection Test & Verification
Forthe ( Existing ) Electrical Installation . anPeriodic Inspection report & Test .

Listall the relevant documentation that need to be competed for this ElectricalInstallation . ??

( 2392-10 ) ► New ◄ Electrical Installation
i)Electrical InstallationCertificate .
ii)Schedule of Inspections .
iii)Schedule of Test results .

thistake you into 2391-10 ) ( Existing ) Electrical Installation
i)Periodic Inspection Report for an Electrical Installation
ii)Schedule of Inspections .
iii)Schedule of Test results .

 

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Twopossible conditions that may have an Effect on the Measured value of ( EarthFault Loop Impedance ) :vanish:

i)Ambient Temperature in the locality of thecables Under Test .
ii)The Actual Temperature of the Cables Under Test .

 

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Whatfunction does a Switch perform in the Circuit . :willy_nilly:

ASwitch ( Opens ) or ( Closes) an electrical circuit .

Ifthe Circuit is ( Open ) – NO electricityflows . “ Circuit is broken “

Ifthe Circuit is ( Closed ) - Electricity Flowsthrough the Circuit . 2392-10 ( Can we confirmPolarity ) “ Circuit is complete “

Sowhat is a Polarity Test .

Basicallyit is a Test that creates a Circuit using the Line conductor & the singlepole device in question . Breaking the circuit when Operating the device . means that the reading on theInstrument will Change .

&thus confirming that device must be connected in the Line Conductor .

Test to Ensure that all singlepole devices ( Fuses . Switches . & Circuit Breaker ) are connected inthe Line Conductor Only .


Switch: Permanent Live. Switch Live .

Switching T&E . You should sleeve the Blue Conductor / with brown sleeve .to say it is being used as a Switch Wire .

UsefulJunk .

Metal is Used in Switches because most Metals are Good Conductors of “ Electricity “

 

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DomesticInstallation

Whatis - Circuit Control from a Testers point of View ? :willy_nilly:

Itis Vital that any Circuit . NO matter its levelof complexity . ( Can be Controlled )

Asimple Switch :- allowing us to turn the CircuitOn or Off .

 

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Regulation/ Definitions – p/26 . Wording

Isolation :- “ Function “ .

Afunction intended to Cut Off for reasons of safety the supply from all . or a discretesection . of the Installation by separating the Installation or Section fromevery source of Electrical Energy .

Isolator :- “ Disconnector “

AMechanical Switching Device which . in the ( OpenPosition ) complies with the requirements specified for the Isolationfunction in part 2 of BS-7671 .

Note : it also States a“ a “ Isolator “ is otherwise known as an “ Disconnector “

 

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Sourceof Supply .

Oftena circuit is Thought of as being the ( Load )& the conductors supplying the ( Load )
InReality they are only ( Part of the Circuit ) inorder for Current to flow two conditions have to be met .

i)There has to be a ( Potential Difference ) applied across the circuit ( Voltage )
ii)There has to be a ( Complete Circuit ) “Circle “ for current to flow around .

A Source of Supply can be ( A.C. or D.C. )

 

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Apprentices.

Determiningthe Protective Device characteristics forCircuits . ??

ElectricalInstallation Design .
Allyou need to know is to use the given details to select the Appropriate Protective Device for a given circuit .

Themain formula we can start with .

Iz > = In > = Ib .

Ib (- design current of circuit ( A ) The loadto be connected to the circuit .
In (- Nominal current of the Protective Device ( A )
Iz (- Effective current carrying capacity ( A )

Thismeans that in every circuit at all times the current carrying capacity of thecables must be ( Higher ) than the highestpossible load current .
Also. the Protective Device rating must be ( Higher) than the load current but lower than the maximum current carrying capacity ofthe cable installed .

 

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As-&-s are touchy about RCDs .

AResidual Current Device ( RCD ) will . ??
i)Operate if a Short Occurs between Line & Neutral
ii)Operate if a Small Overload occurs .
iii)Operate if Line& Neutral are out of Balance . ◄◄ “ Resi/ dual “ . Two Words - Line & Neutral

 

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2392-10(- Q )- . An Overload Current is . ?? Regulation p/27 – A )- . An overcurrentoccurring in a circuit which is Electrically Sound . :6:

 

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Thetype of Fault which occurs when a Line conductorcomes in contact with a Neutral conductor is called? ( ShortCircuit Fault ) :willy_nilly:

 

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Regulation Red book 2008 : Chapter 53 :speechless:

Protection. Isolation . Switching . Control & Monitoring .

Regulation:- The “ Term “ Isolation & Switching – as used in BS-7671 ( Refers to four distinct functions )

i) Isolation
ii)Switching OFF for mechanical Maintenance .
iii)Emergency Switching &
iv) Functional Switching

Referto Table 53.2 – p/117

 

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2392-10:- “ Consumer Controlgear “ 530.3.4. :speechless:

ForInstallation with a 230V single-phase supplyrated up to 100A that are under the control of ( Ordinary Persons) should either .

• Comply with BS-EN 60439-3 & Regulation 432.1. of BS-7671.
• Be a Consumer Unit – incorporating components complyingwith BS-EN 60439-3

 

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Regulation Chapter 532 . :- Devices for Protectionagainst the Risk of Fire . :whatchutalkingabout

532.1.Where in accordance with the requirement of Regulation 422.3.9. it is necessaryto limit the Consequence of ( Fault Currents )in wiring system from the point of view of ( Fire Risk ) THE Circuitshall be either .

i)Protected by an RCD complying with Regulation 531.2. for ( Fault Protection ) &
- The RCD shall be installed at the ( Origin of the Circuit ) to be Protected . &
- TheRCD shall Switch all Live Conductors . &
-The rated Residual Operating Current of the RCDshall Not Exceed ( 300mA) ETC .

 

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Chapter537 . Isolation & Switching .

Isolation– ( 537.2.
Switchingoff for Mechanical Maintenance - ( 537.3.
EmergencySwitching - ( 537.4 .
FunctionalSwitching - ( 537.5.

Fire-fighter’s Switches – ( 537.6 . :speechless:

A Fire-fighter’sswitch shall be provided in the Low Voltage circuit supplying :

i)( Exterior ◄ ) electrical installationsoperating at a Voltage Exceeding Low Voltage . &

ii)( Interior ◄ ) Discharge lighting installations operating ata Voltage ( ExceedingLow Voltage )

(Temporary Installation ) Fire-fighter’s Switches – ( 537.6.1.

Forthe purpose of this regulation . An installation in a CoveredMarket . Arcadeor Shopping Mall is considered to be an (Exterior Installation )
• A Temporary Installation in a Permanent Building used for Exhibitions isconsidered ( NOT to be an Exterior Installation )

 

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“ Isolation“ :icon14:

Itemsto be Considered in Relation to “ Isolation “ devices.

• Number of poles to be “ Isolated“
• “ Isolation “ of groups &inconvenience .
• Devices designed and / orinstalled to prevent Unintentional or Inadvertent closure .
• “ Isolation “ devices to beClearly Indentified and Indicate the Installation or Circuit it Isolates .

 

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Isolationrequirements for ( TN-S ) Systems .

Single-phase. L ≈ N - Circuit Protective Conductor .

Three-phase. L1 . L2 . L3. ≈ N . - Circuit ProtectiveConductor .