***Useful Information for Apprentices***

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“ General Health and Safety at Work “

Question 1.1
What do the letters CDM stand for ?
A: Control of Demolition and Management Regulations
B: Control of Dangerous Materials Regulations
C: Construction (Demolition Management) Regulations
D: Construction (Design and Management Regulations ) Answer: D )
Question 1.2
Identify one method of enforcing regulations that are
available to the Health and Safety Executive:
A: Health Notice
B: Improvement Notice
C: Obstruction Notice
D: Increasing insurance premiums
Answer: B Improvement notices require action to achieve standards which meet health and safety law :
Question 1.3
What happens if a Prohibition Notice is issued by an
Inspector of the local authority or the HSE ?
A: The work in hand can be completed, but no new work started
B: The work can continue if adequate safety precautions are put in place
C: The work that is subject to the notice must cease
D: The work can continue, provided a risk assessment is carried out,
Answer: C The work covered by a prohibition notice must cease until the identified danger is removed.
Question 1.4
Health and Safety Executive Inspector can ?
A: Only visit if they have made an appointment
B: Visit at any time
C: Only visit if accompanied by the principal contractor
D: Only visit to interview the site manager
Answer: B Inspectors have a range of powers, including the right to visit premises at any time.
Question 1.5
A Prohibition Notice means:
A: When you finish the work you must not start again
B: The work must stop immediately
C: Work is to stop for that day only
D: Work may continue until the end of the day
Answer: B The work activity covered by the prohibition notice must cease, until the identified danger is removed ,
Question 1.6
In what circumstances can an HSE Improvement Notice be issued ?
A: If there is a breach of legal requirements
B: By warrant through the police
C: Only between Monday and Friday on site
Answer: A Improvement notices require action to achieve standards which meet health and safety law .
Question 1.7
What is an “Improvement Notice”?
A: A notice issued by the site principal contractor to tidy up the site
B: A notice from the client to the principal contractor to speed up the work
C: A notice issued by a Building Control Officer to deepen foundations
D: A notice issued by an HSE/local authority Inspector to enforce compliance with health
Answer: D Improvement notices require action to achieve standards which meet health and safety law .
Question 1.8
If a Health and Safety Executive Inspector issues a“ Prohibition Notice”, this means that:
A: the Site Manager can choose whether or not to ignore the notice
B: specific work activities, highlighted on the notice, must stop
C: the HSE must supervise the work covered by the notice
D: the HSE must supervise all work from then on
Answer: B Prohibition notices are intended to Stop activities which can cause serious injury.
Question 1.9
Which one of the following items of information will you find on the Approved Health and Safety Law poster?
A: Details of emergency escape routes
B: The location of the local HSE office
C: The location of all fire extinguishers
D: The identity of the first aiders
Answer: B The poster also lists the persons with health and safety responsibilities, but not first aiders.
Question 1.10
Who is responsible for signing a Company Safety Policy ?
A: Site Manager
B: Company Safety Officer
C: Company Secretary
D: Managing Director
Answer: D The Health and Safety at Work Act requires the most senior member of management to sign the health and safety policy
statement.

Question 1.11
Which one of the following must be in a company’s written Health and Safety Policy:
A: Aims and objectives of the company
B: Organisation and arrangements in force for carrying out the health and safety policy
C: Name of the Health and Safety Adviser
D: Company Director’s home address
Answer: B This requirement appears in the Health and Safety at Work Act.
Question 1.12
Employers have to produce a written Health and Safety Policy statement when:
A: A contract commences
B: They employ five people or more
C: The safety representative requests it
D: The HSE notifies them
Answer: B This is a specific requirement of the Health and Safety at Work Act.
Question 1.13
Companies employing five or more people must have a written Health and Safety Policy because:
A: The principal contractor gives them work on site
B: The HSAWA 1974 requires it
C: The Social Security Act requires it
D: The trade unions require it
Answer: B
Question 1.14
What do the letters HSC stand for ?
A: Health and Safety Contract
B: Health and Safety Consultant
C: Health and Safety Conditions
D: Health and Safety Commission Answer: D
Question 1.15
Which ONE of the following statements is correct ? The Health and Safety Executive is:
A: a prosecuting authority
B: an enforcing authority
C: a statutory provisions authority
Answer: B The Health and Safety Executive enforces health and safety legislation.
Question 1.16
The Health and Safety at Work Act requires employers to provide what for their employees?
A: Adequate rest periods
B: Payment for work done
C: A safe place of work
D: Suitable transport to work
Answer: C This is a specific requirement of Section 2 of the Health and Safety at Work Act.
Question 1.17
The Health and Safety at Work Act 1974 and any regulations made under the Act are:
A: Not compulsory, but should be complied with if convenient
B: Advisory to companies and individuals
C: Practical advice for the employer to follow
D: Legally binding Answer: D
Question 1.18
Under the Health and Safety at Work Act 1974, which of the following have a duty to work safely?
A: Employees only
B: The general public
C: Employers only
D: All people at work
Answer: D Employers, employees and the self-employed all have a duty to work safely under the Act.
Question 1.19
What is the MAXIMUM penalty that a Higher Court, can currently impose for a breach of the Health and Safety at Work Act?
A: £20,000 fine and two years imprisonment
B: £15,000 fine and three years imprisonment
C: £1,000 fine and six months imprisonment
D: Unlimited fine and two years imprisonment
Answer: D A Lower Court can impose a fine of up to £20,000 and/or up to six months imprisonment for certain offences. The potential fine in a Higher Court, however, is unlimited and the term of imprisonment can be up to 2 years.
Question 1.20
What do the letters ACoP stand for ?
A: Accepted Code of Provisions
B: Approved Condition of Practice
C: Approved Code of Practice
D: Accepted Code of Practice
Answer: C An ACOP is a code of practice approved by the Health and Safety Commission.

Question 1.21
Where should you look for Official advice on health and safety matters?
A: A set of health and safety guidelines provided by suppliers
B: The health and safety rules as laid down by the employer
C: Guidance issued by the Health and Safety Executive
D: A professionally approved guide book on regulations
Answer: C The HSE is the UK enforcing body and its guidance can be regarded as ‘official’
Question 1.22
Regulations that govern health and safety on construction sites:
A: apply only to inexperienced workers
B: do not apply during ’out of hours’ working
C: apply only to large companies
D: are mandatory ( that is, compulsory )
Answer: D The requirements of health and safety law are mandatory, and failure to follow them can lead to prosecutions.
Question 1.23
Which of the following statements is correct ?
A: The duty for health and safety falls only on the employer
B: All employees must take reasonable care, not only to protect themselves but also their colleagues
C: Employees have no responsibility for Health and Safety on site
D: Only the client is responsible for safety on site
Answer: B The responsibility for management of Health and Safety Act at Work rests with the employer
Question 1.25
Which of the following is correct for risk assessment?
A: It is a good idea but not essential
B: Only required to be done for hazardous work
C: Must always be done
D: Only required on major jobs
Answer: C There is a legal requirement for all work to be suitably risk assessed.
Question 1.26
In the context of a risk assessment, what do you understand by the term risk?
A: An unsafe act or condition
B: Something with the potential to cause injury
C: Any work activity that can be described as dangerous
D: The likelihood that harm from a particular hazard will occur
Answer: D Hazard and risk are not the same. Risk reflects the chance of being harmed by a hazard
Question 1.27
Who would you expect to carry out a risk assessment on your working site?
A: The site planning supervisor
B: A visiting HSE Inspector
C: The construction project designer
D: A competent person
Answer: D A risk assessment must be conducted by a 'competent person’.
Question 1.28
What is a HAZARD ?
A: Where an accident is likely to happen
B: An accident waiting to happen
C: Something with the potential to cause harm
D: The likelihood of something going wrong
Answer: C Examples of hazards include: a drum of acid, breeze blocks on an elevated plank; cables running across a floor.
Question 1.29
What must be done before any work begins ?
A: Emergency plan
B: Assessment of risk
C: Soil assessment
D: Geological survey
Answer: B This is a legal requirement of the Management of Health and Safety at Work Regulations.
Question 1.30
Complete the following sentence: A risk assessment
A: is a piece of paper required by law
B: prevents accidents
C: is a means of analysing what might go wrong
D: isn’t particularly useful
Answer: C Risk assessment involves a careful review of what can cause harm and the practical measures to be taken to reduce the risk of harm.

 

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A combined resistive/reactive circuit dissipates more power than it returns to the source. The reactance dissipates no power; though, the resistor does.
As with any reactive circuit, the power alternates between positive and negative instantaneous values over time. In a purely reactive circuit that alternation between positive and negative power is equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed resistance and reactance like this one, the power waveform will still alternate between positive and negative, but the amount of positive power will exceed the amount of negative power. In other words, the combined inductive/resistive load will consume more power than it returns back to the source.
Looking at the waveform plot for power, it should be evident that the wave spends more time on the positive side of the center line than on the negative, indicating that there is more power absorbed by the load than there is returned to the circuit. What little returning of power that occurs is due to the reactance; the imbalance of positive versus negative power is due to the resistance as it dissipates energy outside of the circuit (usually in the form of heat). If the source were a mechanical generator, the amount of mechanical energy needed to turn the shaft would be the amount of power averaged between the positive and negative power cycles.
Mathematically representing power in an AC circuit is a challenge, because the power wave isn't at the same frequency as voltage or current. Furthermore, the phase angle for power means something quite different from the phase angle for either voltage or current. Whereas the angle for voltage or current represents a relative shift in timing between two waves, the phase angle for power represents a ratio between power dissipated and power returned. Because of this way in which AC power differs from AC voltage or current, it is actually easier to arrive at figures for power by calculating with scalar quantities of voltage, current, resistance, and reactance than it is to try to derive it from vector, or complex quantities of voltage, current, and impedance .

• REVIEW:
• In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
• In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90o out of phase with each other.
• In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0o and 90o.

Circuits :
A circuit is a loop of wire with its ends connected to an energy source such as a cell* or a battery *
One end of the wire is connected to the positive terminal , the other end of the wire is connected to the negative terminal , the wire is connected in this way so a current * can flow through it ,
Advantages of Parallel circuits :

Parallel circuits have two advantages when compared with series circuits .
The first advantage of a parallel circuit is that a failure of one component does not lead to the failure of the other components . this is because a parallel circuit consists of more than one loop and has to fail in more than one place before the other components fail .
The second advantage of parallel circuits is that more components may be added in parallel without the need for more voltage *

It is essential that delegates are familiar with the different types of earthing system likely to be
encountered (TN-C-S, TT and TN-S ).

Electricity System Earthing Arrangements : revision ,
First letter:
T The live parts in the system have one or more direct connections to earth.
I The live parts in the system have no connection to earth, or are connected only through a high impedance.
Second letter:
T All exposed conductive parts are connected via your earth conductors to a local ground connection.
N All exposed conductive parts are connected via your earth conductors to the earth provided by the supplier.

Remaining letter(s):
C Combined neutral and protective earth functions (same conductor).
S Separate neutral and protective earth functions (separate conductors).
TN-C No separate earth conductors anywhere - neutral used as earth throughout supply and installation
TN-S Probably most common, with supplier providing a separate earth conductor back to the substation.
TN-C-S [Protective Multiple Earthing] Supply combines neutral and earth, but they are separated out in the installation.

TT No earth provided by supplier; installation requires own earth rod (common with overhead supply lines).
IT Supply is e.g. portable generator with no earth connection, installation supplies own earth rod.
TN-S The earthing conductor is connected to separate earth provided by the electricity supplier. This is most commonly done by having an earthing clamp connected to the sheath of the supply cable.
TN-C-S The earthing conductor is connected to the supplier's neutral. This shows up as the earthing conductor going onto the connection block with the neutral conductor of the supplier's meter tails. Often you will see a label warning about "Protective Multiple Earthing Installation - Do Not Interfere with Earth Connections" but this is not always present.

TT The earthing conductor goes to (one or more) earth rods, one of them possibly via an old Voltage Operated ELCB (which are no longer used on new supplies). 17th edition
There are probably other arrangements for these systems too. Also, a system may have been converted, e.g. an old TT system might have been converted to TN-S or TN-C-S but the old earth rod was not disconnected.

-&- like to get some off Q/A in : LOOK in the regs , p32/34 , p/187 ( * 32 ) read a must !!!!! earthing

Cables in contact with polystyrene : ←←←←←←
Do not let electrical cables come into contact with polystyrene. It slowly leaches the plasticiser out of the PVC, so that it becomes stiff and brittle. Sometimes it looks like the PVC has melted and run a little.

Question :

Something has failed in this circuit, because the light bulb does not light up when the switch is closed:
Transformer 120V / P : 15V / S .
What type(s) of transformer fault(s) would cause a problem like this, and how might you verify using a multimeter?
A : The most common type of transformer fault causing a problem like this is an open winding. This is very easy to check using a multimeter

Question
Calculate all listed values for this transformer circuit:
48VAC . 13000 turns . ││ 4000 turns . R load . 150Ω .
* V. Primary = / V. Secondary = ,
* I. Primary = / I. Secondary = ,
Explain whether this is a step-up, step-down, or isolation transformer, and also explain what distinguishes the "primary" winding from the secondary" winding in any transformer.
→ A , V. primary = 48Volts ,
V. secondary = 14.77Volts ,
I. primary = 30.3mA ,
I. secondary = 98.5mA ,
This is a step-down transformer.

* the first step in preparing an installation design is to identify the electrical loads, the physical position of the load is required as well as the kVA demand , power factor, voltage , frequency etc ,

Festoon Cable – Internal : Minimum Bending Radius: 10 x cable diameter
SWA Cable - BS-5467 Steel Wire Armoured PVC : Minimum Bending Radius: 1.5mm² - 16mm²: 6 x overall diameter
SWA Cable - BS-6724 Steel Wire Armoured : Minimum Bending Radius: 1.5mm² - 16mm²: 6 x overall diameter
Alarm Cable: Minimum Bending Radius: 12 x overall diameter
Telephone Cable: Minimum Bending Radius 10 x overall diameter
Internal Telecom Cable : Minimum Bending Radius: 8 x overall diameter
( you have to keep in mind the Bending Radius of cables )

 

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Transformer Fault Current Calculation : ( revision )
Transformer Impedance
Transformer Impedance is measured in percent Impedance , this is the percentage of rated Primary Voltage applied to the Transformers’
Primary windings in order for the rated Secondary full load current to flow in the Secondary windings → ( this test is preformed with the Primary winding connected to a variac or variable supply and the Secondary windings shorted-out ) ←
( Transformer Impedance values may vary but typically electrics Transformers will be between 4 to 5% of Primary voltage

Impedance Voltage Vz = Primary Volts (V) x Percent Impedance Z% ----- 100 :
Primary Volts (V) 230 ( Vp
Secondary Volts (V) 110 ( Vs
Transformer Rating (VA) 2000 VA
Transformer Configuration ( CTE )
Percent Impedance ( Z% ) 4
Impedance Voltage ( Vz ) 9.20
Transformer Maximum Earth Fault Current
Maximum Fault Current A = 100 --- Impedance Volts Vz% x Secondary Full Load Current I :
Maximum Earth Fault Current (A) 455.0 * ( Percent Impedance ( Z% ) 4 *
Maximum Earth Fault Current ( kA ) 0.46 * ( Secondary Full Load Current ( I ). 18.18 *
This is the maximum current be achieved at the Transformers Secondary Terminals and doesn’t allow for any Impedance in the
Primary or Secondary circuitry to and from the Transformer .
Transformer Maximum Earth Fault Loop Impedance :
To Calculate the Maximum Earth Fault Current that can be achieved in a circuit fed by a Transformer .
Calculates the Loop Impedance at the end of a circuit fed from the Secondary winding of a Transformer .
When calculating Earth Fault Loop Impedance select the correct Transformer type above
Three-phase Transformers divide the values for ( Vs secondary voltage by √3 and the ( VA by 3 for centre tapped to Earth ( CTE )
Transformers halve the values for ( Vs ) and ( VA ) for RLV / 110V CTE disconnection times & ( Zs figures refer to BS-7671 : 411.8 – table 41.6

Zsec = Zp x }Vs – Vp}2 + {Z% -- 100} x {Vs – VA} + {R1+R2)
0.49Ω ( Zp ) = loop impedance of primary circuit including Source { Ze + 2R1 }
230V ( Vp ) = primary voltage .
55V ( Vs ) = secondary voltage .
1000VA ( VA ) = Transformer rating .
4% ( Z% ) Transformer percentage impedance .
0.2Ω ( R1 + R2 ) = résistance of secondary circuit Phase & Protective conductor .
Primary Circuit Impedance ( Ze + 2R1 ) 0.49 . Zp ( Ze + 2R1 ) Ohms .
Secondary Circuit Impedance ( R1 + R2 ) 0.2 Ω
Total Impedance ( Zsec ) 0.3490 Ωs
Transformer Actual Earth Fault Current :
To calculate actual earth fault current in a transformer we can refer to the below formula taken from BS-7671: 2008 . ( 411.4.5 )
And use the above figure for the transformers total Impedance ( Zsec )

Earth Fault Loop Impedance ( Zs ) Voltage to Earth ( Uo ) ---- Fault Current ( Ia ) :
Fault Current ( Ia ) = Voltage to Earth ( Uo ) ---- Loop Impedance ( Zs ) :
Actual Secondary Fault Current ( Ia ) 157.6 Amps :
Primary Fault Current ( I ) Secondary Fault Current ( Ia ) x Secondary Voltage ( Vs ) ---- Primary Voltage ( Vp ) :
Actual Primary Fault Current ( I ) 37.7 Amps :

Single-Phase Current ( I ) = kVA --- Volts ( V ) 5.0 ÷ 110.0 = ( Current . Amps 45.45 )
Single-Phase ( kVA ) = Volts x Current ( I ) --- 1000 = ( 110.0 x 45.0 ÷ 1000 = ( kVA 4.95 )
Single-Phase Current ( I ) = Power ( kW ) / Power factor ( pf ) --- Volts ( V ) ↔ ( 4.00kW ÷ 0.800pf ÷ 110.0V = 45.45 current (Amps )
Dissipated Power ( W ) = Current ( I2 ) x Résistance ( R ) ↔ 0.010 current (Amps ) 12000.00 Ω = 120* Power ( Watts )
Inductance ( L ) = Volts ( V ) --- ( 2∏ x frequency ( f ) x Current ( I ) ↔ 240V / 50Hz x 18.2A = 0.0420 ( inductance ( Henries ) ←
3-Phase kVA = Volts ( V ) Current ( I ) x √3 ÷ 1000 = 110.0V x 50.0A x √3 ÷ 1000 = 9.53kVA :
3-Phase Current ( I ) kVA / √3 --- Volts ( V ) 10.0 ÷ √3 ÷ 110.0 = 52.49Amps ←
3-Phase Current ( I ) = Power factor ( pf ) --- ( Volts ( V ) x √3 ) 8.00kW ÷ 0.800pf = 10 : 110.0 x √3 = 190.5 : 10 ÷ 190.5 = 52.49 ←

* Frequency : the voltage or current changes from maximum ( plus ) in one direction , through zero to a maximum ( minus ) in the other direction .
This occurs at , f is the frequency in Hertz . ( 1 Hz = 1 cycle / second ,
* Periodic Time :
The time it takes to complete 1 cycle is T seconds ( the periodic time ) it follows that ( T = 1 / f )
* Angular Frequency :
If we think of the voltage and current being generated by a machine that rotates one revolution per cycle , the 1 cycle corresponds to 360° or 2π radian .
( ɷ = 2 π / T rad/s )
* 15 V r.m.s applied across an inductance of 4 µH . calculate the r.m.s . current when the frequency is 200Hz and 200 MHz .
Solution :
20 Hz Xc = 2π fL = 2π x 20 x 4 x 10-6* = 0.5027 mΩ ( Irms = V -- Xc = 15 --- 0.527 x 10-3* = 29.84 kA . )
200 kHz Xc = 2π fL = 2π x 200 x 4 x 10-6* = 5.03Ω ( Irms = V -- Xc = 15 --- 5.03 = 2.984A )
200 MHz Xc = 2π fL = 2π x 2000 x 4 x 10-6* = 5027Ω ( Irms = V -- Xc = 15 --- 5027 = 2.98mA )
* 15 V r.m.s applied across a capacitance of 4.7µf . calculate the r.m.s. current when the frequency is 20Hz & 2000Hz .
Solution :
20Hz Xc = 1 -- 2π fC = 1 -- 2π x 20 x 4.7 x 10-6* = 1693Ω ( Irms = V -- Xc = 15 -- 1693 = 0.00886A )
200Hz Xc = 2π fC = 2π x 200 x 4.7 x 10-6* = 169.3Ω ( Irms = V -- Xc = 15 – 1693 = 0.0886A )
2000Hz Xc = 1 -- 2π fC = 2π x 2000 x 4.7 x 10-6* = 16.93Ω ( Irms = V -- Xc = 15 -- 1693 = 0.886A )
( 6* or 3* small on the number 10 ) ←←←

Phase displacement
Phase displacement\ (Elec.) A charge of phase whereby an alternating current attains its maximum later or earlier. An inductance would cause a lag, a capacity would cause an advance, in phase.

Select Transformer required by multiplying lamp wattage by the number of lamps used i.e. ( 100 x 5w lamps = 500VA transformer )

A string of eighteen identical Christmas tree lights are connected in series to a 120V source, the string dissipates 64.0W.

a) What is the equivalent resistance of the light string ?

P = IV
64W ÷ 120V ( 0.533 )
I = 0.533 A
R = V/I
120V ÷ 0.533A ( 225 )
R = 225 ohms

Equivalent resistance is 225 ohms ?

b) What is the resistance of a single light ?

As it is connected in series 225 ohms ÷ 18 lights = 12.5 ohms

Why is it that when you put two electric lamps into a circuit in parallel with one another, the current through the circuit increases, while when you put those two lamps in series with one another, the current through the circuit decreases ?

When the two lamps are in parallel with one another, they share the current passing through the rest of the circuit. Current arriving at the two lamps can pass through either lamp before continuing its trip around the circuit. The two lamps operate independently and each one draws the current that it normally does when it experiences the voltage drop provided by the rest of the circuit. With both lamps providing a path for current, the current through the rest of the circuit is the sum of the currents through the two lamps.

But when the two lamps are in series with one another, each lamp carries the entire current passing through the circuit. Current arriving at the two lamps must pass first through one lamp and then through the other lamp before continuing its trip around the circuit. There is no need to add the currents passing through the lamps because it is the same current in each lamp. Moreover, the voltage drop provided by the rest of the circuit is being shared by the two lamps so that each lamp experiences roughly half the overall voltage drop. Since lamps draw less current as the voltage drop they experience decreases, these lamps draw less current when they must share the voltage drop. Thus the current passing through the circuit is much less when the two lamps are inserted into the circuit in series than in parallel

Landscape Lighting

Since landscape lights use low voltage, you will need a transformer. How big a transformer you need will be based on how many lights you are planning to use. It's important to know that the farther away lights are from the transformer, the higher the chance of a voltage drop occurring. Voltage drop will cause the lights farther away to be a little less bright than the ones closest to the transformer. How do you prevent the noticeable signs of voltage drop ? By using math, of course! Here’s a formula for figuring out voltage drop:

0.0011 x Total Watts on Cable x Length of Cable = Voltage Drop in %

For example: 8 fixtures x 8 Watts = 64 Watts
0.0011 x 64 watts x 50 feet of cable = 3.52% Voltage Drop

As long as the voltage drop is lower than 8 percent per fixture, you won't notice any dimming. Next, determine what size transformer you will need to power the lights. For instruction purposes, lets say you want to power eight 8-watt path lights, three 50-watt spot lights, and two 12-watt wall lights. First, calculate the total wattage.

8 x 8 = 64 watts
3 x 50 = 150 watts
2 x 12 = 24 watts

64 + 150 + 24 = 238 total watts

You will want to get a 300 watt transformer for this project. Always get a transformer that can handle the addition of any lights down the line. Do not exceed double the load wattage,

Now let’s put it all together. Place the lights where you want them to illuminate. Run the wire from the transformer to the last light in the line

 

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Transformer Calculations
Wattage refers to the amount of power (electricity) consumed by your appliances and equipment. Sounds simple, doesn't it ? Actually it is very simple and it is important to know a little about it because you need to know wattage in order to choose a Converter or a Transformer. The wattage of most appliances and equipment is found on decals or labels of some sort on the appliance or equipment or in the owner's manual. In addition, the decals or labels also usually give the voltage and/or amperage of the appliance.

If the wattage isn't given you can still figure it out if you know the voltage and amperage. If you multiply the voltage times the amperage, the result equals the wattage of the appliance. For example, the decal on the appliance you want to take overseas doesn't list the wattage but gives the voltage as 120 volts and the amperage is 1.1. To find the wattage of that particular appliance, multiply the voltage 120 times the amperage 1.1, ( 120 x 1.1 = 132 watts. ) Convert to UK )

300 watts, times the hours they run per day, times 30 days, divided by 1000, times the cost per kilowatt hour your power company charges. 8 to 18 !!!! depending where you are. ( T wick the pounds around )

300W x 12M x 30D ÷ 1000 = 108 kWH x 10 ? = £ 10.80 per month

First work out the Wattage... that’s easy as a straight V*I on the output side...
You have 12 volts and 1 ampere. Multiplying those together you get 12VA.
First work out the Wattage... that’s easy as a straight V*I on the output side...
12 * 1 = 12 Watts
Now apply the PFC to see what the input side sees by dividing your output wattage by the chosen value of the PFC ( lets use 0.7 )
( 12/0.7 = approximately 17VA ) ↔ 12 ÷ 0.7 = 17.14285714 )

North American 110-120 volt electricity is generated at 60 Hz. (Cycles) ← Alternating Current. Most foreign 220-240 volt electricity is generated at 50 Hz. (Cycles) ← Alternating Current. This difference in cycles may cause the motor in your 60 Hz. North American appliance to operate slightly slower when used on 50 Hz. foreign electricity. This cycle difference will also cause analog clocks and timing circuits that use Alternating Current as a timing base to keep incorrect time. Most modern electronic equipment including battery chargers, computers, printers, stereos, tape and CD players, VCR/DVD players, etc. will not be affected by the difference in cycles.

that being the ~25mA .This figure is the magnetising current, that current being the current required to magnetise the transformer core. The magnetising current depends on the core material, size, shape and makeup.
Then how do you find the total amps out by the other figures ? lets assume we don’t have amps, basically id like to work out the total amount of amps from a 700VA ups ?
V * A = VA
Therefore, A = VA / V

VA = 700 and V = 230, hence A =

700 ÷ 230 = 3A

Your UPS is capable of supplying 3A continually or, in hardware terms, a couple of computers with monitors, your modem and hub/switch without becoming overloaded.

Working on amperes is a bad thing for a UPS as UPSes are rated in VA rather than watts. Working in amps assumes watts and can lead you to overloading your UPS and possibly causing damage.

Solar Panels
Rule of Thumb for sizing Solar Panels: figure out the average current draw of the remote device that you need to power in Milliamps (MA). We're assuming a 12 VDC system. Pick a solar panel with a current rating of at least 10 times this number. Example: your remote Solar Panels device draws 6 MA average. Pick a solar panel of at least 60MA output current, or to calculate power (watts) would be 60MA x 12 VDC = 720 MW which is 7.2 watts so a 10 watt panel will work great for this system.

Rule of Thumb for Batteries: you will want your device to run for 3-5 days without any sun, depending on your weather conditions. (Actually the solar panel will charge a little bit every day, even under sunless conditions.) Example: if your device draws 100MA, figure .1A x 24 hr/day x 5 days = 12 AH (Ampere-Hours). You can only use about 1/2 of the rated battery output so pick a battery rated at least 24 AH.

while marine batteries range from 60 AH to 105 AH. You can connect batteries in parallel ↔ ( + to + and - to - ) to double or triple the available output.

Basic SI units and formulas associated with Electricity :
As mentioned elsewhere, the Watt (W) is the derived SI unit of power, and is defined as one joule per second. When talking about electricity, it is a measure of energy used or generated at a point in time. 1 Kilowatt (kW) = 1000 Watts.
It is often represented by the letter P.
The Ampere (A) is the base SI unit of current. It is normally represented by the letter I.
Voltage (V) is the derived SI unit of electric potential. It is normally represented by the letter V.
Formulas’:
These three units can be tied together in the formula
P = V * I where P is measured in Watts, V is in Volts, & I in Amps. For example. if your current was initially in ( mA ) this must be converted to Amps before using in the formula (e.g. 300mA = 0.3 Amps)
Example 1
A tape player connected to a 12V car battery draws 800mA of current. How much power is it consuming ?
P = V * I = 12 * 0.8 = 9.6 Watts ( 12 x 0.8 = 9.6Watts )
Example 2
A mains operated electric jug draws 9.5 Amps while operating. How much power is it consuming ? Assume the mains is 240V AC.
P = V * I = 240 * 9.5 = 2280 Watts or 2.28 kW ( 240 x 9.5 = 2280. )
Note: The formula for measuring power when dealing with AC is P = V * I * pf. In the above example, ( pf ) would be extremely close to (1) so it can be ignored.
Appliances which plug into standard 240V power outlets are supposed to have a maximum power rating of 2400 Watts. This equates to a maximum current of 10 Amps at 240 Volts.
Measurements of Power :
* The Watt ( symbol W ) is the derived SI unit of power, and is defined as one joule per second. When talking about electricity, it is a measure of energy used or generated at a point in time. 1 Kilowatt (kW) = 1000 Watts.
* The Watt-hour (symbol ↔ Wh ) is a measure of energy used or generated over a period of time. Our electricity providers normally bill us in units of Kilowatt-hours (kWh), one kWh simply being 1000 Wh. Note that the kilowatt-hour is a measure of the instantaneous power multiplied by time and not divided as we are probably more familiar with. e.g. km/h, being kilometres travelled in an hour. The time period is always related to an hour.
- An appliance drawing 1000 Watts (1 kW) for 1 hour is said to have consumed 1 kWh of electricity.
- An appliances drawing 2 kW for a period of 2 hours has consumed 4 kWh of electricity.
- A clock radio rated at 2 W will draw 48 Wh or 0.048 kWh over a 24 hour period.
- A 1000 Watt toaster takes 3 minutes to cook toast. The electricity consumed will be 1000 Watts * (3 minutes / 60 minutes) = 50 Wh or 0.05 kWh. ( 1000 x 3 ÷ 60 = 50 Wh )

 

[amberleaf]

Hot Water Power Consumption Formula Derivation : The following two formula are used to derive the resultant third formula: Formula 1 ( Without going into too much detail here ) Q = m x C x T : Where . Q is the amount of heat energy (Joules) . m is the mass of the substance (kg) . C is the specific heat capacity of the substance ( joule / kg / °K ) . T is the change in Temperature ( K or °C ) The mass ( m ) of water is 1 kg per litre, so the number of litres to be heated can be substituted here. The specific heat ( c ) of water is 4200 . Formula 2 ( 1 KWh = 3.6 MJ ) ( By definition, 1 Joule = 1 Watt per second. Knowing that there are 1000 Watts in a kW and 3600 seconds in an hour, the above formula can be derived ). Resultant formula . By combining these two formulas and substituting the specific heat of water value for the constant, the following formula is arrived at: P = ( 4.2 * L * T ) / 3600 . Where . P is the power used in KWh . L is the number of litres of water heated . T is the Temperature difference between the hot water ended up with and the cold water started with in °C. Voltage & Frequency : Abbreviation ( AC ) : Definition ( Alternating Current ) Common Sources ( Household –Mains ) . Abbreviation ( AC ) : Definition ( Direct Current ) Common Sources ( Batteries , Solar Photovoltaic Cells / Panels AC mains in UK is usually specified as 230 or , with a frequency of 50 Hz. ( cycles per second ). This specified voltage is what is called an RMS ( Root Mean Square ) voltage. It's not a constant voltage like DC. Voltage and current are constantly varying around a reference point, following a sinewave pattern. (The waveform on the mains is rarely a pure sinewave. It is often distorted or 'dirty', due in part to some equipment drawing power in spurts rather than at a constant rate). 50 cycles of this sinewave are completed every second. The peak-to-peak voltage of this sinewave is the RMS voltage multiplied by 1.414 ( the square root of 2 ). If we assume the RMS voltage to be 230 Volts, the voltage from one peak of the sine wave to the other is 230 * 1.414 or ??? Volts. This actual RMS voltage from the mains is not usually very steady. It varies with loads on the circuit, location and the time of day. The voltage at my house varies from around 230V to over 248V.

 

[amberleaf]

Calculating the amount of electricity consumed to heat water : The following formula can be used to calculate the amount of power required to heat a quantity of water. It assumes 100% efficiency, with no losses. * The start and finish temperatures of the water in °C must be known . If heating water in a jug, adding around 10% to the total should compensate for inefficiencies to see how the formula was derived . Pt = (4.2 * L * T ) / 3600 . where . Pt is the power used in kWh . L is the number of litres of water heated . T is the Temperature difference between the hot water ended up with and the cold water started with in °C . Example 1: Calculate the amount of electricity required to boil 1.5 litres of water in an electric jug if the starting temperature of the water is 20°C. ( P = (4.2 * 1.5 * ( 100-20 )) / 3600 = 140 Wh. ) If we add ~10% to overcome inefficiencies then the figure becomes 154 Wh . Example 2: Calculate the amount of electricity consumed by an electric hot water service to provide 45 litres of water at 50°C for a shower, assuming the water started off at 20°C . P = ( 4.2 * 45 * (50-20)) / 3600 = 1575 Wh or 1.575 kWh. ) This figure isn't allowing for any heat losses in the storage tank which can be quite significant (40% or more) if the water is heated and then stored for later use. - It is also possible to work out how much time it should take to heat water: After coming back from holidays, the 250 Litre hot water system was turned on after being off for a few weeks. How long would you have to wait for the water to heat to say, 50°C so you could enjoy a hot shower ? The starting temperature of the water is 20°C, and the heater element is rated at 3.6 kW (15 Amps). First calculate the kWh required to heat the water from the formula above: Pt = (4.2 * 250 * (50-20)) / 3600 = 8.75 kWh. To work out the time taken, divide Pt (kWh) by the element rating (in kW ) Heating Time = 8.75 kWh / 3.6 kW = 2.43 hours. Again, this figure is assuming 100% efficiency. A more realistic guess might be around 2.8 hours. Deriving instantaneous power (Pi): This method involves timing how long the spinning disc on the meter takes to rotate through a given number of divisions or rotations. The formula given below can be used: Pi = ( 3600 * N) / (T * R ) where: Pi = Real power being used at that point in time in kW . N = The Number of full rotations counted. When measuring smaller loads, it's more appropriate to calculate this as a fraction of a rotation. Normally one rotation consists of 100 divisions . T = Time (in seconds) for the disc to rotate through the N rotations or part of a rotation . R = The number of revolutions per Kilowatt hour (rev/kWh) of the meter being used. This is normally printed on the meter. A few values I've seen are 133.3, 266.6 & 400.

Calculate the power being consumed in the house at present if the meter takes 2 min 15 sec (135 seconds) to rotate through 50 divisions, using a meter with a rev/kWh value of 133.3, and total number of divisions per revolution of 100. P = (3600 * 50/100) / (135 * 133.3) = 0.1 KW or 100 Watts. ( 3600 x 50 ÷ 100 = 1800 ) ↔ ( 135 x 133.3 = 17995.5 ) ↔ ( 1800 ÷ 17995.5 = 0.1 )

 

[reedsinthewater]

I've printed all the info and I'll happily share at college with those who are interested

 

[amberleaf]

Impedance of supply :
Voltage falls with an increase in current ( 230V . 40A flows )

Thus Zs system volt drop . 230 228 --- 40 = 2Ω --- 40 = Ohms
Answer : 230 sub 228Ω = 2 ÷ 40 = 0.05Ω
Then : PSSC = Uo2 --- Zs = 230 --- 0.05 Ωs ( 230 ÷ 0.05 = 4600 )
PSSC = Uo2 230 ÷ 0.05 = 4600Amps or 4600kW ,
230 sub 228 = 2 ÷ 40 = 0.05 . ( PSSC = Uo 230 ÷ 0.05 = 46Amps )

 

[amberleaf]

Fuses are placed in a circuit to act as a weak point which will melt if things get too hot. This stops fires happening after a cable overheats or an appliance explodes. If the current flowing through a cable is too great for either the size of the cable , or the appliance it is feeding, something has to give....The idea is the fuse will go first....., its important the right fuse is put in !

A normal house hold fuse is rated at 13 amps. An amp, short for ampere is a unit of electrical current. An electrical appliance needs to have electricity flowing to it to work. How much electricity it needs to work, is measured in amps. It is very important to know how many amps each piece of electrical equipment in your home draws from the power supply because there is only a finite amount of power available before things start to cook!

If you know how many amps an appliance needs, you know which part of the system to get the electricity from and which fuses to apply and which cables and wires to use to stop it, the fuses or wires overloading or overheating.
To work out how many amps an appliance needs to draw from the supply you need to know the Wattage of the appliance. A watt is the way of measuring the rate at which an appliance uses the power available to it.
Appliances, even light bulbs , are marked with a Wattage. To stop the numbers getting too big, 1000 Watts equals
1 kiloWatt. ( Watt being the name of the guy who sorted it all out in the first place). To find out what size fuse to use to protect an appliance you need to know the wattage and the voltage available. Most homes in the UK are served by a 230 volt supply. The super speedy fast kettle we have in the office is rated at 2500 - 3000 Watts.(2.5 -3kW) We must take the larger figure for safety. 3000 kwatts’ is then divided by the voltage to give us the current rating of the cable and fuse. ( 3000kW ÷ by 230 = 13.04amps. )

As 3000 Watts is the maximum it is safe to put a 13amp fuse in the plug. The cable size is determined by looking at our other projects. This is how amps are worked out. If you know how many amps are available, e.g. you have a fuse, MCB or RCD in your consumer unit but you are unsure of what you can put on this circuit, you can do the calculation the other way round. Multiply the voltage by the amps and this will give you the maximum number of Watts you can place on the circuit.

If your lighting circuit at home is protected by a typical 5 amp fuse you can multiply this by the voltage to get 1150 Watts. Now you can work out how many bulbs, and of what size, it is safe to have in your fittings. For low voltage lighting see another project. ( 230V ÷ 5A = 1150Watts )

A cooker circuit is slightly different from this however. Should you be trying to work out the size of cooker you can buy, a cooker, of 12kW or 12,000 Watts, with all rings blazing and the oven and grill firing out juicy steaks can potentially use a whole shed full of electric and should need a 52amp fuse to protect it . ( 12,000kW ÷ by 230volts = 52amps )
Cookers : 12000kW . ( 12000 ÷ 230 = 52A ) Ib = P = 12000 . ( Ib = P/12000 --- U/230 = 52A ) first 10Amp : = 42Amp ← ( 42 x 30% = 12.60 ) ↔ ( first 10Amp + 12.60 + 5 = 27.6Amp ) Allowance for socket outlet / 5 Amp : Actual circuit current 12.6Amps : Without socket-outlet 22.6Amps . ( * Protective Device 32Amps * ) ←

But, the chances of all every part of the cooker being on at one time are very small and a principle, called The Diversity Principle, is applied to the cooker circuit. For the diversity principle to be understood you must assume that the first 10 amps of current are always needed by the cooker. Probably only 30% of the remaining, available current will be used at any one time however. The actual demand will then be, 10 amps (used all the time) added to 30% of the remaining 42amps (= 13amps) making a total of 23 amps. If the cooker control unit has a 13 amp socket as well as the cooker switch ( another 5 amps must be added, making 28 amps. It would then be appropriate to place the cooker on a 30 or 32 amp circuit :

• Cables are used for different applications because they are differing thicknesses and can cope with differing currents safely with differing amounts of electrical resistance. A simple way to explain resistance is to see it as electrical friction. The cable will slow down some of the energy in the current. This means a little less current will reach the target than was actually sent.

Electric cookers, electric showers and Immersion heaters use a great deal of current and therefore require thicker cables. If a cable is too thin for the job it is being asked to do, it will get too hot and catch fire. Generally it can be said that any cable carrying current to an appliance that is intended to produce heat will have a bigger current rating and bigger cable.

Also to be taken into consideration is the distance of the appliance from the electrical source. The greater the distance, the greater the resistance and the less current that will be available at the other end.
Cables must be placed in a situation where they will not be overheated. If cables are run in a loft they must not be placed under insulation and if they are run in insulated walls they will carry a different rating. Please check the tables below :

The tables below show two different cable and amp ratings. The first table is for cables installed by what is known as Method 4. This is cables enclosed in an insulated wall. The second table is for cables fixed using method 1.2 . This is called clipped direct.

Table : Cables enclosed in an conduit in an insulated wall: Method 7 : Regs : 4D5 / p-282 . Cable size : 1mm : Rating in Amps 11.5 : 1.5mm : 14.5A : 2.5mm : 20A 4.00mm : 26A : 6.00mm : 32A : 10.00mm : 44A :

Table 4D5 - 70°C : Cables which are clipped direct: Method C method - 6 : Cable size : 1mm : Rating in Amps 16 : 1.5mm : 20 A : 2.5mm : 27A : 4mm : 37A : 6mm : 47A : 10mm : 64A : 16mm : 85A :

 

[amberleaf]

A current of 8A flows in an A.C. magnet circuit connected to a 230V supply .
( Determine the impedance of the magnet coil )
U = I x Z ( 230V = 8A x Z ) 230 ÷ 8 = 28.75
Z = 230 --- 8 = 28.75Ω

Percentages & Efficiency
Efficiency : the output of a generator ( 1500W ) 1500 ÷ 1900 = 0.789.
The intake is equivalent to ( 1900W ) 0.789 x 100 = 78.9%
Percentage efficiency : ( 78.9%

Efficiency of Motor loading :
Motor produces output power ( 2000W ) 2000 ÷ 2800 = 0.714.
Electrical input ( 2800W ) 0.714 x 100 = 71.4%
Efficiency of motor at loading : = % = 71.4%

Determine Percentages Efficiency of Generator
The power output from a generator is ( 2600W ) 2600 ÷ 3500 = 0.742
The power required to drive it is equivalent to ( 3500W ) 0.742 x 100 = 74.2%

Determine Percentage voltage drop : 400 – 385V ---- 400 x 100 ( 400 sub 385 = 15. )
( 15 ---- 400 x 100 ) 15 ÷ 100 = 0.15 ↔ 0.15 ÷ 400 = 3.7%

Calculate percentage volt drop :
The voltage at the terminals of a motor is 223V ( 223 sub 230 = 7 ) ↔ 0.07 ÷ 230 = 3%

Power circuits :
A Ring final circuit :
In general domestic premises there should be a separate ring
Final circuit for every 100m2 of floor area. We must remember
That the maximum load that can be connected to a ring protected
By a 30A / 32A fuse is just over 7 kW, and the number of circuits Selected accordingly .

Power - DC Circuits : Watts = E x I Amps = W / E
E = Voltage / I = Amps / W = Watts

What is Power Factor ?
Power Factor is a characteristic of alternating current, and can be defined as the ratio of working power to total power.
Alternating current has the following components
* Real Power ( - Power which produces work ( kW )
* Available Power ( -Power delivered or total volt amps ( kVA )
* Reactive Power ( - Power needed to generate magnetic fields required for the operation of inductive electrical equipment. (kVAR) No useful work is performed with reactive power.
Therefore the unitless Power Factor is obtained from
* Power Factor = ( Real Power ----- Available Power = kW --- kVA

Power Factor is generally represented as a percentage or a decimal. Perfect power factor, often referred to as unity power factor would be 100% or 1.0.

What is Power Factor Correction ?
All flowing current causes losses in the supply and distribution system. A load with a power factor of 1.0 results is the most efficient loading for the supply and a load with a power factor of 0.6 will have much higher losses in the supply system. These loses have to be paid for, and result in higher utility bills. It is possible to modify the supply and distribution system to bring the power factor closer to unity. This is called power factor correction.

Correcting Power Factors
The simplest form of power factor correction, sometimes referred to as static correction, is by the addition of capacitors in parallel with the connected inductive load. The resulting capacitive current is a leading current and is used to cancel the lagging inductive current flowing from the supply. The capacitors can be applied at the starter, or the switchboard or at the distribution panel. Note that power factor correction should not be used when a motor is controlled by a variable speed drive.
Rather than correcting each individual load, the total current supplied to the distribution board can be monitored by a controller which switches capacitor banks to maintain the power factor at its predetermined setting. The controller switching in capacitors as new loads come on line, and switching out capacitors as loads go off line. This type of correction is sometimes referred to as bulk correction.

Common Inductive Loads
Commonly used electrical equipment that provide an inductive load include lighting circuits, heaters, arc welders, distribution transformers and electric motors.

 

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RCDs for protecting people have a rated tripping current (sensitivity) of not more than 30 milliamps (mA). Remember:

What is PUWER?
PUWER replaces the Provision and Use of Work Equipment Regulations 1992 and carries forward these existing requirements with a few changes and additions, for example the inspection of work equipment and specific new requirements for mobile work equipment. Many aspects of PUWER should therefore be familiar to you.

The Regulations require risks to people’s health and safety, from equipment that they use at work, to be prevented or controlled. In addition to the requirements of PUWER, lifting equipment is also subject to the requirements of the Lifting Operations and Lifting Equipment Regulations 1998.

Extension leads
The use of extension leads should be avoided where possible. If used, they should be tested as portable appliances. It is recommended that 3-core leads (including a protective earthing conductor) be used.

A standard 13 A 3-pin extension socket-outlet with a 2-core cable should never be used even if the appliance to be used is Class IT, as it would not provide protection against electric shock if used at any time with an item of Class I equipment.

Class III equipment
Equipment, for example for medical use, in which protection against electric shock relies on supply at SELV (Safety extra low voltage) and in which voltages higher than those of SELV are not generated. Class III equipment must be supplied from a safety isolating transformer.

Portable Appliance Testing ↔ (PAT) For Short !!!

The Dangers of Earth Leakage : Modern equipment causes a standing current in the protective earth conductor. This is not a fault, but a functional requirement of the equipment. Because of this standing current earth integrity becomes very important; when a fault develops the voltage on exposed metal work will rise to half the supply voltage with possibly lethal consequences.

* For added safety trailing sockets should be used in conjunction with a residual current device (RCD).
Cable Plug Length ( 2 ) Cable Specification : 1.25mm2 <HAR> 3-core PVC insulated cable to BS 6500: 1994: Table 16/IEC227-5. Colour white. Standards and approvals : All 13A Duraplug trailing socket outlets comply with ** BS 1363/A: Part 2: 1995. ** ↔ Regulations / P/229 . ↔ Regulations / P/229 : Replacement fuses where fitted are to BS 1362 : Regulations / P/230 . BS 6500:2000 ( 2005 ) :

you should use the lowest value fuse depending on the wattage as described in The Electrical Equipment (Safety) Regulations so for a 900 watts fixture 3 amp is too small ,13 amp is too high and a 5 amp is just right.

Also don't forget to allow for the switch on surges which some equipment may exhibit, inductive loads such as those fitted with wound components such as transformers, motors and compressors will draw a lot of current at switch on. Ever seen the stage lights dim slightly when you turn on your amplifier? - well that’s as a result of the surge which it draws when the transformer is energised and the large internal psu smoothing caps charge.

Plug top fuses don't have the anti-surge capabilities of the 20mm glass / ceramic type fitted internally into the equipment which feature a little spiral or thicker points within the fuse wire in order to absorb the initial surge, and ordinary fuses can blow as a result of the surge if it is not accounted for. This is why most fridges will come with a 5A or 13A plug fuse in the moulded plug, even if the appliance is only rated at a few 100 watts

If you run several items from one 13a extension lead then always remember to switch on the connected equipment in a orderly manner. Leaving everything in the 'on' position as you throw the switch at the plug end could cause enough surge from several pieces of equipment all energizing together, to blow even a 13A fuse, and sometimes trip a venue RCD

It is vitally important that the flex is rated equal to or above that of the fuse. Otherwise you could have a 3A piece of flex connected to a 13A plug fitted with a 13A fuse. Whilst the current is below 3A the flex is fine, however if the load increases to 10A or even 13A, then the fuse still will not blow, however the 3A flex will be carrying three or four times its design current causing it to get very hot and eventually melt, possibly igniting any floor tiles of furnishings which it is in contact with, and you have a fire risk.

Bear in mind, that because of the design characteristics of fuses, it may be possible to draw a significantly higher current for a short period of time before the fuse finally gives up and fails. I've seen venues with toaster, deep fat fryer and a tea urn all plugged into one 13A 4 gang trailing socket Fair enough if they aren't all used at the same time, but the large brown spot appearing on the outside of the plastic casing stated otherwise, and yes it is possible to draw anything upto about 20A or 21A from a 13A plug top fuse for half an hour or so before the fuse will break, plenty of time to start a fire so a little common sense and mathematics is called for, where one lead may be powering an entire disco - don't think for one minute that the fuse will blow the moment you step over that 13A rating.

------- 3 METRE 4 WAY EXTENSION LEAD

Electrical Equipment (Safety) Regulations 1994 FAIL ( BS- in the UK )
Plugs and Sockets, etc. (Safety) Regulations 1994 FAIL
Failures include access to live parts, cover removable without tools, inadequate conductors (0.11mm ←←← as opposed to 1.25mm ←←← required ), inadequate cord anchorage, inadequate fuse and the product was labelled as 6 way socket, but was only 4 way!

UNKNOWN BRAND 10 METRE, 3 GANG CABLE REEL
Electrical Equipment (Safety) Regulations 1994 FAIL
Plugs and Sockets, etc. (Safety) Regulations 1994 FAIL
Failures include inadequate shutters on sockets, access to live parts, inadequate cable size, inadequate cord anchorage, inadequate fuse

 

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All emergency fittings should be regularly tested, they are normally installed with a test switch near by, which can only be operated by a special key. Each fitting should be inspected daily to check that the LED is on. Tested monthly to see they operate, and every 6 months they should be run for an hour. Every 3 years a full 3 hour duration test should be carried out, preferably until the batteries are run down completely. In this way the batteries which are sealed Nickel Cadmium can maintain a full charge, if you only partially discharge them they develop a "memory" and only part of their power is available. All emergency fittings should have a red LED which shows that the batteries are charged and ready for action, if the LED is not lit the batteries

 

[Stringbags]

I have copy/pasted a lot of the info that you have provided, my students are due for their 2330 unit 301 exam/gola soon and they will benefit from your contribution. thanks Amberleaf for taking the time to share.

Stringbags

 

[amberleaf]

Electrical safety - main guidance
if electrical equipment is unsafe or in a poor condition it could cause personal injury, workplace fires or even kill.
Electrical injury can arise from
• Electric shock
• Electric burns
• Fires of electrical origin

Inspecting and testing electrical equipment ( Pat Testing )
Inspecting and testing equipment involves action at three levels:-
• Checks by the user
• Formal visual inspections :by an appointed competent person or competent contractor
• Combined inspection and tests :by an appointed, competent person or competent contractor

In most cases, staff are likely to use their own equipment including computers and it is in their own interests to ensure it is in a good, safe condition.
They can do this by carrying out ( User Checks *** ) before they start work. : Ps this will come up a lot in -&- Q. Formal visual inspection

** For double-insulated equipment, breakdown is caused by physical damage 99 times out of
100, so visual inspections should identify this.
***If not in a high risk environment and not subject to abuse. If latter occurs then need to
carry out combined inspection and testing more frequently.

Where extension cables or extensions in drum reels are in use, ensure the equipment is not overloaded - check the information on the extension cable and ensure the required current in amps does not exceed the quoted figures. ( This is what -&- are looking for Q/As )

2.5mm2 – 25 metres
2.5mm2 – Extension Leads are too Large for Standard ( 13A ) plugs ,
Although they may be Used with BS-EN 60309 Industrial Plugs . Extension Leads exceeding the above Lengths should be Fitted with a ( 30mA RCD ) ←← a Must -&-
Manufactured to BS-7071 :

Regs : p/21 Class 111 Equipment : ( if in Doubt most of the time you’ll find it in Definitions ) Part 2 ,

Equipment in which protection against electric shock does not relay on basic insulation only .but in which additional safety precautions such as Supplementary Insulation are provided . there being No Provision for the Connection of Exposed Metalwork of the Equipment to a Protective Conductor . and No Reliance upon precautions to be taken in the fixed wiring of the installation ( see BS-EN 61140 )

Class 111 Equipment : Equipment in which protection against Electric Shock Relies on Supply at SELV and in which Voltages higher than those of SELV are Not Generated ( BS-EN 61140 )

‘powertrack system’ means an assembly of system components including a generally linear assembly of spaced and supported busbars by which accessories may be connected to an electrical supply at one or more points
( pre-determined or otherwise) along the powertrack. )

‘protective conductor’ means a conductor used for some measures of protection against electric shock and intended for connecting together any of the following parts:
(i) exposed conductive parts,
(ii) extraneous conductive parts,
(iii) main earthing terminal
(iv) earth electrode(s),
(v) the earthed point of the source, or an artificial neutral

‘SELV’ (Separated Extra-Low Voltage) means an extra-low voltage which is electrically separated from earth and from other systems in such a way that single fault cannot give rise to the risk of electric shock

‘short circuit current’ means an overcurrent resulting from a fault of negligible impedance between live conductors having a difference in potential under normal operating conditions

‘PELV (Protective Extra-Low Voltage)’ means an extra-low voltage system which is not electrically separated from earth, but which otherwise satisfies all the requirements for SELV

permit-to-work’ means an official form signed and issued by a responsible person to a person having the permission of the responsible person in charge of work to be carried out on any earthed electrical equipment for the purpose of making known to such person exactly what electrical equipment is dead, isolated from all live conductors, has been discharged, is connected to earth and on which it is safe to work

Protection
Mechanical protection includes the provision of barriers, enclosures . protective covers, guards and means of identification, the display of warning notices and the placing of equipment out of reach. Where it is necessary to remove barriers or open enclosures, protective
covers, guards, this should be possible only by use of a key or tool. ←← -&- Q/As in the Regs 17th

Every high voltage (H.V.) switchroom/substation, except when manned, should be kept locked. A duplicate key for each H.V. switchroom/substation should be available, for emergency purposes, in a key box at a designated location. All other keys for use in the H.V. switchroom/substation should be kept under the control of a responsible person.

Exceptionally, a key may be held by a person whose duties require him to have frequent access to an H.V. switchroom/substation. In
such a case, that person should obtain a written authorisation from the responsible person stating the duties for which the person is
required to hold the key

Arrangement of entrance/exit
At least one exit of a switchroom/substation should open outwards and this emergency exit should be identified clearly .

Conductors near the entrance/exit of a switchroom/substation must be so arranged or protected that there is no risk of accidental contact of any live metal by any person entering or leaving

In order to provide free and ready access at all times for the maintenance and operation of the electrical equipment contained in a switchroom/substation, every entrance/exit of a switchroom/ substation should be kept free of any obstruction including
** which impedes the access to the switchroom/substation from a public area

* ‘appliance’ means an item of current using equipment other than a luminaire or an independent motor or motorised drive.
* ‘appliance, fixed’ means an appliance which is fastened to a support or otherwise secured or placed at a specific location in normal use.
* ‘appliance, portable’ means an appliance which is or can easily be moved from one place to another when in normal use and while connected to the supply.
* ‘barrier’ means an effective means of physically preventing unauthorised approach to a source of danger.
* ‘basic protection’ means protection against dangers that may arise from direct contact with live parts of the installation
* ‘bonding’ means the permanent joining of metallic parts to form an electrically conductive path which will assure electrical continuity and has the capacity to conduct safely any current likely to be imposed.
* ‘bonding conductor’ means a protective conductor providing Main Protective bonding Conductor . Regs p/32
* ‘bunched’ means two or more cables to be contained within a single conduit, duct, ducting or trunking or, if not enclosed, are not separated from each other.
* ‘busbar trunking system’ means a type-tested assembly, in the form of an enclosed conductor system comprising solid conductors separated by insulating material. The assembly may consist of units such as expansion units, feeder units, tap-off units, bends, tees, etc. Busbar trunking system includes busduct system.
* ‘cable channel’ means an enclosure situated above or in the ground ventilated or closed, and having dimensions which do not permit the access of persons but allow access to the conductors and/or cables throughout their length during and after installation. A cable channel may or may not form part of the building construction.
* material, other than conduit or cable trunking, intended for the protection of cables which are drawn-in after erection of the ducting, but which is not specifically intended to form part of a building structure.

 

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‘dead’ means at or about ↔ Zero Voltage ← and disconnected from any live system.

* Good workmanship should be used in the construction and installation of every electrical installation * All materials chosen and used in an electrical installation should be purposely designed for the intended application and should not cause harmful effects to other equipment, undue fire risk or electrical hazard . * Special consideration should be given in choosing materials purposely designed for electrical installations which are :- (i) exposed to weather, water, corrosive atmospheres or other adverse conditions ; (ii) exposed to flammable surroundings or explosive atmosphere
* Electrical protection includes the provision of isolation, protective devices and earthing facilities as well as equipotential bonding of all the exposed conductive parts and extraneous conductive parts. * Electrical equipment should be so selected and erected that its temperature in normal operation and foreseeable temperature rise during a fault cannot cause a fire. * Suitable precautions should be taken where a reduction in voltage or loss and subsequent restoration of voltage, could cause danger . “Maintenance” In the design, construction and installation of an electrical installation, consideration must be given
to its subsequent maintenance. It should be noted that electrical equipment must not only be so constructed and protected as to be suitable for the conditions under which they are required to operate, but must also be installed to be capable of being maintained, inspected and tested with due regard to safety .
* An assessment should be made of any characteristics of equipment likely to have harmful effects upon other electrical equipment or other services, or impair the supply. Those characteristics include the following :- • overvoltages : • undervoltages : • fluctuating loads : • unbalanced loads : • voltage drop ( Vd ) : • power factor : • starting currents : • harmonic currents : • d.c. feedback : • necessity for additional connection to earth :

* Isolate and Lockout: The circuit / equipment under maintenance should be isolated as far as practicable. The relevant isolator should be locked out. A suitable warning notice should be placed close to the isolator . * De-energize The circuit/equipment to be worked on should be checked to ensure that it is dead . ←←←
* CIRCUIT ARRANGEMENT : • Basic Requirements of Circuits (1) Protection (2) Control (3) Identification (4) Electrical separation for essential circuits (5) Load distribution

* Basic Requirements of Circuits “ Protection “ Each circuit should be protected by an overcurrent protective device with its operating current value closely related to the current demand of the current using equipment connected or intended to be connected to it and to the current carrying capacity of the conductor connected. This arrangement will avoid danger in the event of a fault by ensuring prompt operation of the protective device at the appropriate current value which will otherwise cause damage to the cable or the current using equipment .

* Maintenance, Inspection and Testing :- (1) Identification ←← A Must ** (2) Maintainability (3) Inspection and testing

* Provision of Earthing :- The socket outlet should be so constructed that, when inserting the plug the earth connection is made before the current carrying pins of the plug become live. When withdrawing the plug, the current carrying parts should separate before the earth contact is broken.

 

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Calculation of current, voltage , résistance and power in simple series and parallel 12 and 24 volts circuits:

All these calculations are based on Ohm’s : In an electrical circuit, a voltage of one volt will pass a current of 1 ampere across a résistance of 1Ω . Ohm.
This law can be expressed mathematically:
The résistance of a circuit can be calculated by dividing the voltage of the circuit by the amount of current : R=V/I

The power of a circuit can be calculated by multiplying the current in a circuit by the voltage : W = I.V
As an apprentice, you need to commit these two formulae to memory. They are fundamental to all electrical calculations. Once you know these two formulae, you can re-arrange them to solve various different problems.

Some worked examples:
* In a plant vehicle fitted with a 12 volts electrical system, what is the résistance of the parking lamp circuit if 2 amperes of electricity flows ? : ( R = V/I = 12/2 = 6 Ohms. < )
* In a plant vehicle fitted with a 24 volt electrical system, what is the résistance of a headlamp circuit if an ammeter inserted in the circuit reads 10 amperes when the headlamp switch is closed ? ( R = V/I = 24/10 = 2.4 Ohms. )
* In a plant vehicle fitted with a 12 volt electrical system, how much current will flow in a warning beacon circuit, if the résistance of the beacon is 4 Ohms ? ( R = V/I; I = V/R = 12/4 = 3 amperes. )
* What is the voltage of a system if 4 amperes of flows in a circuit with a résistance of 3 Ohms ? ( R = V/I; V = R.I = 3*4 = 12 volts
* What is the power of a headlamp bulb that consumes 5 amperes when connected to a 12 volts supply ?
( W = I.V = 5 * 12 = 60 Watts . )
* How much current would a 60 Watt bulb consume when connected to a 24 volt source ? ( W = I.V; I =W/V = 60/24 = 2.5 amperes.

Great care must be exercised when using electrical equipment in high earth leakage areas such as cold rooms, washing up rooms, and in medical/biological laboratories where "wet" experiments are often in progress.

The reason for a fuse blowing, or a circuit breaker tripping, must always be investigated by a competent person and replacement fuses must always be of the correct rating.

Both Classes of equipment require an insulation test. Only Class 1 can be tested for earth bond resistance.

Only those who have both the knowledge and experience to make the right judgements and decisions and the necessary skill and ability to carry them into effect should undertake work subject to this Code. A little knowledge is often sufficient to make electrical equipment function but a much higher level of knowledge and experience is usually needed to ensure safety.

Manufacturers of any electrical equipment have a legal obligation to ensure that it is safe when properly used. Provided that the correct fuse has been fitted,

Equipment to be used out of doors should be fitted with a residual current device (RCD) to provide optimum protection. It should be noted, however, that the use of RCDs cannot be regarded as replacing primary safety features