Magnetism & Electromagnetism: Answers
A basic knowledge of magnetism and electromagnetism is an essential
requirement for understanding electrical machines such as generators and
motors. The following revision exercise should help develop your understanding.
1) Complete the following statement:
In magnetism Like Poles
and Unlike Poles
Repel
Attract
2) Show the magnetic lines of force between each of the following permanent
magnet poles:
a)
b)
N
N
N
Answer: Like poles repel
S
Unlike poles attract
3) Indicate the direction of magnetic field around the following current carrying
conductors:
a)
b)
Clockwise field
Current flowing into conductor
Anti-clockwise field
Current flowing out of the conductor
4) State the rule that is associated with magnetic fields surrounding current
carrying conductors:
Answer: The Screw Rule
A woodscrew screwed into a piece of wood is turned Clockwise, if related to the
direction in which current flows in a conductor, when current flows away from the viewer
the magnetic field will be turning clockwise just as the screw is turned.
© Sparks Magazine
1
2
5) If a current carrying conductor is placed at 90º to a magnetic field what is the
effect on the conductor?
Magnetic field is directed over the current carrying conductor
S
N
Answer:
A downward force is exerted on the current carrying conductor
6) If a conductor is passed through a magnetic field in the direction shown what
is the effect on the conductor?
N
S
Answer: A voltage will be induced in the conductor, and if the conductor is part
of a circuit or closed loop then a current will flow.
7) What is the effect on the conductor in question 6 if it is moved upwards
through the magnetic field?
Answer: Current will flow in the opposite direction.
© Sparks Magazine
8) State two hand rules that can be used to determine the direction of force or
current flow in a conductor when moved through a magnetic field.
Answer:
i) Fleming’s Left Hand Rule is used to indicate direction of force on a current
carrying conductor when lying in a magnetic field.
ii) Fleming’s Right Hand Rule is used to indicate direction of current flow in a
conductor when it is moved through a magnetic field.
© Sparks Magazine
Defining terms: Answers. (Terms in bold are defined elsewhere in the list)
The correct use of terms is an important part of your training to be an electrician.
Try providing your own definition to the following terms.
i) Earthing
The connection of the exposed conductive parts of an electrical installation to
the main earthing terminal of the installation
ii) Equipotential bonding
The electrical connection that maintains exposed conductive parts and
extraneous conductive parts to earth
iii) Exposed-conductive-part
A conductive, (metallic) part of an item of electrical equipment which can be
touched and which is not normally be live but could become live in a line to earth
fault condition
iv) Extraneous-conductive-part
That part of an electrical installation that may become live in the event of a fault but
which is not part of the electrical installation, for example a metal radiator on a
central heating system
v) Main earthing terminal
The terminal or bar provided for the connection and termination of protective
conductors, protective bonding conductors and conductors for functional earthing,
to the actual means of earthing
vi) Protective conductor
A conductor used for the protection against electric shock, and intended for
connecting together exposed and extraneous conductive parts, the main
earthing terminal and the earthed point of the power supply source which may be
an earth electrode at the premises
© Sparks Magazine
Look at the Key and then add the correct labels to the following block diagram
D
A
A
G
G
G
B
C
C
H
F
E
Key:
A: Class I electrical equipment, e.g metal-enclosed washing
machine, electric motor, any equipment requiring a circuitprotective-conductor
B: Main earthing terminal, (MET) located in the Consumers
Control Unit or CCU
C: Extraneous conductive part, e.g. water pipe or gas pipe
D: portable class I equipment
E: Consumers means of earthing
F: Earthing conductor
G: Circuit protective conductor
© Sparks Magazine
Mechanics:
The electrical engineer may be required to use his / her mechanical skills to
overcome various problems such as finding a suitable method of securing a trunking
or cable tray to a wall or suspended from a roof structure.
The following problems introduce you to basic mechanics:
1) Identify the direction in which the following levers will move when pressure is
applied in the direction of the arrow.
2) Identify the direction in which the final wheel will turn.
3) Indicate the direction of rotation for each of the pulleys shown below
4) Show the resulting direction of movement for the following crank levers
© Sparks Magazine © Sparks Magazine Health & Safety Quiz Part 1:
Use the letter and number codes to match the abbreviation with the correct
meaning:
1
HASAWA
A
Health and Safety Executive
2
CDM
B
Mobile Elevating Work Platform
3
PUWER
C
Reporting of Injuries, Diseases
and Dangerous Occurrences
Regulations
4
HSE
D
Personal Protective Equipment
5
RIDDOR
E
Construction (Design and
Management) Regulations
6
COSHH
F
Approved Codes of Practice
7
MEWP
G
Provision and Use of Work
Equipment Regulations
8
PPE
H
Respiratory Protective
Equipment
9
RPE
I
Control of Substances
Hazardous to Health
10
ACoP
J
Health & Safety at Work Act
1974
© Sparks Magazine
Number
Answer
letter
1
2
3
4
5
6
7
8
9
10
J
E
G
A
C
I
B
D
H
F
Check
ü
© Sparks Magazine
Electrical supply systems: Selecting data using the IET On-Site Guide
(Covering Outcome 3 of Unit 304 ‘Understand principles for selecting cables and circuit protective
devices’ (Level 3 NVQ Diploma in Installing Electro-Technical Systems and Equipment 2357-13 / 91)
For each of the following state the specific function of the components:
1) Distributor’s cut-out
The distributors cut-­‐out provides fault – current protection of the supply cables, ‘tails’ from the cut-­‐out to the electricity meter and from the meter to the consumers unit main switch. Removal of the fuse by an authorised person will result in isolation of the premises 2) Electricity meter
The means by which electrical energy used by the consumer can be accurately measured and the consumer be billed for the energy consumed. 3) Consumers Control Unit
The CCU is the point in an electrical installation where final circuits are connected to the electricity supply and protected from overcurrent by the use of suitably rated fuses or circuit-­‐breakers or RCBO’s. 4) Electricity isolator switch
An electricity isolator switch is a means of isolating the power supply to an electrical installation. The isolator will be located between the meter and the consumer’s unit. The isolator will be double-­‐pole and will allow the supply to the installation to be isolated without the need to withdraw the distributor’s cut-­‐out fuse. 5) Main earthing Terminal
The MET is the point at which all circuit protective conductors; main and supplementary bonding conductors and other protective conductors are terminated. The MET is normally located in the CCU or distribution board. © Sparks Magazine Three-phase alternating current: Answers
(Covering Unit 309 ‘Understand the electrical principles associated with the design, building,
installation and maintenance of electrical equipment and systems’ (Level 3 NVQ Diploma in
Installing Electro-Technical Systems and Equipment 2357-13 / 91 or EAL equivalent)
The following questions concern three-phase AC supply systems.
Question 1: Complete the missing words and symbols on the following diagram
U1 or BLACK phase
230 V
Neutral 0 Volts
230 V
U2 or BROWN phase
400 V
U3 or GREY phase
Sometimes the symbols: L1, L2 and L3 rather than U1, U2 and U3 are used to
identify phases.
Question 2: Complete the missing word or words in the following statements
a) The diagram in question 1 shows the Secondary windings of a three-phase
transformer.
b) The voltage measured between a phase and the neutral conductor is called
the Phase Voltage and is identified by the symbol Up or Vp
c) The voltage measured between any pair of phases is called the Line Voltage
and is identified by the symbol UL or VL
© Sparks Magazine
Question 3:
a) State the formula used to convert line voltage to a phase voltage in a threephase system.
Up or Vp = UL or VL ÷ √3
b) The voltage measured between two lines of a three-phase supply system is
400V, what is the voltage measured between one line and neutral of the same
supply?
Up = UL ÷ √3
Up = 400 ÷ √3
Up = 230V
c) The phase voltage of a three-phase system is 240V, what is the line voltage of
the same system?
Up = UL ÷ √3
Rearrange the formula so that UL = √3 x Up
UL = √3 x Up
UL = √3 x 240
UL = 415.68V
__________________________________________
© Sparks Magazine
Practice multiple-choice questions for Environmental Legislation, Working
Practices & Principles of Environmental Technology Systems:
(Covering Unit 302 ‘Understanding environmental legislation, working practices and the
principles of environmental technology systems…’ (Level 3 NVQ Diploma in Installing ElectroTechnical Systems and Equipment 2357-13 / 91 or EAL equivalent)
Attempt ALL questions:
No.
1
a
b
c
d
No.
2
a
b
c
d
No.
3
a
b
c
d
No.
4
a
b
c
d
No.
5
a
b
c
d
Question.
The disposal of transformer oil with general landfill waste
material may result in the contamination of:
Atmosphere and water courses
Land and water courses
Air and water
Land and atmosphere
Answer
Question.
One factor that determines whether a product is classified as
hazardous waste is if it is:
Non-combustible
Corrosive
Fine particles
Inert
Answer
Question.
The Act that defines limits for emissions of products into the
environment is the:
Building Act
Prevention of Polluting Substances Act
Environmental Protection Act
Health & Safety at Work Act
Answer
Question.
The WEEE directive applies to electrical equipment that
operates at voltages of up to and including:
1000V a.c.
500V a.c.
230V a.c.
100V a.c.
Answer
Question.
Non- hazardous waste materials include:
Batteries
Plastic
Lead
Paper
Answer
X
X
X
X
X
© Sparks Magazine
No.
6
a
b
c
d
Question.
A material known as a ‘heavy’ metal is:
Lead
Copper
Mercury
bronze
Answer
No.
7
Question.
An essential requirement when working with PVC conduit
adhesive is to:
Keep the lid on the adhesive container at all times
Only use in outdoor locations
Ensure it is compatible with the conduit
Provide good ventilation
Answer
Question.
The electrical contractor is responsible for waste disposal and
recycling:
Up to the end of the contract
Up to the location allocated for collection from site
Until the material is collected from site
Unless stated in the contract that waste will not be an issue
Answer
Question.
In old industrial locations where rewireable fuses were used one
hazardous material likely to be found is:
Rubber
Porcelain
Lead
Asbestos
Answer
No.
10
a
b
c
d
Question.
Waste generated on a construction site can be reduced by:
Using sub-contractors
Working quickly and efficiently
Careful planning
Removing waste to a designated storage area
Answer
No.
11
Question.
An serious oil leak on site that is likely to have an impact on the
environment should be reported:
Immediately
In writing
At the next site meeting
After attempts to reduce the impact have been taken
Answer
a
b
c
d
No
8
a
b
c
d
No.
9
a
b
c
d
a
b
c
d
X
X
X
X
X
X
© Sparks Magazine
No.
12
a
b
c
d
No.
13
a
b
c
d
No.
14
a
b
c
d
No.
15
a
b
c
d
Question.
To reduce the impact on the environment caused by fire one
method is to install cables with:
PVC insulation
Low smoke and fume insulation
Silicon rubber sheath
Polyethylene sheath
Answer
Question.
To ensure the generation of waste material is kept to a minimum
one action to be taken is:
Store all products safely so they are not damaged and become
waste products
Delivered to site as required
Stored on site until needed
Delivered at the start of the contract
Answer
Question.
The output power from a solar PV installation can be connected
to the public power supply through a:
Static converter
Rotary converter
Static inverter
Rotary inverter
Answer
Question.
Answer
In a hydro generator system the water pressure is used to drive
a:
Generator
Alternator
Motor
turbine
X
X
X
X
No.
16
a
b
c
d
Question.
The most common type of wind turbine in use today is the:
Vertical shaft type
Split shaft type
Horizontal shaft type
Dynamic shaft type
Answer
No.
17
a
b
c
d
Question.
One disadvantage of a solar heating system is the:
Low running cost
High emission
Renewable energy
Long payback time
Answer
X
© Sparks Magazine
No.
18
a
b
c
d
Question.
The output of a ground source heat pump is suitable for:
Electricity generation
Under-floor heating
Hot water
General wet type central heating systems
Answer
No.
19
a
b
c
d
Question.
Potable water is:
Clean water
Pure water
Wholesome water
Safe water
Answer
No.
20
Question.
Solar PV panel outputs are small, so to provide a useful voltage
panels are connected in:
Parallel
Series
Star
Delta
Answer
No.
21
a
b
c
d
Question.
A heat pump produces:
Electricity
Warm air
Gas
Hot water
Answer
No.
22
Answer
a
b
c
d
Question.
A heating system that derives its source fuel from landfill sites is
called:
Biomass
Bioliquid
Biogas
biofuel
No.
23
a
b
c
d
Question.
Micro CHP systems generally have an output of approximately:
1kW
2kW
5kW
10kW
Answer
a
b
c
d
X
X
X
X
X
X
© Sparks Magazine
No.
24
Question.
The installation of environmental technology systems in existing
premises requires consultation before installation, with the:
Health & Safety Executive
Local Planning Authority
Water utility company
Local residents
Answer
No.
25
a
b
c
d
Question.
Grey water can be collected from a range of sources except:
Shower
Toilet
Laundry
Bath
Answer
No.
26
a
b
c
d
Question.
Hazardous materials must be disposed of by:
Main contractor
Authorised contractor
Sub-contractor
Client
Answer
No.
27
Question.
One method of reducing power consumption for lighting is to
install lamps with a lumen per watt efficacy of:
40
30
20
10
Answer
Question.
One factor that determines whether a substance is hazardous
waste is if it is:
Flammable
Chemical
Organic
metallic
Answer
No.
29
a
b
c
d
Question.
A wind turbine converts wind energy into:
Electrical energy
Rotating mechanical energy
Kinetic energy
Heat energy
Answer
No.
30
a
Question.
A hydro-generator producing an output of 80kW is classed as a:
Mega – hydro system
Answer
a
b
c
d
a
b
c
d
No.
28
a
b
c
d
X
X
X
X
X
X
© Sparks Magazine
b
c
d
Micro-hydro system
Hydro system
Large hydro-system
X
© Sparks Magazine
Cable supports:
Electric cable can be constructed in many forms, non-sheathed, sheathed,
armoured, metal covered and so on. In respect to the way in which a cable
should be installed depends on the wiring system that has been chosen by the
designer.
Question 1:
Complete the following table:
Type of cable
Method of installation
Precautions when
installing or in use
Non – sheathed
Enclosed in conduit, trunking
or ducting
Single layer of insulation
can be damaged and so
care must be taken when
drawing-in cables to
conduit or ducting, or
laying cables in trunking
Sheathed
Clipped direct or enclosed as
necessary
Care should be taken to
ensure suitably spaced
clips are provided for the
support of the cable.
Care must be taken when
drawing – in cables to
conduit, ducting or laying
in trunking
Laid in ducting or supported
by clips, cleats, brackets,
tray, basket or ladder
Sufficient cable clips or
cleats to be used to
support surface mounted
cables.
Armoured
Flexible
Metal sheathed
Laid directly on the ground, in Care should be taken to
ducting, trunking or other
ensure cable is not in a
mechanical enclosure
position where it may be
damaged. Suitable
mechanical protection
should be provided when
cables installed at floor
level
Clipped direct or supported
on tray, basket or ladder
MICC cable in particular
should not be overworked as copper will
work-harden and crack
© Sparks Magazine
Question 2:
Complete the following table of cable clip spacings:
Cable
Horizontal spacing
Vertical spacing
300mm
400mm
300mm
400mm
450mm
N/A
900mm
1200mm
Non-armoured sheathed
Diameter: 1cm
Flat twin & earth
9mm
18mm
PVC SWA 25mm dia.
45º
cable
MICC 9.5mm dia.
dia
© Sparks Magazine
Heater circuits: 3 – Heat Switching:
Where different heat settings may be required, for example in a convector heater or electric
hot-plate, one method is to switch heating elements. If the convector heater or hot-plate is
fitted with two sets of resistance elements it will be possible to switch them either in series,
parallel or as a single resistance using a 3-heat switch.
3 – Heat switch:
Medium
Low
High
A three-heat switch will have an OFF position, a LOW, MEDIUM
and HIGH heat setting. Each position is selected by turning the
knob to the required position.
The internal switching mechanism will connect line and neutral to
specific terminals, (labelled 1, 2, 3, and 4 in the circuit diagram).
Off
More than one switch connection can be made in any position,
e.g. contact 1 may connect to contact 4 and at the same time
contact 3 may connect with contact 2, and so on
Task 1:
Complete the circuit diagram showing how the 3 – heat switch should be connected to obtain
Low, Medium and High heat settings.
Element 1 Element 2 1 2 3 4 L N Task 2:
Complete for each setting, the following information table.
Setting
Switch terminal connections
Heating element connection
arrangement
(Series, Parallel, single resistance)
Low
3 switches to 1
Series (low heat)
© Sparks Magazine Medium
3 switches to 2
Single (medium heat)
High
3 switches to 2
4 switches to 1
Parallel (high heat)
© Sparks Magazine Matching Quantity and Unit:
(Covering Unit 309 ‘Understand the electrical principles associated with the design, building,
installation and maintenance of electrical equipment and systems’ (Level 3 NVQ Diploma in Installing
Electro-Technical Systems and Equipment 2357-13 / 91)
The following terms all relate to electrical Quantity and its corresponding Unit. Try to
match the following quantities to their respective units.
Quantity
Unit
1
Frequency
A
Farad
2
Capacitance
B
Newton
3
Energy
C
Ohm
4
Force
D
Newton-metre
5
Charge
E
Weber
6
Reactance
F
Hertz
7
Torque
G
Henry
8
Resistivity
H
Ohm / metre
9
Magnetic flux
I
Joule
10
Mutual Inductance
J
Coulomb
© Sparks Magazine Solutions:
1–
F
2-
A
3-
I
4–
B
5-
J
6-
C
7–
D
8-
H
9-
E
10 -
G
© Sparks Magazine Cable capacities of conduit and trunking:
For these questions refer to the IET On-Site Guide BS7671:2008 (2011)
Appendix E:
If thermosetting cables are installed in the same trunking or conduit as
thermoplastic insulated cables, the conductor operating temperature of any cable
must not exceed that stated for the thermoplastic insulated cables
Question 1a:
What is the difference between thermosetting and thermoplastic PVC insulation?
PVC (Poly Vinyl chloride), insulation is thermoplastic
XLPE (Cross linked Poly Ethylene) is thermosetting plastic.
Notes:
PVC and XLPE are two plastic materials used as insulation for electric cables.
XLPE has more tolerance of high temperature and will not deteriorate until it reaches
90ºC. Even above this temperature it will not instantly be destroyed, but the overall life
of the cable will be reduced.
PVC can withstand temperatures only up to 70ºC.
Question 1b:
Appendix E concerns cable capacities for three specific cases:
i.
Straight runs of conduit not exceeding 3 m in length
ii.
Straight runs of conduit exceeding 3 m in length, or in runs of any length
with bends or sets
iii.
Trunking
Question 2:
Find and state the cable factors for each of the following cable types and
installation methods:
i.
2.5 mm2 PVC insulated cable with solid conductors installed in a straight
length of galvanised steel conduit. 39 Table E1
ii.
10 mm2 PVC insulated cables with stranded conductors installed in a 2m
length of galvanised steel conduit. 146 Table E1
iii.
10 mm2 PVC insulated cables with stranded conductors installed in a 4m
length of galvanised steel conduit.
105 Table E3
© Sparks Magazine
Question 3:
Determine a suitable size of conduit to accommodate the following sized cables
in a 2.8 m straight run of steel conduit:
3 x 1.5 mm2 (stranded conductors)
3 x 4.0 mm2
1 x 10.0 mm2
2 x 16.0 mm2
•
•
•
•
Answer:
Use Tables E1 and E2 to find a suitable size of conduit.
•
•
•
•
Cable factor (Table E1) Total cable factor
3 x 1.5 mm
27
27 x 3 = 81
2
3 x 4.0 mm
58
58 x 3 = 174
1 x 10.0 mm2
146
= 146
2
2 x 16.0 mm
202
202 x 2 = 404
Total factor = 805
2
From Table E2 a total factor of 805 is not listed so therefore the next rating up
must be chosen, in this case the next factor is 1400 for a 32 mm Dia. conduit
Question 4:
Determine the number of 25 mm2 cables that can be installed in a 25 mm Dia.
conduit.
Answer:
From Table E2 the total cable factor for a 25 mm conduit is 800
Next find the cable factor for a 25 mm2 cable. Table E1 gives this as 385
Finally divide 800 by 385 and the answer is: 2 cables
© Sparks Magazine
Question 5: The next questions concern cables installed in a steel conduit that is
5 m in length with one 90º bend and two off-sets.
Note: one off-set is equivalent to one bend
Question 5:
Determine a suitable size of conduit to accommodate the following sized cables
in the conduit:
•
•
•
3 x 1.5 mm2 (solid conductors)
3 x 4.0 mm2 (solid conductors)
2 x 10.0 mm2 (stranded conductors)
Answer:
From Table E3 the following cable factors are obtained:
Note: cables can be stranded or solid conductor in Table E3
•
•
•
3 x 1.5 mm2
3 x 4.0 mm2
2 x 10.0 mm2
Cable factor (Table E1) Total cable factor
30
3 x 30 = 90
43
3 x 43 = 129
105
2 x 105 = 210
Total factor = 429
From Table E4 find the conduit length in the left – hand column, (5 m), read
across the row to the section headed 3 bends (i.e. one 90º and 2 off-sets), the
nearest rating is 474 and this is for a 32 mm Dia. conduit.
Question 6:
Determine the number of 16 mm2 with thermosetting insulation that can be
installed in a 50 mm x 50 mm galvanised steel trunking.
Answer:
From Table E6 for a 50 mm x 50 mm trunking the total factor is 1037
From Table E5 the cable factor for a 16 mm2 thermosetting cable is 50.3
The total number of cables that can be installed is: 1037 = 20.61 cables
50.3
© Sparks Magazine
In practice a total of 20 cables will be assumed.
Question 7:
What is the minimum inner bend radius for a 25 mm conduit?
2.5 x the outside diameter of the conduit.
2.5 x 25 = 62.5 mm
© Sparks Magazine
Electrical Instrument readings: Answers
In the following exercise match the actual value to the Standard Form for that value
No.
Value as Standard Form
Measured Unit
1
Current
2 x 102
A
0.030A
2
Voltage
33x103
B
50000Ω
3
Resistance
200 x 106
C
6000W
4
Current
30 x 10-3
D
200000000Ω
5
Resistance
50 x 103
E
6000000W
6
Power
6 x 103
F
33000V
7
Power
6 x 106
G
200A
8
Voltage
400 x 106
H
400000000V
Solutions:
1
G
5
B
2
F
6
C
3
D
7
E
© Sparks Magazine 4
A
8
H
© Sparks Magazine Illumination: Point-to-Point Lighting problems:
Revision:
Task 1: Artificial light can be produced in one of two ways:
•
By passing a current through a filament of tungsten wire
•
By passing an electric current through a gas such as mercury vapour
The passage of current through a tungsten filament causes the filament to heat-up to
about 2500ºC, at this temperature a considerable amount of heat is produced and an
amount of light. This is called ‘incandescence’
If current is made to pass through a gas or metallic vapour, such as mercury, light
will be produced. This is called ‘discharge’
Point – to - Point
Point-to Point calculations are concerned with the amount of light at the lamp, the
distance the light source, or lamp, is located away from the surface to be illuminated
and the amount of light actually falling on the surface.
Lighting terminology:
Question 1:
Define the flowing terms; show the symbol that represents the unit and / or quantity:
Term
Luminous Intensity
Definition
Unit/s
The intensity of the light at the source
lamp or luminaire
Candela (Cd)
The flow of light passing from the source
to the surface to be illuminated
Lumen (Lm)
The luminous flux density at the surface
to being illuminated
Lux (Lx)
Luminaire
A complete light fitting with control gear,
(where required), and lamp or lamps.
N/A
Luminous efficacy
This is the ratio of luminous flux emitted
by a lamp to the power, (in Watts), taken
by the lamp
Lumens per Watt
Luminous Flux
Illuminance
Symbol (I)
Symbol (Φ) or (F)
Symbol (E)
Lm / W
© Sparks Magazine The Inverse Square Law:
Light source
The illuminance on a surface produced from a
single light source reduces the further away from
the light source the surface is situated.
The actual reduction is such that it reduces by the
square of the distance as shown by the formula:
Flow of light
E = I / d²
This is called the Inverse Square Law and is
applicable to a point immediately beneath the light
source as shown
Illumination
Example:
The intensity of a light source is 300cd and is located perpendicular to a surface 4m
from it, what is the illuminance at the surface?
300 cd
E = I / d²
E = 300 / 42
4m
E = 18.75 lx
E
Question 2:
Calculate the illuminance if the surface is 5m away from the source.
300 cd
5m
© Sparks Magazine E
The Cosine Law:
Light source
The illuminance at point other than that
immediately below the light source can
be found by applying the Cosine Law.
θ
Flow of light
EB = (I / h2) x Cos θ
h
To find h use Pythagoras Theorem:
Height (d)
h = √ d2 + x2
Illumination
To find the Cosine of the angle (θ):
Cos θ = d / h
Distance (x)
Example:
EB = (I / h2) x Cos θ
To find h use Pythagoras Theorem:
h = √ d2 + x2
300 cd
h = √ 32 + 42 = √ 25
θ
h = 5m
3m
To find the Cosine of the angle (θ):
Cos θ = d / h
4m
EB
Cos θ = 3 / 5
Cos θ = 0.6
EB = (I / h2) x Cos θ
EB = (300 / 52) x 0.6
EB = 7.2 lx
© Sparks Magazine Question 3:
Calculate the illuminance at point EB
1000 cd
θ
5m
6m
Question 2: Solution:
EB
Question 3: Solution:
E = I / d²
Calculate the illuminance at point EB
E = 300 / 52
EB = (I / h2) x Cos θ
E = 12 lx
1st find h
Although the distance has only
increased by 1m the lux level has
dropped considerably
h = √ d2 + x2
h = √ 52 + 62
h = 7.8m
Next find Cos θ
Cos θ = d / h
Cos θ = 5 / 7.8
Cos θ = 0.64
Now find the illuminance at EB
EB = (I / h2) x Cos θ
EB = (1000 / 7.82) x 0.64
EB = 10.52 lx
© Sparks Magazine Revision practice for City & Guilds 2357-13 Unit 309 (or EAL equivalent)
The following multiple – choice questions relate to electric motors.
No 1
a
b
c
d
No 2
a
b
c
d
No 3
a
b
c
d
No 4
a
b
c
d
No 5
a
b
c
d
No 6
a
The terms, ‘stator’ and ‘rotor’ are normally associated with
which one of the following?
Direct current generators
Alternating current generators
Direct current motors
Alternating current motors
Answer
A motor data plate indicates that it is an eight – pole
machine, which one of the following statements is correct?
The motor has four pairs of poles
The motor can be connected to a d.c. supply
The motor needs a single-phase power supply
The motor has eight magnetic fields
Answer
Which of the following statements best describes an AC
induction motor?
The motor has an armature and field windings
The motor requires a variable power supply
The motor has a stator and a rotor
The motor always has a wound rotor
Answer
To reverse the direction of rotation of a three-phase
induction motor which of the following actions is required?
Reverse connect any two lines of the three phase supply
Reverse connect one of the stator windings
Reverse connect the rotor
Reverse connect (+) and (-) of the supply
Answer
A three-phase induction motor has 4 – poles and is
connected to a supply at 50Hz. What is the synchronous
speed of the motor?
3000 rpm
1500 rpm
1000 rpm
750 rpm
Answer
The slip speed of a single-phase induction motor is stated to
be 0.02p.u. What does this mean in real terms?
0.2% slip
Answer
X
X
X
X
X
© Sparks Magazine
b
c
d
No 7
a
b
c
d
No 8
a
b
c
d
No 9
a
b
c
d
No 10
a
b
c
d
2.0% slip
20 rpm
2 rpm
X
To increase motor torque without increasing the physical
size of the rotor, which one of the following is
recommended?
A wound rotor
A cage rotor
A double-cage rotor
A solid rotor core and copper windings
Answer
Which one of the following rules can be used to describe the
principle of electro-magnetic induction in the rotor of an ac
motor?
Flemings Left – Hand rule
Flemings Right – Hand rule
The Right – Hand grip rule
The Right - Hand grasp rule
Answer
The reason for laminating the stator and rotor cores of
induction motors is to:
Help reduce torque
Allow the escape of heat
Reduce the current induced in the rotor
Reduce eddy currents
Answer
The wound – rotor is connected to externally located
resistances by the use of?
A commutator
Split-copper conductors
Slip-Rings
Steel end rings
Answer
X
X
X
X
© Sparks Magazine
NUMBERSEARCH – British and European Standards
Try to locate the following BS and BS EN numbers in the grid. (Numbers can be
found vertically downwards and horizontally from left to right)
BS67 Ceiling rose
BS88 fuses
BS951 bonding clamp
BS1361 cartridge fuse
BS1362 fuse links
1
2
3
4
5
6
8
3
9
2
3
5
4
7
8
9
0
3
9
8
7
7
2
1
1
2
9
4
0
4
0
2
0
6
7
7
2
9
5
1
0
0
2
6
6
1
5
5
8
0
9
8
9
2
4
1
3
3
6
1
8
5
1
3
4
9
8
1
3
7
1
6
7
3
8
6
8
7
4
1
0
7
2
3
0
8
4
2
1
3
BS1363 13A sockets
BS3036 fuses
BS7671 IET Regs
BS7540 Cables
BSEN60309 sockets
6
6
4
1
3
6
3
8
2
4
1
2
7
1
6
7
6
2
2
8
1
3
2
5
0
1
6
0
9
8
3
8
2
5
7
7
1
2
7
4
8
1
3
6
2
1
4
3
1
5
7
4
3
3
0
1
0
2
4
7
6
7
1
5
0
6
1
0
0
0
8
1
BSEN61000 EMC
BSEN60898 CB’s
BSEN60601 Medical
BSEN61009 RCD
BSEN50266 fire test
3
0
9
7
3
8
6
2
0
2
4
0
9
1
7
5
9
1
6
0
1
2
0
6
7
1
7
6
0
3
6
0
2
6
3
9
2
4
6
6
5
5
0
6
7
3
6
0
8
9
8
5
4
3
3
0
4
8
0
1
2
6
1
0
0
9
5
3
2
1
0
0
4
5
3
6
5
0
1
0
2
1
7
4
2
4
9
0
1
4
BSEN61558
transformers
7
3
7
1
2
3
7
6
3
7
5
1
8
9
5
1
3
0
5
1
2
1
7
8
5
0
2
6
6
7
5
0
7
5
1
5
2
5
4
7
4
2
9
1
3
0
5
9
1
6
6
8
0
2
6
0
6
7
4
3
1
3
2
8
4
3
9
0
3
5
1
9
© Sparks Magazine
Useful Formulae for City & Guilds 2357-13 Unit 309 (or EAL equivalent)
I have often been asked for a summary of formulae needed for the science and
principles unit, here is the list that I provided for my students:
Resistance:
V = IR or U = IR
Series:
R1
I
R2
pd 1
pd2
V or U
RT = R1 + R2
I = V ÷ RT or I = U ÷ RT
pd1 = I R1
pd2 = I R2
V = pd1 + pd2
Parallel:
I1
1/RT = 1/R1 + 1/R2
Or when two resistors are in parallel use
RT = Product ÷ Sum or RT = R1 + R2 ÷ R1 x R2
IT = I1 + I2
IT = V ÷ R T
I1 = V ÷ R 1
I2 = V÷ R2
R1
IT
I2
R2
V or U
Resistivity or Conductor resistance:
R = ρL ÷ a
remember ρ (rho) is the specific resistance of the conductor material, (L) is
length of conductor in metres and (a) is the cross-sectional-area of the conductor
AC circuits:
Resistive
I
R
V
V and I are ‘in-phase’ and I = V ÷ R
Phasor diagram
V
I
Inductive
L
I
I lags V by 90° L is inductance in henry’s
Inductive reactance XL = 2πfL
V
V
I lags V by 90°
C
Capacitive
I
I leads V by 90° C is capacitance in Farads
Capacitive reactance XC = 1 ÷ 2πfC
V
© Sparks Magazine Or when C is in µF then XC = 106 ÷ 2πfC
I leads V by 90°
V
Various combinations can be made resistance and inductance, (R – L), resistance and
capacitance, (R – C), or resistance, inductance and capacitance (R – L – C), series circuits.
R – L circuit:
I
L
R
VR
VL
Note: I is common throughout
the circuit. VR will be in-phase
VR will be in-phase
I will lag VL by 90°
VS
Total impedance (Z) is found from: Z = √R2 + XL2
I = VS ÷ Z
VR = IR
VL = IXL
Phase angle θ = VR ÷ VS
VS
VL
Phase angle θ
I reference line
VR
Impedance triangle:
By taking the Phasor diagram for the R-L circuit and highlighting the sides that make up a
triangle, we can indicate the sides of the triangle that represent R, L and Z in the circuit; this
produces the Impedance Triangle
Phasor diagrams,
VS
impedance triangles
VL
and power triangles,
Z
XL
(see below), all rotate
anticlockwise from
I
their point of origin
VR
R
Phasor diagram
Impedance triangle
Power factor:
It can be seen that power factor (pf) can be determined by:
Or
pf or cos θ = VR ÷ VS from the Phasor diagram
pf = R ÷ Z from the impedance triangle
Power triangle:
If Power (P) can be found from V x I then it is possible to show that power can be calculated
from each side of the impedance triangle, BUT each power will be determined differently and
will represent a different aspect of the circuit.
© Sparks Magazine True Power in Watts is found from P = V x I and is the resistive part of the circuit (R)
Apparent power in VoltAmps (VA) = V x I x cosθ and relates to the overall impedance of the
circuit (Z)
Reactive power in VoltAmpsReactive (VAr) = V x I x tanθ
True Power W
Phase angle θ
Reactive power VAr
Apparent power VA
Note that power factor is normally lagging therefore the triangle is shown in this format with
the Apparent power below the True power
© Sparks Magazine Alternating Current (AC): Inductance
Task 1: Complete the missing words and symbols
Inductive reactance: (symbol __XL__) measured in: (unit: Ohms symbol: Ω)
Remember when a coil of wire such as a winding in a motor, transformer or choke,
(ballast), is connected to an alternating current supply, opposition to current flow is
caused by two conditions, these are?
i) ___Resistance__
ii) __ Inductance__
This combination results in the effect called: _Inductive reactance_ and will depend
on the Frequency of the a.c. supply and the Inductance of the coil.
Task 2: Label the following diagram using correct terms at (i), (ii) and (iii)
(ii - current)
(iii - inductance)
(i – applied voltage)
State the formula for inductive reactance and make a key stating the name and unit
symbol for each relevant part of the formula:
Key:
Formula:
XL = Inductive reactance
XL = 2πfL
f = frequency
L = inductance
© Sparks Magazine 1
2
Task 3: Complete the following practice questions
1) Calculate the inductive reactance of a coil of 0.23H when it is connected into
an a.c. circuit operating at 50Hz.
XL = 2πfL
XL = 2 x π x f x L
XL = 2 x 3.142 x 50 x 0.23
XL = 72.3Ω
2) A coil has 200V measured across it and a current of 1.75A flowing through it
when connected to a 50Hz a.c. supply. Determine the value, (or rating), of the
inductor, in Henries.
XL = U / I
XL = 2πfL
XL = 200 / 1.75
L = XL / 2πf
XL = 114.3 Ω
L = 114.3 / 2 x 3.142 x 50
L = 0.364 H
3) Determine the current flowing through a coil of 0.65H when connected to a
230V 60Hz a.c. supply.
XL = 2πfL
I = U / XL
XL = 2 x π x 60 x 0.65
I = 230 / 245
XL = 245 Ω
I = 0.94 A
4) Calculate the voltage across an inductor of 0.25H when the current flowing
through it is 2.65A at 48Hz.
XL = 2πfL
U = I x XL
XL = 2x 3.142 x 48 x 0.25
U = 2.65 x 75.4
© Sparks Magazine XL = 75.4 Ω
U = 199.8 V (200)
© Sparks Magazine Initial verification of electrical installations: Answers
(Covering Unit 307 ‘Understand principles, practices and legislation for the inspection, testing
commissioning and certification of electro-technical systems and equipment in buildings, structures
and the environment’ (Level 3 NVQ Diploma in Installing Electro-Technical Systems and Equipment
2357-13 / 91 and the EAL equivalent)
Task 1: complete the following questions regarding the principles, practices and
legislation for the initial verification of electrical installations.
Reference (GN) is for the IET Guidance Note 3 document
1) State two occasions when initial verification can be carried out.
Answer: during construction and on completion of the installation (page15 GN3)
2) Identify two aspects of BS7671 that initial verification will confirm.
Answer: design and construction (page15 GN3)
3) When an inspector carries out an initial verification explain why inspection and
testing can only be ‘so far as is reasonably practicable’?
Answer: It would not be possible for an inspector to confirm the correct size cable
has been installed throughout its length due to the cable being buried or enclosed
within the fabric of the building. However he/she may confirm the correct cable size
at a distribution board. (page15 GN3)
4) Identify the three BS7671 generic requirements for items to be verified during the
initial verification procedure.
Answer:
Equipment is correct type and complies with applicable British Standards, or
equivalent
The fixed installation is correctly selected and erected
The fixed installation is not visibly damaged
(page15 GN3)
5) State the correct title of the statutory document that concerns inspection and test
records, and identify the length of time records must be kept.
Answer: The Electricity at Work Regulations 1989, and in particular Regulation 4(2).
The Memorandum of guidance on the Electricity at Work Regulations (HSR25)
recommends records be kept for the lifetime of the installation. (page16 GN3)
6) Briefly explain the meaning of the term ‘Relevant criteria’ in respect to the initial
verification process.
© Sparks Magazine
Answer: The designer may have made specific requirements for a given installation;
as such the inspector will need to ask for these requirements or, if relevant, forward
a copy of the test results to the designer for verification. (page16 GN3)
1
2
7) State who is responsible for comparing and verifying inspection and test results
with relevant criteria for an installation.
Answer: The person responsible for inspecting and testing the installation. (page16
GN3)
8) State the correct titles of three certificates that apply to a new installation or
additions to an existing installation.
Answer:
Electrical Installation Certificate – multiple-signature
Electrical Installation Certificate – single-signature
Minor Electrical Installation Works Certificate
(page16 GN3)
9) List three items of required information regarding the assessment of general
characteristics for an installation.
Answer:
• Maximum demand
• Number and type of live conductors
• Type of earthing
• Nominal voltage
• Supply frequency
• Prospective short-circuit current
• Earth loop impedance Ze
• Type and rating of overcurrent protective devices
See page 17 of GN3
10) The Health and safety at Work etc… Act 1974 requires relevant information to be
available for what purpose?
Answer: safe use / inspection & testing / maintenance (page17 GN3)
_________________________________
© Sparks Magazine
Three – Phase power supplies: Star and Delta connections and formulae:
Three – phase AC motors and transformers can be connected so that the full
mains supply is connected across each winding, or across two windings.
Task 1: Complete the missing words and symbols
Star connection: One end of each winding is connected to a common point
called the star point.
Figure: 1
Star point
VP
VP
IL IP
L1
IL IP
L2
IL IP
L3
VL
VL
Voltage measured across a single winding is the phase voltage (VP), the voltage
measured between (L1, L2 or L3) is the line voltage (VL)
Current flowing in each single winding is the phase current (IP) and the current
flowing in each line conductors is the line current (IL).
Task 2:
Show the line and phase currents and voltages on Figure 1 above.
Question: 1
By using the formula shown in figure 1, complete the following:
i) Calculate the phase voltage when the line voltage is stated to be 400V
VL = √3 x VP rearrange the formula to make (Vp) the subject
VP = VL / √3
VP = 400 / 1.732
VP = 230V
ii) Three 200Ω resistances are connected in star formation. If the phase voltage
is 240V calculate the line current flowing.
IL = IP because the currents are the same in both line and phase conductors it is
only necessary to calculate the phase current only using Ohms Law in this case:
IP = VP / R
IP = 240 / 200
IP = 1.2A (IL is also 1.2A)
© Sparks Magazine
2
Task 3: Complete the missing words and symbols
Delta connection: When connected in delta, each winding is connected to a pair
of lines.
Figure 2:
L1
Formulae:
VL = VP
L2
IL = √3 x IP
L3
Voltage measured across a winding is the phase voltage (VP), the voltage
measured between the lines (L1, L2 or L3) is the line voltage (VL)
Current flowing in each winding is the phase current (IP) and the current flowing
in the line conductors is the line current (IL).
Question: 1
By using the formula shown in figure 2, complete the following:
Example:
i) Calculate the phase current when the line current is found to be 50A
IL = √3 x IP rearrange the formula to make (Ip) the subject
IP = IL / √3
IP = 50 / 1.732
IP = 28.87A
ii) Three 100Ω resistors are connected in delta. If the line voltage is 400V
calculate the phase current and line current that is expected to flow.
First determine the phase current using Ohms Law and the line voltage.
IP = VL / R
IP = 400 / 100
IP = 4 A, now determine the line current from IL = √3 x IP
IL = √3 x 4
IL = 6.93A
© Sparks Magazine
Matching Units and Quantities:
(Covering Unit 309 ‘Understand the electrical principles associated with the design, building,
installation and maintenance of electrical equipment and systems’ (Level 3 NVQ Diploma in Installing
Electro-Technical Systems and Equipment 2357-13 / 91)
The following terms all relate to the quantities and units used in electrical
engineering. Try to identify the correct quantity for each unit given.
Unit name or symbol
Quantity
1
X
A
Impedance
2
E
B
Magnetic field strength
3
Weber
C
Reactance
4
L
D
Resistivity
5
Coulomb
E
Electric field strength
6
H
F
Magnetic flux
7
Z
G
Charge
8
Rho
H
Self - inductance
9
Candela
I
Weight
10
G
J
Luminous intensity
© Sparks Magazine Solution:
1–
C
5-
G
9-
J
2–
E
6-
B
10 -
I
3–
F
7-
A
4–
H
8-
D
© Sparks Magazine