This document provides information on carrying out electrical assembly. It discusses recognizing basic electrical accessories and components, electrical safety in the workplace, and general tool descriptions. It also covers topics like electrical wiring assembly, electrical component symbol recognition, and electrical switches, supply devices, and meter symbols. The document includes examples of using engineering notation with SI prefixes to solve problems converting between units like volts, amps, ohms, and watts. It also provides examples solving circuit problems using Ohm's Law and calculating fuse ratings.

1. Certificate in
Mechatronic
Engineering 2850-107
Carrying Out Electrical
Assembly
By Steven Khoo

2. Unit 107:
Carrying out electrical assembly
 H1: Recognition of basic electric accessories.
 H2: Electrical component symbol recognition.
 H3: Electrical switches, supply, devices and meter
symbols.
 H4: General tool description.
 H5: Electrical Safety at Workplace.
 H6: Electrical Assembly Accessories.
 H7: Electrical Wiring Assembly.

3. Prefix
 Engineering notation with SI prefixes.

4. Example: Direct Sol.
 1.25 𝑘𝑉 = 1.25 × 103 = 1250 𝑉
 0.3815 𝐴 = 381.5 × 10−3 = 381.5 𝑚𝐴
 5555 𝜇𝐴 = 5555 × 10−6
= 5.555 𝑚𝐴
 5555 𝜇𝐴 = 5.555 × 10−3 = 0.005555 𝐴
 1.2 𝑀Ω = 1.2 × 106 = 1200000 Ω
 1.2 𝑀Ω = 1200 × 103 = 1200 𝑘Ω
 0.0047 𝑀𝑉 = 0.0047 × 106 = 4.7 𝑘𝑉
 0.0047 𝑀𝑉 = 4.7 𝑘𝑉 = 4700 𝑉
 0.56 𝑘Ω = 0.56 × 103 = 560 Ω

5. Exercise: Prefix
1. 2.35kV = __________mV
2. 1.56MΩ = _________Ω
3. 3.33pA = __________µA
4. 6.5V = __________MV
5. 10,000µA = _________A
6. 0.56GΩ = _________kΩ
7. 2.6TV = __________kV
8. 0.001,234,5kA = __________mA

6. Exercise: Prefix (Sol)
1. 2.35kV = 2,350V = 2,350,000mV
2. 1.56MΩ = 1,560kΩ = 1,560,000Ω
3. 3.33pA = 0.003,33nA = 0.000,003,33µA
4. 6.5V = 0.006,5kV = 0.000,006,5MV
5. 10,000µA = 10mA = 0.01A
6. 0.56GΩ = 560MΩ = 560,000kΩ
7. 2.6TV = 2,600GV = 2,600,000MV =
2,600,000,000kV
8. 0.001,234,5kA = 1.2345A = 1,234.5mA

7. Example 1: Mathematical Sol.
 2.35 𝑘𝑉 = ? 𝑚𝑉
 2.35 × 103
= 𝑎 × 10−3

2.35×103
1×10−3 = 𝑎
 𝑎 = 2.35 × 106
 2.35 𝑘𝑉 = 2,350,000 𝑚𝑉

8. Example 2: Mathematical Sol.
 3.33 𝑝𝐴 = ? 𝑚𝐴
 3.33 × 10−12
= 𝑎 × 10−3

3.33×10−12
1×10−3 = 𝑎
 𝑎 = 3.33 × 10−9
 𝑎 = 0.000,000,003,33
 3.33 𝑝𝐴 = 0.000,000,003,33 𝑚𝐴

9. Example 3: Mathematical Sol.
 0.56 𝐺Ω = ? 𝑘Ω
 0.56 × 109
= 𝑎 × 103

0.56×109
1×103 = 𝑎
 𝑎 = 560 × 103
 𝑎 = 560,000
 0.56 𝐺Ω = 560,000 𝑘Ω

10. Example 4: Mathematical Sol.
 0.001,234,5 𝑘𝐴 = ? 𝑚𝐴
 0.001,234,5 × 103
= 𝑎 × 10−3

0.0012345×103
1×10−3 = 𝑎
 𝑎 = 1.2345 × 103
 𝑎 = 1,234.5
 0.001,234,5 𝑘𝐴 = 1,234.5 𝑚𝐴

11. Electricity Generation
 Thermal Power Plant
 TNB’s Thermal Power Plant produces power by using
conventional steam turbine and steam generator
principally fired by coal, oil or natural gas (steam
power plant), gas-fired or diesel-fired open cycle gas
turbine generators, and gas-fired or diesel-fired
combined cycle turbine generators.
 Tuanku Jaafar Power Station, Negeri Sembilan with
1500 MW. [Gas-fired]
 Manjung Power Station, Perak with 2295 MW. [Coal-
fired]
 Gelugor Power Station, Penang with 398 MW. [Oil-
fired]

12. Electricity Generation

13. Electricity Generation
 Hydroelectric Power Plants
It requires the means of conveying water to
produce the necessary force to spin a turbine
linked to an electric generator, usually through
a conduit such as a pipeline or tunnel to a
turbine-generator which is spun by the
passing water.
Temenggor Power Station, Perak with 348
MW.
Pergau Dam, Kelantan with 600 MW.
Bakun Dam, Sarawak with 2400 MW.

14. Electricity Generation

15. Power Station in Malaysia

16. Power Station in Malaysia

17. Electricity
 Electricity always follows the path of least
resistance. Electricity always tries to find a
path back to the source.
 Measurement of electricity in kWh.

18. Ohm’s Law
 Ohm Law: the current
through a conductor between
two points is directly
proportional to the voltage
across the two points.

19. Example 1: Ohm’s Law
 A 9-volt battery supplies power to a
cordless blow dryer with a resistance of 18
Ohms. How much current is flowing thru
the blow dryer?
𝐼 =
𝑉
𝑅
=
9 𝑉
18 Ω
= 500 𝑚𝐴

20. Example 2: Ohm’s Law
 A 230-volt wall outlet supplies power to a
black light with a resistance of 4400 Ω.
How much current is flowing through the
black light?
𝐼 =
𝑉
𝑅
=
230 𝑉
4400 Ω
= 52.273 𝑚𝐴

21. Exercise: Ohm’s Law
1. Calculate the circuit current flowing in a
230 V circuit when the load is a heater
element of resistance with 75.6 Ω.
2. Calculate the circuit current flowing in a
110 V circuit when the load is a tungsten
lamp of resistance 211 Ω.
3. Calculate the circuit current flowing in a
240 V circuit when the load is an
immersion heater element of 13.67 Ω.

22. Exercise: Ohm’s Law (Sol)
1. 𝐼 =
𝑉
𝑅
=
230𝑉
75.6Ω
= 3.0423𝐴
2. 𝐼 =
𝑉
𝑅
=
110𝑉
211Ω
= 521.327𝑚𝐴
3. 𝐼 =
𝑉
𝑅
=
240𝑉
13.67Ω
= 17.5567𝐴

23. Power Triangle
 Electrical components are given a “power
rating” in watts that indicates the maximum
rate at which the component converts the
electrical power into other forms of energy
such as heat, light or motion.

24. Example 1: Power
 What is the power consumption in watts
when the current is 3 A and the voltage
supply is 240 V?
𝑃 = 𝑉𝐼 = 240 𝑉 × 3 𝐴 = 720 𝑊

25. Example 2: Power
 A hair dryer rated 2200 W being supplied
with 240 V of voltage, what is the current?
What is the minimum fuse to use?
𝑃 = 𝑉𝐼
2200 𝑊 = 240 𝑉 × 𝐼
𝐼 =
2200 𝑊
240 𝑉
= 9.167 𝐴
𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑓𝑢𝑠𝑒: 13𝐴

26. Fuse Rating
 Use the next highest fuse rating after the
calculation. For example, the calculated
fuse rating is 2.2679 A, use a 3 A fuse.
 𝐹𝑢𝑠𝑒 𝑟𝑎𝑡𝑖𝑛𝑔 =
𝑤𝑎𝑡𝑡𝑠
𝑣𝑜𝑙𝑡𝑠
× 1.25
 The power of the appliance – usually in
the appliance manual.
 The voltage (240 volts in Malaysia or UK).

27. Exercise: Fuse Rating
 The assumption in this calculation is that
there is only 3 fuse ratings available; 3A,
5A and 13A.
1. Power is 500W and the voltage is 240V.
2. Power is 2200W and the voltage is 240V.
3. Power is 900W and the voltage is 240V.
4. Power is 320W and the voltage is 240V.
 Select a suitable fuse to use.

28. Exercise: Fuse Rating (Sol)
1. 𝐹𝑢𝑠𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 =
500𝑊
240𝑉
× 1.25 = 2.6042𝐴.
Suitable fuse = 3A.
2. 𝐹𝑢𝑠𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 =
2200𝑊
240𝑉
× 1.25 = 11.4583𝐴.
Suitable fuse = 13A.
3. 𝐹𝑢𝑠𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 =
900𝑊
240𝑉
× 1.25 = 4.6875𝐴.
Suitable fuse = 5A.
4. 𝐹𝑢𝑠𝑒 𝑅𝑎𝑡𝑖𝑛𝑔 =
320𝑊
240𝑉
× 1.25 = 1.6667𝐴.
Suitable fuse = 3A.

29. Exercise: Power
1. Calculate the resistances of 110-V light bulbs
rated at the following:
a) 25 W,
b) 60 W,
c) 75 W,
d) 100 W.
2. A 250 V bulb passes a current of 0.3 A.
Calculate the power in the lamp.
3. A certain appliance uses 350 W. If it is allowed
to run continuously for 24 days, how many
kilowatt-hours of energy does it consume?

30. Exercise: Power (Sol)
1. 𝐼 =
25𝑊
110𝑉
= 227.2727𝑚𝐴. 𝑅 =
𝑉
𝐼
=
110𝑉
227.2727𝑚𝐴
= 484Ω.
𝐼 =
60𝑊
110𝑉
= 545.4545𝑚𝐴. 𝑅 =
𝑉
𝐼
=
110𝑉
545.4545𝑚𝐴
= 201.6667Ω.
𝐼 =
75𝑊
110𝑉
= 681.8182𝑚𝐴. 𝑅 =
𝑉
𝐼
=
110𝑉
681.8182𝑚𝐴
= 161.3333Ω.
𝐼 =
100𝑊
110𝑉
= 909.0909𝑚𝐴. 𝑅 =
𝑉
𝐼
=
110𝑉
909.0909𝑚𝐴
= 121Ω.
2. 𝑃 = 𝑉𝐼 = 250 × 0.3 = 75𝑊
3. 𝑇𝑜𝑡𝑎𝑙 𝑘𝑖𝑙𝑜𝑤𝑎𝑡𝑡 = 0.35 𝑘𝑊
𝑇𝑜𝑡𝑎𝑙 ℎ𝑜𝑢𝑟 = 24 × 24 = 576 ℎ𝑜𝑢𝑟𝑠
𝑇𝑜𝑡𝑎𝑙 𝑘𝑤ℎ = 201.6 𝑘𝑤ℎ

31. Kilo Watt Hour (kWh)
 A house uses 10 light bulbs and each is 100
watts. These light bulbs has been used for 196
hours in a month. TNB is charging 15 cents for
every kilowatt hour. What is the bill for that
month?
 𝑇𝑜𝑡𝑎𝑙 𝑤𝑎𝑡𝑡 = 10 × 100𝑊 = 1000𝑊.
 𝑇𝑜𝑡𝑎𝑙 𝑘𝑖𝑙𝑜𝑤𝑎𝑡𝑡 = 1 𝑘𝑊.
 𝑇𝑜𝑡𝑎𝑙 𝑘𝑊ℎ = 1𝑘𝑊 × 196 ℎ𝑜𝑢𝑟𝑠 = 196 𝑘𝑊ℎ.
 𝐶𝑜𝑠𝑡 = 196𝑘𝑊ℎ × 0.15 = 𝑅𝑀29.4.

32. Exercise: kWh
1. At the end of a 14 day period, your utility
bill shows that you have used 18 kWh.
What is your average daily power?
2. If you used an electric blender rated 400 W
of power for 30 hours, how many kWh?
3. A hairdryer rated with 2200W and being
used 2 hours per day for 30 days. TNB is
charging 25 cent for every kilowatt hour.
How much the bill will cost in a year?

33. Exercise: kWh (Sol)
1. Average one day usage:
18𝑘𝑊ℎ
14
= 1.2857𝑘𝑊ℎ
2. 400𝑊 × 30ℎ𝑟 = 12𝑘𝑊ℎ
3. 𝑇𝑜𝑡𝑎𝑙 ℎ𝑜𝑢𝑟𝑠 = 60ℎ𝑜𝑢𝑟𝑠
𝑇𝑜𝑡𝑎𝑙 𝑘𝑊ℎ = 2200𝑊 × 60ℎ𝑟 = 132𝑘𝑊ℎ
132𝑘𝑊ℎ × 25𝑐𝑒𝑛𝑡 = 𝑅𝑀33?

34. Current Tariff Calculation:
Domestic
 First 200 kWh (1-200 kWh)/month: 200 × 0.218
 First 100 kWh (201-300 kWh)/month: 100 × 0.334
 First 300 kWh (301-600 kWh)/month: 300 × 0.516
 First 300 kWh (601-900 kWh)/month: 300 × 0.546
 First 901 kWh onwards/month: n × 0.571

35. TNB Bill

36. Current Tariff Calculation:
Commercial
 Tariff B – Low Voltage Commercial Tariff
 First 200 kWh (1-200 kWh)/month: 200 × 0.435
 For the next kWh (201 kWh onwards)/month:
𝑛 × 0.509

37. Socket Overload
 Good practice: use at most
80% of any given power
outlet’s current.
 UK sockets are not actually
rated as 13A but they are
typically on a ‘ring main’
which is fed from a 32A
breaker or fuse.
 The plug is limited to 13A
but most PC rating is 3A or
possibly 5A fuse.

38. Socket Overload
 UK uses 230V (±10%) and highest rated fuse up
to 13A for a plug including multiway adapters.
 𝑃 = 13 × 230 = 2990𝑊, can connect devices
which use up to 2990W to a single UK mains
socket without overloading it.

39. Socket Overload Calculator
 Are you overloading your sockets and
putting your home at risk of fire? Plug in
some appliances to find out!
 http://www.twothirtyvolts.org.uk/socket-
overload/?hdpi=1#
 Plugs for appliances rated up to about
700W should have a 3A fuse.
 Maximum load for a single socket is 13
amps (13 A) or about 3000 watts (3 kW).

40. Type G
 Type G is mainly used in the United Kingdom,
Ireland, Cyprus, Malta, Malaysia, Singapore and
Hong Kong.
 British Standard BS 1362 requires use of a
three-wire grounded and fused plug for all
connections to the power mains.
 Two-wire class II appliances are not earthed and
often have a plastic grounding pin which only
serves to open the shutters of the outlet.

41. Live (L), Neutral (N), Earth (E)
 A lot of mains powered appliances need
three wires to work safely.

42. Live (L), Neutral (N), Earth (E)
 Only two of the
wires are used
when the
appliance works
properly. These
are the live
(brown) and the
neutral (blue)
wires.

43. Live (L), Neutral (N), Earth (E)
 The live wire carries current to the
appliance at a high voltage.
 The neutral wire completes the circuit and
carries current away from the appliance.
 The third wire, called the earth wire
(green/yellow) is a safety wire and connects
the metal case of the appliance to the earth.
 This stops a fault making the case of the
appliance live.

44. Live (L), Neutral (N), Earth (E)
 If a fault occurs where
the live wire connects
to the case, the earth
wire allows a large
current to flow through
the live and earth
wires. This overheats
the fuse which melts
and breaks the circuit.