Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET (0963959194)�

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COURSE CONTENT

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET (0963959194)�

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TEXTBOOK AND REFERENCES

TextBook:

• J. O. Bird, Electrical and electronic principles and technology, Sixth edition. London ; New York: Routledge, Taylor & Francis Group, 2017.

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References:

• Equipment Maintenance Fundamentals Field Manual. Copyright Intel Corporation 2008

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• Mulukutla S. Sarma, Introduction to Electrical Engineering, Oxford University Press, 2001.

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• M.E. El-Hawary, Introduction to Electrical Power Systems, A John Wiley & Sons, 2008.

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Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET (0963959194)�

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ASSESSMENT METHODS

DOCUMENT LINK

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1

UNITS ASSOCIATED WITH BASIC ELECTRICAL QUANTITIES

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1

SI UNITS

The system of units used in engineering and science is the Système Internationale d’Unités (International system of units), usually abbreviated to SI units, and is based on the metric system. This was introduced in 1960 and is now adopted by the majority of countries as the official system of measurement. �

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1

SI UNITS

Why it is important to understand: Units associated with basic electrical quantities�The relationship between quantities can be written using words or symbols (letters), but symbols are normally used because they are much shorter; for example V is used for voltage, I for current and R for resistance. Some of the units have a convenient size for electronics, but most are either too large or too small to be used directly so they are used with prefixes. The prefixes make the unit larger or smaller by the value shown; for example, 25 mA is read as 25 milliamperes and means 25 × 10-3A = 25 × 0.001A = 0.025 A.�Knowledge of this chapter is essential for future studies and provides the basis of electrical units and prefixes; some simple calculations help understanding. �

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1

SI UNITS

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1

Which of the following systems is the most widely used in science and trade? �A. US System�B. UK System�C. Metric and SI�D. None of the above �

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1

CHARGE

The unit of charge is the coulomb∗ (C), where one coulomb is one ampere second (1coulomb = 6.24×10^18 electrons). The coulomb is defined as the quantity of electricity which flows past a given point in an electric circuit when a current of one ampere∗ is maintained for�one second. Thus,

charge, in coulombs Q=It

where I is the current in amperes and t is the time in seconds. �

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CHAPTER 1

CHARGE

I(A)

t(s)

Q(C)

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CHAPTER 1

FORCE

The unit of force is the newton∗ (N), where one newton is one kilogram metre per second squared. The newton is defined as the force which, when applied to a mass of one kilogram, gives it an acceleration of one metre per second squared. Thus,

force, in newtons F=ma

where m is the mass in kilograms and a is the acceleration in metres per second squared. Gravitational force, or weight, is mg, where g = 9.81m/s2. �

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1

FORCE

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CHAPTER 1

WORK

The unit of work or energy is the joule∗ (J), where one joule is one newton metre. The joule is defined as the work done or energy transferred when a force of one newton is exerted through a distance of one metre in the direction of the force. Thus,�work done on a body, in joules, W=Fs�where F is the force in newtons and s is the distance in metres moved by the body in the direction of the force.�Energy is the capacity for doing work.

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CHAPTER 1

POWER

The unit of power is the watt∗ (W), where one watt is one joule per second. Power is defined as the rate of doing work or transferring energy. Thus,

power, in watts P=W/t

where W is the work done or energy transferred, in joules, and t is the time, in seconds. Thus,

energy, in joules W =Pt �

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CHAPTER 1

POWER

EXERCISE:

An electromagnet exerts a force of 12 N and moves a soft iron armature through a distance of 1.5 cm in 40 ms. Find the power consumed. (4.5 W)

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CHAPTER 1

Electrical Potential and E.M.F.

A change in electric potential between two points

in an electric circuit is called a potential difference. The electromotive force (e.m.f.) provided by a source of energy such as a battery or a generator is measured in volts.

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CHAPTER 1

Resistance and Conductance

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CHAPTER 1

EXERCISE:

An e.m.f. of 250 V is connected across a resistance and the current flowing through the resistance is 4 A. What is the power developed? (1 kW)

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CHAPTER 1

Electrical power and energy

When a direct current of I amperes is flowing in an electric circuit and the voltage across the circuit is V volts, then

�power, in watts P=VI�Electrical energy=Power×time�=VIt joules

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Although the unit of energy is the joule, when dealing with large amounts of energy, the unit used is the kilowatt hour (kWh) where

1kWh = 1000watt hour�= 1000 × 3600watt seconds or joules�= 3600000J �

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CHAPTER 1

EXERCISE:

1. 450 J of energy are converted into heat in 1 minute. What power is dissipated? (7.5 W)

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b) A battery of e.m.f. 12 V supplies a current of 5 A for 2 minutes. How much energy is supplied in this time? (7200 J or 7.2 kJ)

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P(W)

I(A)

E(J, Ws)

t(s)

V(V)

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CHAPTER 1 - EXERCISE

1. A resistance of 50 kΩ has a conductance of:

[a] 20 S

[b] 0.02 S

[c] 0.02 mS

[d] 20 kS

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2. Which of the following statements is incorrect?

[a] 1 N = 1 kg m / s²

[b] 1 V = 1 J / C

[c] 30 mA = 0.03A

[d] 1 J = 1 N / m

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3. The power dissipated by a resistor of 10 Ω when a current of 2 A passes through it is:

[a] 0.4 W

[b] 20 W

[c] 40 W

[d] 200 W

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CHAPTER 1 - EXERCISE

4. A mass of 1200 g is accelerated at 200 cm/s² by a force. The value of the force required is:

[a] 2.4 N

[b] 2,400 N

[c] 240 kN

[d] 0.24 N

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5. A charge of 240 C is transferred in 2 minutes. The current flowing is:

[a] 120 A

[b] 480 A

[c] 2 A

[d] 8 A

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6. A current of 2 A flows for 10 h through a 100 Ω resistor. The energy consumed by the resistor is:

[a] 0.5 kWh

[b] 4 kWh

[c] 2 kWh

[d] 0.02 kWh

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CHAPTER 1 - EXERCISE

7. The unit of quantity of electricity is the:

[a] volt

[b] coulomb

[c] ohm

[d] joule

8. Electromotive force is provided by:

[a] resistances

[b] a conducting path

[c] an electric current

[d] an electrical supply source

9. The coulomb is a unit of:

[a] power

[b] voltage

[c] energy

[d] quantity of electricity

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CHAPTER 1 - EXERCISE

10. In order that work may be done:

[a] a supply of energy is required

[b] the circuit must have a switch

[c] coal must be burnt

[d] two wires are necessary

11. The ohm is the unit of:

[a] charge

[b] resistance

[c] power

[d] current

12. The unit of current is the:

[a] volt

[b] coulomb

[c] joule

[d] ampere

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CHAPTER 1 - EXERCISE

1. What force is required to give a mass of 20 kg an acceleration of 30 m/s² ? (600 N)
2. Find the accelerating force when a car having a mass of 1.7 Mg increases its speed with a constant acceleration of 3m/s^2 . (5.1 kN)
3. A force of 40 N accelerates a mass at 5 m/s2. Determine the mass. (8 kg)
4. Determine the force acting downwards on a mass of 1500 g suspended on a string (14.72 N)
5. A force of 4 N moves an object 200 cm in the direction of the force. What amount of work is done ? (8 J)
6. A force of 2.5 kN is required to lift a load. How much work is done if the load is lifted through 500 cm? (12.5 kJ)
7. An electromagnet exerts a force of 12 N and moves a soft iron armature through a distance of 1.5 cm in 40 ms. Find the power consumed. (4.5 W)
8. A mass of 500 kg is raised to a height of 6 m in 30 s. Find (a) the work done, and (b) the power developed (29.43 kNm, 981 W)
9. What quantity of electricity is carried 6.24x10^21 by electrons? (1000 C)
10. In what time would a current of 1 A transfer a charge of 30 C? (30 s)
11. A current of 3 A flows for 5 minutes. What charge is transferred? (900 C)
12. How long must a current of 0.1 A flow so as to transfer a charge of 30 C? (5 minutes)

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Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 1 - EXERCISE

13. Rewrite the following as indicated:

  (a) 1 000 pF = ……… nF (b) 0.02 μF = ………. pF

(c) 5 000 kHz = ……… MHz (d) 47 kΩ = …….. MΩ

(e) 0.32 mA = ……. μA

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PRESENTATION TOPIC:

1. Describe the history of the discovery of electric current.
2. Who is Joule? present his contributions to science.

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 2

AN INTRODUCTION TO ELECTRIC CIRCUITS

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CHAPTER 2

Electrical/electronic system block diagrams

An electrical/electronic system is a group of components connected together to perform a desired function.

A sub-system is a part of a system which performs an identified function within the whole system; the amplifier in Fig. 4.1 is an example of a sub-system.�A component or element is usually the simplest part of a system which has a specific and well-defined function – for example, the microphone in Fig. 4.1. �

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CHAPTER 2

Standard symbols for electrical components

Symbols are used for components in electrical circuit diagrams

(autocad electrical drawing)

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CHAPTER 2

IEC electrical symbols

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CHAPTER 2

Electric current and quantity of electricity

All atoms consist of protons, neutrons and electrons.�The protons, which have positive electrical charges, and the neutrons, which have no electrical charge, are contained within the nucleus. Removed from the nucleus are minute negatively charged particles called electrons. Atoms of different materials differ from one another by having different numbers of protons, neutrons and electrons. An equal number of protons and electrons exist within an atom and it is said to be electrically balanced, as the positive and negative charges cancel each other out.

All atoms are bound together by powerful forces of attraction existing between the nucleus and its electrons. Electrons in the outer shell of an atom, however, are attracted to their nucleus less powerfully than are electrons whose shells are nearer the nucleus.

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CHAPTER 2

Electric current and quantity of electricity

It is possible for an atom to lose an electron; the atom, which is now called an ion, is not now electrically balanced, but is positively charged and is thus able to attract an electron to itself from another atom. Electrons that move from one atom to another are called free electrons and such random motion can continue indefinitely. However, if an electric pressure or voltage is applied across any material there is a tendency for electrons to move in a particular direction. This movement of free electrons, known as drift, constitutes an electric current flow. Thus current is the rate of movement of charge.

�Conductors are materials that contain electrons that are loosely connected to the nucleus and can easily move through the material from one atom to another.

Insulators are materials whose electrons are held firmly to their nucleus.

The unit used to measure the quantity of electrical charge Q is called the coulomb∗ (where 1 coulomb=6.24×10^18 electrons). If the drift of electrons in a conductor takes place at the rate of one coulomb per second the resulting current is said to be a current of one ampere.

1ampere = 1coulomb per second or 1A = 1C/s

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CHAPTER 2

Electric current and quantity of electricity

Now try the following Practice Exercise �1. In what time would a current of 10A transfer a charge of 50C?�2. A current of 6A flows for 10 minutes. What charge is transferred?�3. How long must a current of 100mA flow so as to transfer a charge of 80C? �

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CHAPTER 2

Potential difference and resistance

For a continuous current to flow between two points in a circuit a potential difference (p.d.) or voltage, V, is required between them; a complete conducting path is necessary to and from the source of electrical energy.�The unit of p.d. is the volt, V (named in honour of the Italian physicist Volta∗. �The flow of electric current is subject to friction. This friction, or opposition, is called resistance, R, and is the property of a conductor that limits current. The unit of resistance is the ohm; 1ohm is defined as the resistance which will have a current of 1ampere flowing through it when 1volt is connected across it.� �

Figure 4.5 �

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CHAPTER 2

Basic electrical measuring instruments

An ammeter is an instrument used to measure current and must be connected in series with the circuit. Fig. 4.5 shows an ammeter connected in series with the lamp to measure the current flowing through it. Since all the current in the circuit passes through the ammeter it must have a very low resistance.�A voltmeter is an instrument used to measure p.d. and must be connected in parallel with the part of the circuit whose p.d. is required. In Fig. 4.5, a voltmeter is connected in parallel with the lamp to measure the p.d. across it. To avoid a significant current flowing through it a voltmeter must have a very high resistance.�An ohmmeter is an instrument for measuring resistance.�A multimeter, or universal instrument, may be used to measure voltage, current and resistance. An ‘Avometer’ and ‘Fluke’ are typical examples. �

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CHAPTER 2

The oscilloscope may be used to observe waveforms and to measure voltages and currents. The display of an oscilloscope involves a spot of light moving across a screen. The amount by which the spot is deflected from its initial position depends on the p.d. applied to the terminals of the oscilloscope and the range selected.�The displacement is calibrated in ‘volts per cm’. For example, if the spot is deflected 3cm and the volts/cm switch is on 10V/cm then the magnitude of the p.d. is 3cm×10V/cm, i.e. 30V. �

Basic electrical measuring instruments

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CHAPTER 2

A wattmeter is an instrument for the measurement of power in an electrical circuit. �Continuity testing is the measurement of the resistance of a

cable to discover if the cable is continuous, i.e. that it has no

breaks or high-resistance joints.

Insulation resistance testing is the measurement of resistance

of the insulation between cables, individual cables to earth or

metal plugs and sockets, and so on.�An insulation resistance in excess of 1MOhm is normally

acceptable. �A tachometer is an instrument that indicates the speed, usually

in revolutions per minute, at which an engine shaft is rotating.

Basic electrical measuring instruments

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CHAPTER 2

Ohm’s law∗ states that the current I flowing in a circuit is directly proportional to the applied voltage V and inversely proportional to the resistance R, provided the temperature remains constant. Thus, ��

Ohm’s law

Linear and non-linear devices

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CHAPTER 2

Multiples and sub-multiples

Currents, voltages and resistances can often be very large or very small. Thus multiples and sub-multiples of units are often used, as stated in chapter 3. The most common ones, with an example of each, are listed in Table 4.1.�

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CHAPTER 2

Multiples and sub-multiples

Problem 6. A 100V battery is connected across a resistor and causes a current of 5mA to flow. Determine the resistance of the resistor. If the voltage is now reduced to 25V, what will be the new value of the current flowing (20k Ohm, 1.25mA) � �

Problem 7. What is the resistance of a coil which draws a current of (a) 50mA and (b) 200μA from a 120V supply?

(2.4 k Ohm, 0.6M Ohm)�

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CHAPTER 2

Multiples and sub-multiples

(1k Ohm)

(3.2k Ohm)

Practice Exercise Ohm’s law�1. The current flowing through a heating element is 5A when a p.d. of 35V is applied across it. Find the resistance of the element.�2. A 60W electric light bulb is connected to a 240V supply. Determine (a) the current flowing in the bulb and (b) the resistance of the bulb.�3. Determine the p.d. which must be applied to a 5kOhm resistor such that a current of 6mA may flow.�4. A 20V source of e.m.f. is connected across a circuit having a resistance of 400Ohm. Calculate the current flowing. ��

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CHAPTER 2

Conductors and insulators

A conductor is a material having a low resistance which allows electric current to flow in it. All metals are conductors and some examples include copper, aluminium, brass, platinum, silver, gold and carbon.�An insulator is a material having a high resistance which does not allow electric current to flow in it. Some examples of insulators include plastic, rubber, glass, porcelain, air, paper, cork, mica, ceramics and certain oils. �

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CHAPTER 2

Electrical power and energy

Electrical power�Power P in an electrical circuit is given by the product of potential difference V and current I, as stated in chapter 3. The unit of power is the watt, W.∗

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CHAPTER 2

Electrical power and energy

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 2

Electrical power and energy

Electrical energy�Electrical energy = power × time If the power is measured in watts and the time in seconds then the unit of energy is watt-seconds or joules.∗ If the power is measured in kilowatts and the time in hours then the unit of energy is kilowatt-hours, often called the ‘unit of electricity’. The ‘electricity meter’ in the home records the number of kilowatt-hours used and is thus an energy meter. �

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CHAPTER 2

Electrical power and energy

COE = E x Price

E (kWh); Price (VND/kWh) => COE (VND)

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CHAPTER 2

Electrical power and energy

COE = E x Price

E (kWh); Price (VND/kWh) => COE (VND)

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CHAPTER 2

Electrical power and energy

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 2 - EXERCISE

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CHAPTER 2 - EXERCISE

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 2 - EXERCISE

Introduction to Specialized Engineering�Phong Nguyen Hoai – FEET �

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CHAPTER 2 - EXERCISE

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CHAPTER 2 - EXERCISE

Practice Exercise�1. The hot resistance of a 250V filament lamp is 625Ohm. Determine the current taken by the lamp and its power rating.�2. Determine the resistance of a coil connected to a 150V supply when a current of (a) 75mA, (b) 300μA flows through it.�3. Determine the resistance of an electric fire which takes a current of 12A from a 240V�supply. Find also the power rating of the fire and the energy used in 20 h.�4. Determine the power dissipated when a current of 10mA flows through an appliance�having a resistance of 8kOhm.�5. 85.5 J of energy are converted into heat in 9 s. What power is dissipated?�6. A current of 4A flows through a conductor and 10W is dissipated. What p.d. exists across the ends of the conductor �7. Find the power dissipated when: (a) a current of 5mA flows through a resistance of 20kOhm(b) a voltage of 400V is applied across a 120kOhm resistor (c) a voltage applied to a resistor is 10 kV and the current flow is 4m

8. A battery of e.m.f. 15V supplies a current of 2A for 5min. How much energy is supplied in this time?

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1. Introduction about the modern electrical measuring instruments.
2. Who is Faraday? present his contributions to science.
3. What are Overcurrent Protection Devices? Types of Overcurrent Devices
4. Introduction about insulation and the dangers of constant high current flow
5. Introduction about: Resistance variation, Batteries, Series and parallel networks
6. Introduction about: Magnetic circuits, Electromagnetism, Electromagnetic induction, Electrical measuring instruments and measurements.
7. Introduction about: Semiconductor diodes, Transistors ���� � �

CHAPTER 2 - EXERCISE

Practice Exercise�10. A p.d. of 500V is applied across the winding of an electric motor and the resistance of the winding is 50Ohm. Determine the power dissipated by the coil.�11. In a household during a particular week three 2 kW fires are used on average 25 h each and eight 100W light bulbs are used on average 35 h each. Determine the cost of electricity for the week if 1 unit of electricity costs 15 p.�12. Calculate the power dissipated by the element of an electric fire of resistance 30Ohm when a current of 10A flows in it. If the fire is on for 30 hours in a week determine the energy used. Determine also the weekly cost of energy if electricity costs 13.5p per unit. �

PRESENTATION TOPIC:

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CHAPTER 2 - TEST

1. Introduction about the modern electrical measuring instruments. (2đ)�2. Determine the resistance of an electric fire which takes a current of 12A from a 240V supply. Find also the power rating of the fire and the energy used in 20 h. (1đ)�3. Determine the power dissipated when a current of 10mA flows through an appliance having a resistance of 8kOhm. (1đ)�4. A current of 4A flows through a conductor and 10W is dissipated. What p.d. exists across the ends of the conductor. (1đ) �5. Find the power dissipated when: (a) a current of 5mA flows through a resistance of 20kOhm(b) a voltage of 400V is applied across a 120kOhm resistor (c) a voltage applied to a resistor is 10 kV and the current flow is 4mA. (1đ)

6. A battery of e.m.f. 15V supplies a current of 2A for 5min. How much energy is supplied in this time? (1đ)

7. A p.d. of 500V is applied across the winding of an electric motor and the resistance of the winding is 50Ohm. Determine the power dissipated by the coil. (1đ)�8. In a household during a particular week three 2 kW fires are used on average 25 h each and eight 100W light bulbs are used on average 35 h each. Determine the cost of electricity for the week if 1 unit of electricity costs 15 p. (1đ)�9. Calculate the power dissipated by the element of an electric fire of resistance 30Ohm when a current of 10A flows in it. If the fire is on for 30 hours in a week determine the energy used. Determine also the weekly cost of energy if electricity costs 13.5p per unit. (1đ)

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CHAPTER 2 - TEST

Determine the resistance at terminals A-B �

(3kOhm)

(22kOhm)

(5kOhm)

(12kOhm)

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CHAPTER 2 - TEST

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CHAPTER 2 - TEST

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CHAPTER 2 - TEST

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CHAPTER 2 - TEST

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CHAPTER 2 - TEST

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CHAPTER 2 - TEST

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CHAPTER 3

DC CIRCUIT THEORY

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CHAPTER 3 - INTRODUCTION

A single equivalent resistance can be found when two or more resistors are connected together in series, parallel or combinations of both, and that these circuits obey Ohm’s Law.

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However, sometimes in more complex circuits we cannot simply use Ohm’s Law alone to find the voltages or currents circulating within the circuit. For these types of calculations we need certain rules which allow us to obtain the circuit equations and for this we can use Kirchhoff’s laws.

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In addition, there are a number of circuit theorems – superposition theorem, Thévenin’s theorem, Norton’s theorem – which allow us to analyse more complex circuits. In addition, the maximum power transfer theorem enables us to determine maximum power in a d.c. circuit.

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�

An electrical/electronic engineer often needs to be able to analyse an electrical network to determine currents flowing in each branch and the voltage across each branch.

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CHAPTER 3 - KIRCHHOFF’S LAWS

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CHAPTER 3 - KIRCHHOFF’S LAWS

Applying Kirchhoff’s current law:

For junction B: 50=20+I1 ; Hence I1 =30A�For junction C: 20+15=I2 ; Hence I2 =35A

i.e. 30=I3 +120 ; Hence I3 =−90A

For junction E: I4 +I3 =15 ; i.e. I4 =15−(−90)

Hence I4 =105A

For junction F: 120=I5 +40 ; Hence I5 =80A

E1 - E2 - IR1 - IR2 - IR3 = 0

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CHAPTER 3 - KIRCHHOFF’S LAWS

step 1: draw current direction loop (clockwise)

step 2: write K2 laws

I same direction of clockwise => IR

I same direction of anti-clockwise => -IR

clockwise flow (-)E => -E

clockwise flow (+)E => E

(I*1+I*2+I*2.5+I*1.5) + (4-3-6-E)=0

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CHAPTER 3 - KIRCHHOFF’S LAWS

(I3= 2 A, I4= − 1 A, I6= 3 A)

1. I1= 4 A, I2= − 1 A, I3= 13 A
2. I1= 40 A, I2= 60 A, I3 = 120 A,

I4= 100 A, I5= − 80 A

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CHAPTER 3 - KIRCHHOFF’S LAWS

Use Kirchhoff’s laws to find the current flowing in the 6Ohm resistor of Fig. 15.14 and the power dissipated in the 4Ohm resistor.

I1

I2

I3

K1: I1-I2-I3=0

K2(a):40-5*I1-4*I2=0

K2(b): 4*I2-6*I3=0

K2(a)

K2(b)

I1= 1.259 A,I2= 0.752 A,I3= 0.153 A,I4= 1.412 A, I5= 0.599 A �

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CHAPTER 3 - GENERAL D.C. CIRCUIT THEORY

The following points involving d.c. circuit analysis need to be appreciated before proceeding with problems using Thévenin’s and Norton’s theorems:�(i) The open-circuit voltage, E, across terminals AB in Fig. 15.29 is equal to 10V, since no current flows through the 2Ohm resistor and hence no voltage drop occurs. �

(ii) The open-circuit voltage, E, across terminals AB in Fig. 15.30(a) is the same as the voltage across the 6Ohm resistor. The circuit may be redrawn as shown in Fig. 15.30(b)�E = 6*50/(6+4), by voltage division in a series circuit, i.e. E=30 �

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CHAPTER 3 - GENERAL D.C. CIRCUIT THEORY

K2 => open circuit voltage

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CHAPTER 3 - GENERAL D.C. CIRCUIT THEORY

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CHAPTER 3 - GENERAL D.C. CIRCUIT THEORY

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CHAPTER 3 - THÉVENIN’S THEOREM

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CHAPTER 3 - THÉVENIN’S THEOREM

E

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CHAPTER 3 - THÉVENIN’S THEOREM

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CHAPTER 3 - THÉVENIN’S THEOREM

(current in the 0.8 resistor = 1.5A )

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CHAPTER 3 - THÉVENIN’S THEOREM

current in the 4 resistor = 0.571 A

P = 1.304W

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CHAPTER 3 - THÉVENIN’S THEOREM

�

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CHAPTER 3 - SUPERPOSITION THEOREM

The superposition theorem states:

In any network made up of linear resistances and containing more than one source of e.m.f., the resultant current flowing in any branch is the algebraic sum of the currents that would flow in that branch if each source was considered separately, all other sources being replaced at that time by their respective internal resistances. �

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CHAPTER 3 - SUPERPOSITION THEOREM

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CHAPTER 3 - NORTON’S THEOREM

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CHAPTER 3 - NORTON’S THEOREM

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CHAPTER 3 - NORTON’S THEOREM

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CHAPTER 3 - NORTON’S THEOREM

(I = 0.571 A)

KCL: I1 + I2 – Isc = 0

KVL(a): -4 + 2*I1 = 0

KVL(b): -2 + I2 = 0

​

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CHAPTER 3 - NORTON’S THEOREM

Use Norton’s theorem to find the current in each branch of the arrangement shown.

the current in the 20 Ω resistor, I = 1.616 (A)

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CHAPTER 3 - NORTON’S THEOREM

Use Norton’s theorem to find the current in each branch of the arrangement shown.

the current in the 20 Ω resistor, I = 1.616 (A)

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CHAPTER 3 - Thévenin and Norton equivalent networks

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CHAPTER 3 - Thévenin and Norton equivalent networks

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CHAPTER 3 - Thévenin and Norton equivalent networks

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CHAPTER 3 - Thévenin and Norton equivalent networks

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CHAPTER 3 - EXERCISE

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CHAPTER 3 - EXERCISE

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CHAPTER 3 - EXERCISE

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CHAPTER 3 - EXERCISE

(a) Convert the network to the left of terminals AB in the diagram below to an equivalent Thevenin circuit by initially converting to a Norton equivalent network.

(b) Determine the current flowing in the 1.8 Ω resistance connected between A and B in the circuit shown.

E = 18 V and r = 1.2 Ω

For the bridge network shown below, find the current in the 5 Ω resistor, and its direction, by using Norton’s theorem.

0.154 A

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CHAPTER 3 - EXERCISE

Use Norton’s theorem to find the current flowing in the 14 Ω resistor of the network shown below. Find also the power dissipated in the 14 Ω resistor.

I = 0.434 A, P = 2.64 W

In the network shown below, the battery has negligible internal resistance. Find, using Norton’s theorem, the current flowing in the 4 Ω resistor.

I = 0.918 A

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CHAPTER 3 - EXERCISE

Problem A Wheatstone Bridge network is shown in Fig. 15.47. Calculate the current flowing in the 32Ohm resistor, and its direction, using Thévenin’s theorem.

Assume the source of e.m.f. to have negligible resistance.�

1. Name two laws and three theorems which may be used to find unknown currents and p.d.s in electrical circuits.
2. State Kirchhoff’s current law and voltage law.
3. State, in your own words, the superposition theorem.
4. State, in your own words, Thévenin’s theorem and Norton’s theorem.
5. State the maximum power transfer theorem for a d.c. circuit.

PRESENTATION TOPIC:

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CHAPTER 3 - EXERCISE

Use Kirchhoff’s laws to find Vo in the circuit �

Use Kirchhoff’s laws to find Io in the circuit �

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CHAPTER 3 - EXERCISE

Use Norton’s/Thevenin theorem to find Io in the circuit �

Use Norton’s/Thevenin theorem to find current flow in the 2.2kOhm resistor �

Use Norton’s/Thevenin theorem to find current flow in the 2Ohm resistor �

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CHAPTER 3 - EXERCISE

Use Norton’s/Thevenin theorem to find Io in the circuit �

Use Norton’s/Thevenin theorem to find current flow in the 2kOhm resistor �

Use Norton’s/Thevenin theorem to find current flow in the 4kOhm resistor �

Io

Io

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CHAPTER 4

ALTERNATING VOLTAGES AND CURRENTS

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CHAPTER 4 - INTRODUCTION

Electricity is produced by generators at power stations and then distributed by a vast network of transmission lines (called the National Grid system) to industry and for domestic use. It is easier and cheaper to generate alternating current (a.c.) than direct current (d.c.) and a.c. is more conveniently distributed than d.c. since its voltage can be readily altered using transformers.

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CHAPTER 4 - THE AC GENERATOR

Let a single turn coil be free to rotate at constant angular velocity symmetrically between the poles of a magnet system as shown in Fig. 16.1. �

An e.m.f. is generated in the coil (from Faraday’s laws) which varies in magnitude and reverses its direction at regular intervals.

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CHAPTER 4 - WAVEFORMS

If values of quantities which vary with time t are plotted to a base of time, the resulting graph is called a waveform. Some typical waveforms are shown in Fig. 16.3. Waveforms (a) and (b) are unidirectional waveforms, for, although they vary considerably with time, they flow in one direction only (i.e. they do not cross th time axis and become negative). Waveforms (c) to (g) are called alternating waveforms since their quantities are continually changing in direction (i.e. alternately positive and negative).�

A waveform of the type shown in Fig. 16.3(g) is called a sine wave. It is the shape of the waveform of e.m.f. produced by an alternator and thus the mains electricity supply is of ‘sinusoidal’ form.

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CHAPTER 4 - WAVEFORMS

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CHAPTER 4 - WAVEFORMS

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CHAPTER 4 – AC VALUES

Instantaneous values are the values of the alternating quantities at any instant of time. They are represented by small letters, i, v, e, etc., (see Fig. 16.3(f) and (g)). The largest value reached in a half-cycle is called the peak value or the maximum value or the amplitude of the waveform. Such values are represented byV m, Im, Em, etc.(see Fig. 16.3(f) and(g)). A peak-to-peak value of e.m.f. is shown in Fig. 16.3(g) and is the difference between the maximum and minimum values in a cycle.

The average or mean value of a symmetrical alternating quantity (such as a sine wave), is the average value measured over a half-cycle (since over a complete cycle the average value is zero).

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CHAPTER 4 – AC VALUES

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CHAPTER 4 – AC VALUES

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CHAPTER 4 – AC VALUES

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CHAPTER 4 – AC VALUES

(50Hz, 5.5A, 3.1A

avg 2.8A, rms 4A)

3. (a) 150 V (b) 170 V �

(a)(i) 100 Hz (ii) 2.50 A (iii) 2.87 A (iv) 1.15 (v) 1.74

(b)(i) 250 Hz (ii) 20 V (iii) 20 V (iv) 1.0 (v) 1.0

(c)(i) 125 Hz (ii) 18 A (iii) 19.56 A (iv) 1.09 (v) 1.23

(d)(i) 250 Hz (ii) 25 V (iii) 50 V (iv) 2.0 (v) 2.0

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ELECTRICAL SAFETY – INSULATION AND FUSES

Insulation is used to prevent ‘leakage’, and when determining what type of insulation should be used, the maximum voltage present must be taken into account.�

​

​

For this reason, peak values are always considered when choosing insulation materials. Fuses are the weak link in a circuit and are used to break the circuit if excessive current is drawn. Excessive current could lead to a fire. Fuses rely on the heating effect of the current, and for this reason r.m.s. values must always be used when calculating the appropriate fuse size.

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

(a) 13.18 (b) 15.17 A (c) 52.63 ◦ lagging

(d) 772.1 V (e) 603.6 V �

R = 131, L = 0.545H �

(a) 11.12 (b) 8.99 A (c) 25.92 ◦ lagging

(d) 53.92 V, 78.53 V, 76.46 V �

V1 = 26.0V at67.38◦ lagging

�

V2 = 67.05V at72.65◦ leading

V = 50V,53.14◦ leading

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

(180kVA, 156kVAr)

(6.25 Ohm, 15 Ohm, 13.64 Ohm, 0.4167 , 65.37 lagging)

Problem 30. A circuit consisting of a resistor in series with a capacitor takes 100 watts at a power factor of 0.5 from a 100V, 60Hz supply. Find (a) the current flowing, (b) the phase angle, (c) the resistance, (d) the impedance and (e) the capacitance.�

(61.26μF)

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CHAPTER 4 – SINGLE PHASE AC CIRCUITS

530.5μF �

https://grok.com/share/c2hhcmQtMg%3D%3D_d59b8de9-d721-4592-8ff1-7ca6761f3e9c

(188.2μF) �

CHAPTER 4 – THREE PHASE SYSTEMS

The voltage induced by a single coil when rotated in a uniform magnetic field is shown in Fig. 23.1 and is known as a single-phase voltage. Most consumers are fed by means of a single-phase a.c. supply. Two wires are used, one called the live conductor (usually coloured red) and the other is called the neutral conductor (usually coloured black). The neutral is usually connected via protective gear to earth, the earth wire being coloured green. �

A three-phase supply is generated when three coils are placed 120◦ apart and the whole rotated in a uniform magnetic field as shown in Fig. 23.2(a). The result is three independent supplies of equal voltages which are each displaced by 120◦ from each other, as shown in Fig. 23.2(b).

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

CHAPTER 4 – THREE PHASE SYSTEMS

�

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CHAPTER 4 – THREE PHASE SYSTEMS

1. (a) 231 V (b) 4.62 A (c) 4.62 A

2. (a) 212 V (b) 367 V

3. 165.4 μF 4. 16.78 mH

5. IR = 64.95A, IY = 86.60A,

IB = 108.25A, IN = 37.50A �

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

1. (a) 400 V (b) 8 A (c) 13.86 A

2. (a) 415 V (b) 3.32 A (c) 5.75 A �3. 55.13 μF 4. 73.84 mH

5. (a) 219.4 V (b) 65 A (c) 37.53 A

6. 8 μF �

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

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CHAPTER 4 – THREE PHASE SYSTEMS

1. (a) 9.68 kW (b) 29.04 kW

2. 1.35 kW 3. 5.21 kW

4. (a) 0.406 (b) 10 A (c) 17.32 A (d) 98.53 V

5. 0.509

6. (a) 13.39 kW (b) 21.97 A (c) 12.68 A �

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CHAPTER 4 – EXERCISE

13.86 A, 5.76 kW, 9.60 kVA

8.027 A, 13.90 A, 5.797 kW, 9.630 kVA, 52.99◦ leading �

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CHAPTER 4 – EXERCISE

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CHAPTER 4 – EXERCISE

Three loads, each of 10Ohm resistance, are connected in star to a 400V, three-phase supply. Determine the quantities stated in questions 1 to 5, selecting answers from the following list: �

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CHAPTER 4 – EXERCISE

1. 20

2. 78.27 , 2.555 A, 39.95◦ lagging

3. (a) 40 (b) 1.77 A (c) 56.64 V (d) 42.48 V

4. (a) 4 (b) 8 (c) 22.05 mH

5. 30 V, 53.13◦ lagging

6. (a) 200 (b) 223.6 (c) 1.118 A (d) 111.8 V,

223.6 V (e) 63.43◦ lagging

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CHAPTER 4 – EXERCISE

28 V �

(a) 93.98 (b) 2.128 A (c) 57.86 ◦ leading �

(a) 160 Hz (b) 90 V (c) 120 V �

(a) 39.05 (b) 4.526 A (c) 135.8 V (d) 113.2 V (e) 39.81◦ leading

�

37.5 , 28.61 μF �

(a) 7 A (b) 53.13◦ lagging (c) 4.286 (d) 7.143

�

(e) 9.095 mH

(a) 4 (b) 7 (c) 5.745 (d) 11.08 μF (e) 0.571

�

(f) 55.15◦ leading

60 , 255 mH �

CHAPTER 4 – EXERCISE

In the Figure a, b, c, d, e, find:

1. The mean, rms value of the current waveform.
2. If the current is passed through a 2Ω, find the average power absorbed by the resistor.

(c)

(d)

(e)

Calculate the average power delivered to a 2.2Ω load by the voltage vs equal to (a) 5 V; (b) 4 cos 80t - 8 sin 80t V; (c) 10 cos 100t + 12.5 cos (100t + 19°) V.

Irms=8.165(A)

Iavg=1.2A; Irms=1.36A

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CHAPTER 5

WAYS OF GENERATING ELECTRICITY

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CHAPTER 5

WAYS OF GENERATING ELECTRICITY

Why it is important to understand: Ways of generating electricity – the present and the future

In 1831, Michael Faraday devised a machine that generated electricity from rotary motion – but it took almost 50 years for the technology to reach a commercially viable stage. In 1878, in the USA, Thomas Edison developed and sold a commercially viable replacement for gas lighting and heating using locally generated and distributed direct current electricity. The world’s first public electricity supply was provided in late 1881, when the streets of the Surrey town of Godalming in the UK were lit with electric light. This system was powered from a water wheel on the River Wey, which drove a Siemens alternator that supplied a number of arc lamps within the town.

Now, some 80 years later, we have become almost totally dependent on electricity! Without electricity we have no lighting, no communication via mobile phones/smartphones, internet or computer, television, iPods or radios, no air-conditioning, no fans, no electric heating, no refrigerator or freezer, no coffee maker, no kitchen appliances, no dishwasher, no electric stoves, ovens or microwaves, no washing machines or tumble dryers, problems with drinking water if it comes from a system dependent on electrical pumps … Our whole lifestyles have become totally dependent on a reliable electricity supply.

In the future, civilization will be forced to research and develop alternative energy sources. Our current rate of fossil fuel usage will lead to an energy crisis this century. In order to survive the energy crisis many companies in the energy industry are inventing new ways to extract energy from renewable sources and while the rate of development is slow, mainstream awareness and government pressures are growing. Traditional methods of generating electricity are unsustainable, and new energy sources must be found that do not produce as much carbon.

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CHAPTER 5 - INTRODUCTION

In this chapter, methods of generating electricity using coal, oil, natural gas, nuclear energy, hydro power, pumped storage, wind, tidal power, biomass and solar energy are explained.

Coal, oil and gas are called ‘fossil fuels’ because they have been formed from the organic remains of prehistoric plants and animals. Historically, the transition from one energy system to another, as from wood to coal or coal to oil, has proven an enormously complicated process, requiring decades to complete. In similar fashion, it will be many years before renewable forms of energy – wind, solar, tidal, geothermal, and others still in development– replace fossil fuels as the world’s leading energy providers. This chapter explains some ways of generating electricity for the present and for the future.

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CHAPTER 5

Generating electrical power using coal

Generating electrical power using coal

A coal power station turns the chemical energy in coal into electrical energy that can be used in homes and businesses. �

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CHAPTER 5

Steam Turbine

• Grinding the coal to a fine powder and blown into the boiler
• Burning coal produces ash and exhaust gases. The ash falls to the bottom of the boiler and is removed by the ash systems
• The exhaust stacks of coal power stations are built tall so that the exhaust plume can disperse before it touches the ground.

New technology called Carbon Capture and Storage (CCS) is being developed to remove up to 90% carbon dioxide from power station emissions and store it underground. �

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CHAPTER 5

ADVANTAGES OF COAL

​

• easily combustible, and burns at low temperatures

​

• coal-fired boilers cheaper and simpler

​

• inexpensive to buy, simple to mine

​

• the least expensive fossil fuel

​

​

DISADVANTAGES OF COAL

​

• coal is non-renewable and fast depleting

​

• has the lowest energy density of any fossil fuel

​

• coal has high transportation costs, storage costs are high

​

• coal dust is an extreme explosion hazard

​

• releases carbon dioxide

​

Generating electrical power using coal

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CHAPTER 5

Generating electrical power using oil

An oil power station turns the chemical energy in oil into electrical energy that can be used in homes and businesses. �

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CHAPTER 5

ADVANTAGES OF OIL

​

• high energy density

​

• easy availability and infrastructure for transport

​

• easy to produce and refine

​

• the least expensive fossil fuel

​

​

DISADVANTAGES OF OIL

​

• cause pollution of water and earth

​

• sulphur dioxide, carbon monoxide and acid rain

​

• lead to production of very harmful and toxic materials during refining

​

Generating electrical power using oil

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CHAPTER 5

A gas power station turns the chemical energy in natural gas into electrical energy that can be used in homes and businesses. �

Generating electrical power using natural gas

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CHAPTER 5

ADVANTAGES

• Less harmful than coal or oil

​

• Easy to store and transport

​

• Burns cleaner without leaving any smell, ash or smoke

​

DISADVANTAGES

• Gas is toxic and flammable, can damage the environment

​

• Non-renewable and is expensive to install.

​

Generating electrical power using natural gas

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CHAPTER 5

A nuclear power station turns the nuclear energy in uranium atoms into electrical energy that can be used in homes and businesses. �

Generating electrical power using nuclear energy energ

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CHAPTER 5

ADVANTAGES

• Nuclear energy is reliable, has low fuel costs

​

• Low electricity costs, no greenhouse gas emissions/air pollution

​

• Has a high load factor and huge potential.

DISADVANTAGES

• The fear of nuclear and radiation accidents, the problems of nuclear waste disposal

​

• the low level of radio activity from normal operations, the fear of nuclear prolife ration

​

• High capital investment, cost overruns and long gestation time

​

• the many regulations for nuclear energy power plants, and fuel danger – uranium is limited to only a few countries and suppliers.

Generating electrical power using nuclear energy

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CHAPTER 5

�A hydroelectric power station converts the kinetic – or moving – energy in flowing or falling water into electrical energy that can be used in homes and businesses. �

Generating electrical power using hydro power

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CHAPTER 5

ADVANTAGES

• With hydro power there are no fuel costs

​

• low operating costs and little maintenance, low electricity costs

​

• no greenhouse gas emissions/air pollution

​

• energy storage possibilities, small size hydro plants possible,reliability

​

• a high load factor and long life.

DISADVANTAGES

• Disadvantages of hydro power includes environmental, dislocation and tribal rights difficulties, wildlife and fish being affected

​

• the possibility of earthquake vulnerability, siltation, dam failure due to poor construction or terrorism

​

• the fact that plants cannot be built anywhere, and long gestation times.

Generating electrical power using hydro power

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CHAPTER 5

�Pumped storage reservoirs provide a place to store

energy until it’s needed.

Generating electrical power using pumped storage

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CHAPTER 5

ADVANTAGES

• Pumped storage provides a way to generate electricity instantly and quickly

​

• no pollution or waste is created, and there is little effect on the landscape, as typically pumped storage plants are made from existing lakes in mountains.

DISADVANTAGES

• Pumped storage facilities are expensive to build, and once the pumped storage plant is used

​

• it cannot be used again until the water is pumped back to the upper reservoir.

Generating electrical power using pumped storage

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CHAPTER 5

�Wind turbines use the wind’s kinetic energy to generate electrical energy �

Generating electrical power using wind

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CHAPTER 5

ADVANTAGES

• Wind energy has no pollution and global warming effects

​

• low operating costs, a large industrial base, no fuel costs, and offshore advantages.

DISADVANTAGES

• Wind energy has low persistent noise

​

• can cause a loss of scenery

​

• requires land usage and is intermittent in nature.

Generating electrical power using wind

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CHAPTER 5

�The tide moves a huge amount of water twice each day, and harnessing it could provide a great deal of energy�

Generating electrical power using tidal power

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CHAPTER 5

ADVANTAGES

• Tidal power is renewable, non-polluting and carbon negative

​

• predictable

​

• needs no fuel, has low costs

​

• long life, high energy density and high load factor.

DISADVANTAGES

• Tidal power has high initial capital investment, limited locations

​

• detrimental effects on marine life, immature technology, long gestation time

​

• some difficulties in transmission of tidal electricity

​

• and weather effects that can damage tidal power equipment.

Generating electrical power using tidal

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CHAPTER 5

�Biomass is fuel that is developed from organic materials, a renewable and sustainable source of energy used to create electricity or other forms of power ��

Generating electrical power using biomass

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CHAPTER 5

ADVANTAGES

• Biomass energy is carbon neutral, uses waste efficiently

​

• is a continuous source of power, can use a large variety of feedstock, has a low capital investment

​

• can be built in remote areas and on a small scale, reduces methane,

​

• is easily available and is a low cost resource.

DISADVANTAGES

• With biomass energy, pollution can occur where poor technology is used

​

• continuous feedstock is needed for

​

• efficiency, good management of biomass plants are required

​

• it has limited potential compared to other forms of energy like solar, hydro etc., and biomass plants

​

• are unpopular if constructed near homes.

Generating electrical power using biomass

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CHAPTER 5

�Solar panels turn energy from the sun’s rays directly into useful energy ��

Generating electrical power using solar energy

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CHAPTER 5

ADVANTAGES

• Solar power is environmentally friendly

​

• has declining costs, no fuel, low maintenance, no pollution

​

• has almost unlimited potential, has the advantage that installations

​

• can be any size, and installation can be quick.

DISADVANTAGES

• Solar power has higher initial costs than fossil energy forms

​

• is intermittent in nature and has high capital investment.

Generating electrical power using solar energy

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CHAPTER 5 – EXERCISE

1. Briefly explain with a block diagram how electricity is generated using coal / oil/ natural gas, and state the advantages/disadvantages of coal/ oil/ natural gas?

2. Briefly explain with a block diagram how electricity is generated using nuclear energy, and state the advantages/disadvantages of nuclear energy?

3. Briefly explain with a block diagram how electricity is generated using hydro power/ pumped storage, and state the advantages/disadvantages of hydro power/ pumped storage?

4. Briefly explain with a block diagram how electricity is generated using wind/ solar energy, and state the advantages/disadvantages of wind/ solar energy?

5. Briefly explain with a block diagram how electricity is generated using tidal power, and state the advantages/disadvantages of tidal power?

6. Briefly explain with a block diagram how electricity is generated using biomass, and state the advantages/disadvantages of biomass?

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CHAPTER 5 – EXERCISE

1. Briefly explain with a block diagram how electricity is generated using coal, and state the advantages/disadvantages of coal?

Coal supply

Boiler

Water supply

Exhaust stack

Ash systems

Steam turbine

Exhaust plume

AC generator

Electricity transmission

Consumer

BLOCK DIAGRAM

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CHAPTER 6

PROBLEM SOLVING

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CHAPTER 6

PROBLEM SOLVING

Introduction to Structured Problem Solving

Think about a problem you face in your job. What do you think is causing the problem? These questions, and finding their answers, are at the center of structured problem solving.

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Importance

Why use structured problem solving?

To have everyone using the same approach

To have a common vocabulary around problem solving.

To have a checklist for covering all the necessary steps.

To have consistency.

To have a measurable and repeatable process.

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CHAPTER 6

PROBLEM SOLVING

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CHAPTER 6

Introduction to Step 1: Define the Problem

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The purpose of Step 1, Defining the Problem, is to concisely state the problem and its current impact. The resulting output of performing this step is the problem statement.

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A good problem statement will help focus you in the right direction. A poorly defined problem statement can cause you to waste time and effort.

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"A problem well stated is a problem half-solved." -Charles Kettering, American inventor �

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CHAPTER 6