Diploma in Electrical Installations. (Buildings and Structures) Unit 202 Workbook PRE-ATTENDANCE REVISION V1

Transcription

1 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Workbook PRE-ATTENDANCE REVISION V1

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3 Welcome This booklet is designed to outline the areas that you will study when you complete unit 202 at the centre. Throughout your time in centre you will have a qualified assessor there to help and advise you. We want you to achieve the best possible benefits from your time in centre, therefore before attending we would like you to revise by carrying out the work detailed in this booklet. Studying this workbook will ensure that you are properly prepared before attending centre and therefore help you achieve the best possible results from your time in centre. Please remember that you should not carry out any electrical practical work unless you are in class with one of our qualified assessors. We very much look forward to welcoming you at the centre.

4 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 1 202: Principles of electrical science Handout 1: Principles of electricity Learning outcome The learner will: 1. know the principles of electricity. Assessment criteria The learner can: 1.1. describe the reaction of electrons when charged to form an electric current 1.2. identify sources of an electromotive force 1.3. describe the effects of an electric current 1.4. identify SI Units for various electrical quantities. Range Sources: Chemical, magnetic, thermal. Effects: Chemical, magnetic, thermal. SI Units: Quantity symbol, unit symbol, multiples, sub-multiples. Electrical quantities: Current, potential, resistance, resistivity, temperature, mass, force, magnetic flux, magnetic flux density, period, frequency, power, energy, time, length, area, mass, weight. Electron theory structure of matter Principles of electricity The smallest part of any material is called a molecule, yet the latter is made up of one or more atoms. For example, water is made up of H 2O (two atoms of hydrogen and one atom of oxygen). Basically, the atom is constructed of a central core containing protons surrounded by orbiting electrons. Electrical nature of atoms PROTONS are POSITIVELY CHARGED ELECTRONS are NEGATIVELY CHARGED In an electrically neutral atom the number of PROTONS is equal to the number of ELECTRONS The simplest atom is hydrogen (1 proton and 1 electron) see right. The heaviest atom is uranium (92 protons balanced by 92 electrons). A Helium atom has two protons and two electrons.

5 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 1 Copper atom 29 protons and 29 electrons An electrically neutral atom has as many (+ve positive) protons as there are (-ve negative) electrons. The single electron in the outer orbit of the copper atom is only loosely attached to the atom because: it is so far away from the core (nucleus) inner electrons try to push it off (like charges repel). As a result, this electron may be easily detached. If the balance of protons and electrons is upset and the atom becomes positively charged it will then attract any nearby electron. This process occurs millions of times every second; at any instant in time the material has a large number of free electrons moving in all directions. Random free electron movement If the material is then connected across a battery, the positive plate (or terminal) attracts electrons, whilst the negative plate repels them. The battery provides a source of electromotive force (EMF). The resultant electron flow around the circuit is called an ELECTRIC CURRENT Note: even when an appropriate EMF is applied, there must be a complete circuit for the electrons to flow; a break in the circuit will cause the electron flow to stop. This aspect is useful, as we can use it to control the flow of electricity using, for example, a switch.

6 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 1 Sources of an electromotive force In the example above the device providing the electromotive force that caused the flow of electrons was a battery. This is a device for converting chemical energy into electrical energy. There are various methods of directly transforming other forms of energy into electrical energy, thus providing a source of electromotive force. Electromagnetic induction: where an electrical generator, dynamo or alternator transforms kinetic energy (energy of motion) into electricity. This is the most used form for generating electricity. Electrochemistry: the direct transformation of chemical energy into electricity, as in a battery. Photoelectric effect: the transformation of light into electrical energy, as in solar cells. Thermoelectric effect: the direct conversion of temperature differences to electricity, as in thermocouples. Piezoelectric effect: electricity generated due to the mechanical strain placed on crystals. Effects of an electric current When electricity flows, one or more effects occur as follows: thermal chemical magnetic. These will be covered later in the course. Electrical quantities Many different quantities are used in electrical systems and they need to be standardised. The standardisation system is an international one called the Système International d Unités (abbreviated to SI Units). SI Units are based upon a small number of fundamental units from which all other units may be derived. The table below shows a selection of units appropriate to the electrical industry, including the symbols used in formulae and also their abbreviation. SI Unit Measure of Symbol Abbreviation Ampere Electric Current I A Volt Electric potential/potential difference/electromotive force V V Ohm Electrical resistance R Ω Rho Resistivity Ρ Ohm/m 3 Degrees Celsius Celsius temperature t C Kilogram Mass m kg Newton Force F N Weber Magnetic flux Φ Wb Tesla Magnetic flux density B T or Wb/m 2 Period Duration of one cycle p s Hertz Frequency number of cycles per second f Hz Watts Power P W Joule Energy/Work/Quantity of heat E J Second Time t s Metre Length l m Square metre Area a m 2 m x gravitational constant Weight

7 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 2 202: Principles of electrical science Handout 2: Transposition of basic formulae Learning outcome The learner will: 1. know the principles of electricity. Assessment criteria The learner can: 1.5. transpose basic formulae. Range Basic formulae: Ohm s law, transformer formulae, Magnetism formulae, Pythagoras formulas, trigonometry (for power factor and a.c. theory). Transposition of basic formulae This is also known as changing the subject of the formulae. In the example on the left, I is the subject of the formula. By inserting the values for V and R, the value of the subject I can be calculated. If we need to find V, for example, we must transpose the formula to make V the subject. There is one fundamental rule for transposing a formula, as found below. Whatever you do to one side of the formula, you must do the same to the other side. In other words: add the same quantity to both sides of the formula subtract the same quantity from both sides of the formula multiply both sides of the formula by the same quantity divide both sides of the formula by the same quantity take functions of both sides of the formula; for example, square both sides or find the reciprocal of both sides. Example 1 I = V/R make V the subject of the formula. V I = (I is currently the subject of the formula.) R The question is: Make V the subject of the formula. This means that V must be put on its own on one side of the equals sign and the other terms must be on the other side. In order to do this, first multiply both sides by R. I R = V R R Now cancel through: I R = V R R I R = V Now reverse the formula: VV = II RR

8 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 2 Example 2 I = V/R make R the subject of the formula. V I = (I is currently the subject of the formula.) R The question is: Make R the subject of the formula. This means that R must be put on its own on one side of the equals sign and the other terms must be on the other side. In order to do this, first multiply both sides by R. I R = Now cancel through: I R = Now divide both sides by I: Now cancel through : I R = I R I I R I = = RR = V R R V R R V The formula used in examples 1 and 2 is the Ohm s law formula and is used to calculate the relationship between resistance (R), current (I) and voltage (V). In formulae (the plural of formula) where there are three values, with two divided or multiplied by each other, the values can be put into a triangle that will allow you to determine easily which variation of the formula should be used. V I V I VV II If we need to find a value, simply cover it and what s left gives the formula. If the two values are side by side, multiply them; if the two values are one on top of the other then divide the bottom one into the top one to give the following: VV = II RR II = VV RR RR = VV II

9 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 2 Example 3 E = B x l x v make v the subject of the formula. E = B l v (E is currently the subject of the formula.) The question is: Make v the subject of the formula. This means that v must be put on its own on one side of the equals sign and the other terms must be on the other side. In order to do this, first divide both sides by B x l: Now cancel through: E B l E B l E B l Reverse the formula: vv = = = = v B l v B l B l v B l EE BB ll Useful formulae VOLTAGE = I.R = P/I = P.R CURRENT = V/R = P/V = P/R RESISTANCE = V/I = V 2 /P = P/I 2 POWER = V.I = I 2.R = V 2 /R RESISTORS In series: R t = R 1 + R 2 + R 3 etc KIRCHHOFF S LAWS in parallel: 1 R t = 1 R R R 3 Voltage: VT = V1 + V2 + V3 + etc in a Series circuit Current: IT = I1 + I2 + I3 + etc in a Parallel circuit etc RESISTIVITY R = ρ l a R L R 1/a CHARGE (Quantity) Q = It or Q = VC CAPACITORS in parallel: C t = C 1 + C 2 + C 3 etc in series: 1 = C t C 1 C 2 C 3 etc

10 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 2 ELECTRO- MAGNETISM Φ = B.A B H mmf = N x I H = n l L INDUCED EMF E = B l v (v = velocity) E = Φ2 Φ1 t E = (I2 I1) L t FORCE ON A CONDUCTOR F = B I l Newtons ENERGY in a magnetic field: ½.L.I 2 Joules STORED in a capacitor: ½.C.V 2 Joules EFFICIENCY Efficiency = Output power Input power MECHANICS Force = Mass x Newtons Torque = Force x Distance Work Done (WD) = Force x Distance Energy = Joules Joules = Watts x Seconds Joules = Newton Metre PYTHAGORAS Z = R 2 + X 2 POWER FACTOR p.f. = R Z POWER FACTOR p.f. = P VA

11 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 3 202: Principles of electrical science Handout 3: Basic electrical circuits and Ohm s law Learning outcome The learner will: 2. know the principles of basic electrical circuits. Assessment criteria The learner can: 2.2. apply Ohm s law to electrical circuits. Range Electrical circuits: Series, parallel. Current Basic electrical circuits and Ohm s law The uniform flow of electrons through a conductor is referred to as electric current. The unit of electric current is the AMPERE (A) In formulae the symbol for electric current is I. Electromotive force (EMF) The EMF provides a difference in potential between two open terminals of an electrical circuit. When the circuit is complete, this potential difference causes the electrons to flow in a uniform direction around the circuit and produce a flow of current. The unit of EMF is the VOLT (V) In formulae the symbol for EMF is V. Resistance Every circuit presents some opposition to the flow of current in the electric circuit, which has to be overcome by the electrical pressure applied. This opposition is called resistance. The unit of resistance is the OHM (Ω) In formulae the symbol for resistance is R. Ohm s law In 1827 a man named Georg Ohm published his experiments regarding the relationship between, current, voltage and resistance. His findings, referred to as Ohm s Law, are shown below: The current flowing in any circuit is directly proportional to the applied voltage and inversely proportional to the resistance of the circuit, provided that the temperature of the circuit remains constant.

12 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 3 The simple relationship between symbols is shown on the right. The simple method of transposing the symbols is to use the cover-up method, ie cover the symbol required; the answer is then given by the other two symbols. VV = II RR II = VV RR RR = VV II Example 1 An EMF of 10 volts is applied to a resistance of 20Ω. Calculate the current that will flow. I = V R = = Example 2 Calculate the applied EMF when 2 amperes flows through a resistance of 40Ω. V = I R = 2 40 = 8888 vvvvvvvvvv Example 3 When an EMF of 50 volts is applied to a circuit, a current of 5 amperes flows. Calculate the resistance of the circuit. R = V I = 50 5 = 1111 Ω

13 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 4 202: Principles of electrical science Handout 4: Resistors in series Learning outcome The learner will: 2. know the principles of basic electrical circuits. Assessment criteria The learner can: 2.2. apply Ohm s law to electrical circuits. Range Electrical circuits: Series, parallel. Resistors in series When there is only one resistance in a circuit, the Ohm s law calculation is straightforward. However, when there are two or more resistors in a circuit, the total effective resistance must be calculated first. Resistors can be connected in a number of configurations: series parallel series-parallel. In order to find the total resistance of any series circuit, just add all the resistances together. The formula for calculating the total resistance of resistors connected in series is given below: RR tt = RR 11 + RR 22 + RR 33 Example 1 Calculate the total resistance of the circuit shown below: R t = R 1 + R 2 + R 3 + R 4 = = Method of determining current flow in a series circuit 1) Calculate the total resistance of the circuit, using the series resistor formula. 2) Redraw the circuit diagram, using the equivalent total resistance. 3) Using Ohm s law, calculate the current flowing.

14 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 4 Example 2 Calculate the current that will flow in the circuit shown below: R t = R 1 + R 2 + R 3 Equivalent circuit = = V I = R = 24V 12 = 22 aaaaaaaaaaaaaa The current flowing in any series circuit will be the same wherever you measure it. In other words, the current coming from the supply will flow through resistance 1, then through resistance 2, followed by resistance 3, and so on, until it gets back to the supply terminal. It does not matter where you measure the current in the series circuit, as it will always be the same. Voltage drop If a current is passed through a resistance then, according to Ohm s law, a voltage is produced across it, ie V = I R This voltage is often referred to as a voltage drop but is calculated the same way as any other voltage.

15 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 4 After you have read a question, the first thing that you do is to draw a diagram relating to the question, with all the information on it. Example 3 What potential is produced across a resistance of 23Ω if a current of 10 amps is flowing through it? V = I R V = VV = vvvvvvttaa Kirchoff s Voltage law Kirchhoff's Voltage law states that: the algebraic sum of the voltages around a circuit is equal to zero (or the supply voltage). Putting Kirchhoff's statement into symbols, we get the formula: VV SS = VV 11 + VV 22 + VV 33

16 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 4 Example 4 Calculate V 1, V 2, V 3, hence proving Kirchhoff's Voltage law. R t = R 1 + R 2 + R 3 = = I = V R = 90V 15 = 66 aaaaaaaaaaaaaa Resistor 1 Resistor 2 Resistor 3 V 1 = I R 1 V 2 = I R 2 V 3 = I R 3 = 6 3 = 6 5 = 6 7 = = = V S = V 1 + V 2 + V 3 = = Kirchhoff s law is proved.

17 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 5 202: Principles of electrical science Handout 5: Resistors in parallel Learning outcome The learner will: 2. know the principles of basic electrical circuits. Assessment criteria The learner can: 2.2. apply Ohm s law to electrical circuits. Range Electrical circuits: Series, parallel. Resistors in parallel In order to find the total resistance of any parallel circuit, we must add the reciprocal (1 x) of all the resistances together. The formula for calculating the total resistance of resistors connected in parallel is given below. 11 = RR tt RR 11 RR 22 RR 33 Example 1 Calculate the total resistance of a parallel circuit if: R 1 = 9Ω, R 2 = 12Ω and R 3 = 18Ω. 1 1 = R t R 1 R 2 R 3 1 = Find the lowest common denominator, which is 36: 1 R t = 1 R t = Inverting both sides of the equation will give us R t: R t = The total resistance of the circuit will determine the amount of current that will flow in that circuit.

18 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 5 Example 2 Calculate the total resistance of a parallel circuit if R 1 = 45Ω, R 2 = 90Ω and R 3 = 30Ω 1 1 = R t R 1 R 2 R 3 1 = Find the lowest common denominator, which is 90: 1 R t = 1 R t = Inverting both sides of the equation will give us R t: R t = It can be seen that in all parallel circuits the total resistance of the circuit is always less than the smallest resistance in that circuit. Two resistors in parallel When there are only two resistors in parallel, the equivalent total resistance of the combination may be found by using the product over sum method, as shown below. The product of two numbers is the multiplication of the two numbers. The sum of two numbers is the addition of the two numbers. Example 3 Calculate the total resistance of two resistors connected in parallel if R 1 = 6Ω and R 2 = 4Ω. R t = R 1 R 2 R 1 + R 2 = = This method only works for two resistors in parallel. If all the resistors in parallel are of the same value, then all that has to be done, in order to calculate the total resistance of the circuit, is to take any one resistor and divide its value by the number of resistors that are in the parallel combination.

19 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 5 Kirchoff s Current law The sum of the currents arriving at a point must equal the sum of the currents leaving that point. II tt = II 11 + II 22 + II 33 + eeeeee In any parallel branch/circuit, the voltage will always be the same across each individual resistor, but the current may be different depending on the value of the resistor. Since the voltage is constant, it makes it easy to calculate the current flowing through each individual resistor. Example 4 I 1 = V R 1 = 12 4 = 3 amps I 2 = V R 2 = 12 6 = 2 amps I 3 = V = 12 R 3 12 = 1 amps I t = I 1 + I 2 + I 3 I t = II tt = 66 aaaaaaaa If we calculate the total resistance of the parallel network: 1 1 = R t R 1 R 2 R = R t R t = 1 R t = 12 R t = 6 =

20 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 5 and then calculate the total current using Ohm s law: I t = V R t 12 = 2 = 66 aaaaaaaa Kirchhoff s law is proved. The sum of the currents through the resistors is the same as the current drawn from the supply; therefore, the solution is correct according to Kirchhoff s Current law.

21 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 6 202: Principles of electrical science Handout 6: Power Learning outcome The learner will: 2. know the principles of basic electrical circuits. Assessment criteria The learner can: 2.3. calculate power in a basic electrical circuit. Power Electrical power is the rate of doing electrical work or of expenditure of electrical energy. The formula for Power is obtained by using the power triangle, as shown on the right. From the triangle it can be seen that: PPPPPPPPPP = VVVVVVVVVV AAAAAAAA or PP = VV II (d.c. only) Using Ohm s law and substituting for either voltage or current in the power equation, as shown below, you can find other formulae for power. Since Ohm s law is: V = I R substituting I x R for V in the power equation we get: P = I I R giving: PP = II 22 RR (a.c. or d.c.) By substituting V R for I in the power equation, we get: P = V R V This gives: PP = VV22 RR (a.c. or d.c.)

22 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 6 Example 1 Power dissipated by R 1 = P 1: Total power = V I = = wwwwwwwwww Power dissipated by R 2 = P 2: P 1 = I 2 R 1 = = wwwwwwwwww Alternatively, since: then voltage across R 1 is V 1: P 1 = P 2 = I 2 R 2 = = 8888 wwwwwwwwww Total power = P 1 + P 2 V = = wwwwwwwwww = I R V 1 = I R 1 = 2 30 = 60 volts V 1 2 R 1 or P 1 = V 1 I = = 60 2 = wwwwwwwwww = wwwwwwwwww

23 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 6 Voltage across R 2 is V 2: P 2 = V 2 = I R 2 = 2 20 = 40 volts V 2 2 R 2 or P 2 = V 2 I = = 40 2 = 8888 wwwwwwwwww = 8888 wwwwwwwwww

24 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 7 202: Principles of electrical science Handout 7: Resistivity Learning outcome The learner will: 2. know the principles of basic electrical circuits. Assessment criteria The learner can: 2.1. calculate resistance of a conductor in a basic electrical circuit. Range Conductors 20 celsius): Copper conductor, aluminium conductor, silver conductor, gold conductor, brass conductor. Resistivity Up until now we have assumed that the conductors that form part of the circuit have no resistance. However, every conductor possesses resistance, the value of which depends on four factors: 1. the length of the conductor (L metres) 2. the cross sectional area of the conductors (CSA mm 2 ) 3. the type of conducting material (ρ Rho) 4. the temperature (t C). Length of conductor The conductor length can be considered as being made up of a lot of resistors all in series. If the length of short conductors is L 1, L 2, L 3, etc then the total length L M is made up of short lengths as shown: LL MM = LL 11 + LL 22 + LL 33 eeeeee Also, if the short conductors have resistances R 1, R 2, R 3 and R 4 then the total resistance of the length L M is: RR TT = RR 11 + RR 22 + RR 33 + RR 44 eeeeee It therefore follows that the resistance of a conductor is proportional to its length: RR LL (resistance is proportional to length)

25 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 7 Example 1 If a cable has a resistance of 0.15Ω/m what will be the resistance of 320m of the same cable? Since resistance is proportional length: Resistance = Ω m length = = Example 2 If 50m of a conductor have a resistance of 0.01Ω what would be the resistance of 500m of the same cable? R = = new length old length resistance = Example 3 If 750m of a conductor have a resistance of 0.013Ω what will be the resistance of 150m of the same conductor? Since resistance is proportional to length: new length R = old length resistance 150 = =

26 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 7 Cross sectional area (CSA) of conductors A conductor of cross sectional area (a) can be regarded as being made up of a number of smaller conductors joined together in parallel, as shown. If the areas of the small conductors are a 1, a 2, a 3, a 4, etc then the total area a, made up of small conductors, is calculated by: aa TT = aa 11 + aa 22 + aa 33 + aa 44 If the resistances of the small conductors are R 1, R 2, R 3 and R 4 then the total resistance R T is given by: 11 RR TT = 11 RR RR RR RR 44 Therefore, the resistance of a conductor is inversely proportional to its CSA. RR 11 aa (resistance is inversely proportional to area) Example 4 If the resistance of 100m of a certain conductor is 0.076Ω, calculate the resistance of 100m of the same conductor but with: a) twice the CSA b) four times the CSA c) half the CSA d) one quarter of the CSA. a) Since R is to 1 a, then: b) Since R is to 1 a, then: R = = R = =

27 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 7 c) Since R is to 1 a, then: d) Since R is to 1 a, then: R = = R = = Example 5 If the resistance of 100m of a certain conductor is 1.24Ω, calculate the resistance of 500m of the same conductor but with twice the CSA. But: Type of conducting material R L R = = 6.2Ω = 1 (therefore the resistance must be halved) a = Since different materials have a different number of electrons in their structure, it must follow that they will not all have the same resistance to current flow. This resistance to current flow in a material is known as its resistivity (symbol ρ - RHO). A material that is a good conductor has a low resistivity, while a poorer conductor will have a higher value. The table below gives the resistivity of some commonly used materials in the electrical industry. Material Resistivity at 20 C Copper 1.72 x 10-8 ohm/metre 3 Aluminium 2.65 x 10-8 ohm/metre 3 Silver 1.59 x 10-8 ohm/metre 3 Gold 2.24 x 10-8 ohm/metre 3 Brass (58% copper) 5.90 x 10-8 ohm/metre 3 Brass (63% copper) 7.10 x 10-8 ohm/metre 3 The table lists copper as 1.72 x 10-8 ohm/metre 3 at a temperature of 20 C. This means that the resistance of a cubic metre of copper 1m long by 1m high and 1m deep has a resistance of Ω or 1.72 x 10-8 Ω when measured across two opposite faces.

28 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout iiii tttttt ssssssss aaaa EEEEEE 88 but it will be written as 17.2E-8 for convenience. Since the resistance of any conductor is directly proportional to length and inversely proportional to its CSA, the formula for calculating the resistance of any conductor is: RRRRRRRRRRRRRRRRRRRR = RRRRRRRRRRRRRRRRRRRRRR LLLLLLLLLLLL CCCCCCCCCC ssssssssssssssssss aaaaaaaa or RR = ρρ ll aa where: R = conductor resistance in ohms ρ = cable resistivity in Ω/m l = cable length in metres a = cross sectional area (CSA) in metre 2. Cable CSA is usually quoted in mm 2 and this will need to be converted to m 2. In order to convert: mm 2 to m 2 multiply by 10 6 Example 6 Calculate the resistance of 1,000m of 16mm 2 (CSA) single copper conductor. Take ρ to be 1.72 x 10-8 Ω/metre. R = ρ l a = , =

29 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 7 Example 7 Calculate the resistance of an aluminium wire 100m long and CSA of 25mm 2. Take the resistivity of aluminium to be 2.65 x 10-8 Ω/metre. R = ρ l a = = Example 8 Calculate the resistance of 100m of copper conductor if ρ = 1.72 x 10-8 Ω/metre for the material and the CSA is 2.5mm 2. R = ρ l a = = Example 9 Calculate the resistance of 100m of silver conductor if ρ = 1.59 x 10-8 Ω/metre for the material and the CSA is 2.5mm 2. R = ρ l a = =

30 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 8 202: Principles of electrical science Handout 8: Connection of meters Learning outcome The learner will: 2. know the principles of basic electrical circuits. Assessment criteria The learner can: 2.4. state how instruments are connected into circuits in order to measure electrical quantities. Range Instruments: Voltmeter, ammeter, wattmeter, ohmmeter. Connection of meters It is often necessary to know how much voltage, current or energy is being used in any particular circuit. In order to do this, we use meters. Since current flows in a cable, we must connect the ammeter in series with the circuit to measure this flow. The ammeter, when connected, must not affect the current flow in any way. Because of this, the ammeter must have a very low resistance. As voltage is the pressure in the system, in order to measure voltage we must connect the voltmeter in parallel with the system or load. The voltmeter must also not affect the circuit in any way, when connected. Because of this, the voltmeter must have a very high resistance. NB: +ve (pos) and ve (neg) signs apply for d.c. instruments. A wattmeter is an instrument which measures the amount of power being supplied to a circuit. A wattmeter measures d.c. and a.c. power. The wattmeter is a combination of a voltmeter and an ammeter, and is connected into a circuit as shown. It is sometimes required to take all three readings at the same time, for example to measure the power factor in a circuit. When this is the case, the instruments are connected as shown on the right.

31 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 8 Ohmmeters When measuring resistance, we use an ohmmeter, which is simply connected across the circuit or piece of equipment to be tested. The ohmmeter requires its own internal supply, normally a battery. For this reason, the ohmmeter must never be connected across a circuit or piece of equipment that has a power supply connected to it, as this could cause damage to the meter and possibly the circuit or piece of equipment being tested. When reading low values, a low reading ohmmeter should be used; for higher readings (thousands of ohms and above) a high reading ohmmeter (normally an insulation resistance tester) should be used.

32 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 9 202: Principles of electrical science Handout 9: Electro-magnetism Learning outcome The learner will: 3. know the principles of electro-magnetism. Assessment criteria The learner can: 3.1. describe the magnetic flux patterns of electromagnets. Range Electromagnets: Current carrying conductor, solenoid, inductor, magnetic poles, relays. Magnetic field Electro-magnetism This is the area around a magnet or electromagnet where the effects of the magnetic force produced can be felt. Magnetism is represented by unseen lines of flux that form closed loops, as shown in the following diagrams. Some conventions must be remembered: Lines of flux cannot cross Lines of flux flow externally from the North Pole to the South Pole. Flux patterns for various arrangements of permanent magnets are shown below: Two permanent magnets North Pole to North Pole Like poles repel

33 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 9 Two permanent magnets North Pole to South Pole Unlike poles attract Electro-magnetism Electricity and magnetism are closely related. An electrical current flowing through a conductor produces a magnetic field in the form shown below, around the conductor. In order to help us to establish the direction of the magnetic fields around conductors, we must have a current direction convention, as shown below. Current flowing into the conductor (ie away from the observer) Current flowing out of the conductor (ie towards the observer) The direction of magnetic fields around cables can be found by using: Maxwell s screw rule Maxwell s screw rule shows the relationship between the direction of the current flowing and the magnetic field produced by that current. Imagine that you are driving a screw into the conductor in the direction of the current flow. The magnetic field produced by the current will form circular lines around the conductor, in the direction in which you have to drive the screw to advance it.

34 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 9 Magnetic fields due to electric current In straight conductors In a flat coil In a solenoid Winding the conductors into a coil/solenoid increases the magnetic effect and produces a magnetic field similar to a bar magnet with North and South Poles. A solenoid is a coil wound into a tightly packed helix. In physics, the term solenoid refers to a long, thin loop of wire, often wrapped around a metallic core, which produces a magnetic field when an electric current is passed through it. Solenoids are important because they can create controlled magnetic fields and can be used as electromagnets. The term solenoid refers specifically to a coil designed to produce a uniform magnetic field in a volume of space. A common use for a solenoid is in a relay (or contactor): an electrically operated switch. Relays are used where it is necessary to control a circuit by a low-power signal (with complete electrical isolation between control and controlled circuits) or where several circuits must be controlled by one signal. The first relays were used in long-distance telegraph circuits, repeating the signal coming in from one circuit and re-transmitting it to another. A type of relay that can handle the high power required to directly control an electric motor or other loads is called a contactor.

35 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 10: Generation of an EMF Learning outcome The learner will: 3. know the principles of electro-magnetism. Assessment criteria The learner can: 3.2 apply Fleming s right hand rule to the operating principles of a simple alternator 3.3 calculate magnitudes of a generated EMF. Generation of an EMF When a conductor cuts through a magnetic field, at right angles to the magnetic flux, an EMF (electromotive force) is induced in the conductor. The strength of the induced EMF is determined by: the strength of the magnetic flux density (between pole faces) BB the length of the conductor in the magnetic field ll the velocity or speed of the conductor through the magnetic field vv We can calculate the EMF by using the following formula. ee = BB ll vv Where: ee = iiiiiiiiiiiiii EEEEEE iiii vvvvvvvvvv BB = mmmmmmmmmmmmmmmm ffffffff iiii tttttttttt (TT) ll = llllllllllll oooo cccccccccccccccccc iiii mmmmmmmmmmmm (mm) vv = vvvvvvvvvvvvvvvv oooo cccccccccccccccccc iiii mmmmmmmmmmmm ssssss (mm ss) ee (vvvvvvvvvv) = BB (tttttttttttt) ll (mmmmmmmmmmmm) vv (mmmmmmmmmmmm ssssssssssss)

36 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 10 Example 1 A conductor of 15cm in length is moved at 20 metre/second (m/s) perpendicularly through a magnetic field of flux density 2 teslas. Calculate the induced EMF. e = Blv = = 66 vvvvvvvvvv Example 2 A conductor of 0.5m in length is moved at 30 metre/second (m/s) perpendicularly through a magnetic field of flux density 3 teslas. Calculate the induced EMF. e = Blv The direction of this current can be found by using: = = 4444 vvvvvvvvvv Fleming s right hand generator rule

37 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 11: a.c. generation Learning outcome The learner will: 3. know the principles of electro-magnetism. Assessment criteria The learner can: 3.4 state how an alternator produces a sinusoidal waveform output. a.c. generation When a conductor cuts through a magnetic field, at right angles to the magnetic flux, an EMF (electromotive force) is induced in the conductor. The diagram shows a single loop coil which rotates between the poles of a magnet. Slip-rings and carbon brushes are used to make contact with the coil. The former consist of a brass or copper shell connected to each side of the coil. As the loop rotates, its sides cut through the magnetic flux set up by the poles and so an EMF is induced in the loop. We have already seen that when a conductor moves across a magnetic field, it cuts the flux and an EMF is generated in it. When the conductor moves along or parallel to the magnetic field, no flux is cut and therefore no EMF is induced. If the conductor moves at right angles to the field, flux is cut at the maximum rate and a maximum EMF is induced. If the conductor cuts the field at an angle, the induced EMF will depend on the angle between the line taken by the conductor and the magnetic flux, which means that the induced EMF will be somewhere between 0 and maximum value. Consider the single loop rotated between poles, as in the diagram on the following page; the EMF generated will be as shown.

38 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 11 The single loop with slip-rings is a form of a.c. alternator. The simple alternator generates one cycle of its waveform for every complete revolution that it makes. If more pairs of poles are added, the alternator will produce one cycle for each pair of poles during one revolution. The frequency (symbol f) of an alternator is the number of cycles it produces every second. Frequency is measured in hertz (Hz). The waveform produced is known as a sine wave. A machine having P pairs of poles and running at N rev/sec generates a frequency of f Hertz. Therefore: ff = NN. PP where: ff = FFFFFFFFFFFFFFFFFF (HHHH) NN = SSSSSSSSSS iiii rrrrrrrrrrrrrrrrrrrrrr pppppp ssssssssssss PP = NNNNNNNNNNNN oooo pppppppppp oooo pppppppppp (a North and South are one pair)

39 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 11 Plotting a sine wave to represent an a.c. waveform

40 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 12: Sine wave quantities Learning outcome The learner will: 3. know the principles of electro-magnetism. Assessment criteria The learner can: 3.5 calculate sinusoidal quantities. Range Sinusoidal quantities: R.M.S. voltage, average voltage, instantaneous voltage, peak voltage, peak to peak voltage, periodic time, frequency. What is alternating current? Sine wave quantities Alternating current (a.c.) is the flow of electrons, which rises to a maximum value in one direction and then falls back to zero before repeating the process in the opposite direction. The electrons within the conductor do not just flow in one direction. They move backwards and forwards. The journey taken from start to finish is one cycle and the number of cycles that occur every second is said to be the frequency. When we look at a sine wave, there are several values that can be measured from the alternating waveform. These are shown on the following diagrams. Amplitude is the maximum voltage reached by the signal. It is measured in volts (V). Peak voltage is another name for amplitude. Peak to peak voltage is twice the peak voltage (amplitude). Frequency is the number of cycles per second. It is measured in hertz (Hz). 11 FFFFFFFFFFFFFFFFFF = TTTTTTTT pppppppppppp

41 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 12 Periodic time is the time taken for the signal to complete one cycle. It is measured in seconds (s). TTTTTTTT pppppppppppp = 11 FFFFFFFFFFFFFFFFFF The Instantaneous value of an alternating voltage or current is the value of voltage or current at one particular instant. The value may be zero if the particular instant is the time in the cycle at which the polarity of the voltage is changing. It may also be the same as the peak value, if the selected instant is the time in the cycle at which the voltage or current stops increasing and starts decreasing. There are actually an infinite number of Instantaneous values between zero and the peak value. The Average value of an alternating current or voltage is the average of all the Instantaneous values during one alternation. Since the voltage increases from zero to peak value and decreases back to zero during one alternation, the average value must be some value between those two limits. You could determine the average value by adding together a series of instantaneous values of the alternation (between 0 and 180 ) and then dividing the sum by the number of instantaneous values used. The computation would show that one alternation of a sine wave has an average value equal to times the peak value. The formula for average voltage is: VV aaaaaa = VV ppffaapp Where V avg is the average voltage of one alternation, and V peak is the maximum or peak voltage. Similarly, the formula for average current is: II aaaaaa = II ppffaapp Where I avg is the average current in one alternation, and I peak is the maximum or peak current. Do not confuse the above definition of an average value with that of the average value of a complete cycle. Since the voltage is positive during one alternation and negative during the other alternation, the average value of the voltage values occurring during the complete cycle is zero.

42 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 12 The effective value of an alternating current or voltage is the value of alternating current or voltage that produces the same amount of heat in a resistive component that would be produced in the same component by a direct current or voltage of the same value. The effective value of a sine wave is equal to times the peak value. The effective value is also called the root mean square or RMS value. The formula for RMS voltage is: VV rrrrrr = VV ppffaapp Where V rms is the RMS voltage of one alternation, and V peak is the maximum or peak voltage. Similarly, the formula for RMS current is: II rrrrrr = II ppffaapp Where I rms is the average current in one alternation, and I peak is the maximum or peak current. Also: and VV ppffaapp = VV rrrrrr II ppffaapp = II rrrrrr The RMS value is the effective value of a varying voltage or current. It is the equivalent steady d.c. (constant) value which gives the same effect. For example, a lamp connected to a 6V RMS a.c. supply will light with the same brightness when connected to a steady 6V d.c. supply. However, the lamp will be dimmer if connected to a 6V peak a.c. supply because the RMS value of this is only 4.2V (it is equivalent to a steady 4.2V d.c.). Let s look at the voltage we use on a daily basis. VV rrrrrr = VV pppppppp = = What does 230V a.c. really mean? Is it the RMS or peak voltage? If the peak value is meant then it should be clearly stated, otherwise assume it is the RMS value. In everyday use a.c. voltages (and currents) are always given as RMS values because this allows a sensible comparison to be made with steady d.c. voltages (and currents), such as from a battery. For example, a 6V a.c. supply means 6V RMS ; the peak voltage is 8.6V. The UK mains supply is 230V a.c.; this means 230V RMS so the peak voltage of the mains is about 325V! VV pppppppp = VV rrrrrr = =

43 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 13: a.c. distribution Learning outcome The learner will: 3. know the principles of electro-magnetism. Assessment criteria The learner can: 3.6 state the reason for a.c. distribution. Range a.c. distribution: Step up transformers, step down transformers. a.c. distribution Why use alternating current (a.c.) instead of direct current (d.c.) With direct current the electromotive force (EMF) is always in one direction only. The resultant flow of electrons (electric current) will also be in one direction only and therefore current flow is always in one direction. However, with alternating current the EMF periodically reverses its direction. The resultant flow of electrons will also periodically reverse its direction and therefore current flow periodically reverses its direction. Alternating current is used for the electricity supply network in Britain and most countries of the world. The main reasons for this are as follows: a.c. can be easily transformed from one voltage to another a.c. produces less arcing and sparking on contacts, etc If electricity is transmitted and distributed nationally at very high voltages, the current will be low and, hence, transmission losses will be low. As voltage drop in a cable can be found by multiplying the cable resistance by the current, with low current the voltage drop will also be low. This means we can use cables with a smaller cross-sectional area, thus reducing costs. The high voltages used for transmission are far too dangerous for consumer use so must be reduced down to the levels that we are familiar with (230 volt single phase; 400 volt three phase). In the UK, for example, generators in power stations produce electricity at around 25,000 volts (25kV). By using a step-up transformer, this is then increased to 400,000 volts (400kV) for transmission around the country from the power stations to the load centres. By using a step-down transformer when these transmission lines are in the vicinity of the load centres, the voltage is stepped down for secondary transmission (132kV; 66kV). When the load centre is reached, it is stepped down again for local distribution at 33kV and 11kV. Supplies to individual users will see a further step-down to 400V for commercial and industrial users (heavy industry will be supplied at 33kV or 11kV depending on demand) and 230V for domestic users. The device used to transform the voltages is a transformer, which will only work on alternating current not direct current and this is why we do not transmit and distribute electricity using d.c.. See the diagram on the following page for the transmission and distribution supply system.

44 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 13

45 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 14: Star Delta configurations Learning outcome The learner will: 5. know electrical quantities in Star Delta configurations. Assessment criteria The learner can: 5.1. differentiate between voltages and currents in Star configured loads 5.2. differentiate between voltages and currents in Delta configured loads 5.3. describe why single phase loads should be balanced across a three line supply. Range Voltages: Line voltage, phase voltage. Currents: Line current, phase current. Balanced: No neutral current, neutral current values (simple methods). Star Delta configurations Three phase supplies In Unit 11 we saw that rotating a coil in a magnetic field produced an EMF and that if this EMF is fed out via slip-rings then an alternating current is produced. If three separate windings are equally positioned around the rotor then three separate alternating currents will be produced, each 120 out of phase with each other, as shown in the diagram below. Depending on how these windings are connected together, a range of voltages are available and these can be used to power a variety of different loads. Connection of three phase supplies There are two main methods of connecting the phases to produce a three phase system. These are called star connection and delta connection (also known as mesh).

46 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 14 Star connection It can be seen that this connection produces a voltage that is 3 times larger between any two connecting lines than between any line and neutral (or earth). The current in each phase coil is the same as the current in the line connected to it. VV LL = 33 VV PP & II LL = II PP Example 1 A three phase, four wire, star connected transformer has a line voltage of 520 volts and supplies a line current of 25 amps. Calculate: a) the phase voltage of the transformer b) the phase current in each winding. a) IIII ssssssss VV LL = 33 VV PP VV LL VV pp = 33 = = vvvvvvvvvv b) IIII ssssssss II LL = II PP II PP = 2222 aaaaaaaaaaaaaa

47 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 14 Delta connection When this connection is used, it can be shown that the line voltage is the same as the phase voltage and that the line current is 3 times larger than the phase current. VV LL = VV PP & II LL = 33 II PP Example 2 A three phase, balanced, delta connected resistive load is supplied from a transformer at a line voltage of 400 volts and draws a line current of amps. Calculate: a) the phase voltage b) the phase current c) the resistance of the load. a) In delta V p = V L V P = 400 volts b) In delta I L = 3 I P I P = I L 3 = = 8 amps c) R = V P I P = = 50Ω

48 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 14 Balancing loads All local distribution in England and Wales is by underground cables from sub-stations placed close to the load centre and supplied at 11kV. Transformers in these local sub-stations reduce the voltage to 400 volts; three phase and neutral distributor cables connect this supply to the consumers. Connecting to one phase and neutral of a three phase 400 volt supply gives a 230 volt single phase supply suitable for domestic consumers. When single phase loads are supplied from a three phase supply, as shown in the diagram below, the load should be balanced across the phases. That is, the load should be equally distributed across the three phases so that each phase carries approximately the same current. This prevents any one phase being overloaded. On the standard supply system in Great Britain, the secondary of the supply transformer is connected in star, as shown in the diagram below. If the loads on each phase are balanced, ie the same, then the neutral current at the supply transformer will be zero; this is the ideal condition. However, if the individual phase currents are not the same because the system is not balanced then there will be some current flow in the neutral at the supply transformer. This neutral current can be found using a number of ways but the simplest is to draw the three individual currents to scale in a phasor diagram. Example 3 A three phase, unbalanced, star connected system has the following loads connected to each phase: brown phase 10 amperes black phase 20 amperes grey phase 15 amperes. Using the graphical method (phasors), determine the neutral current. NNNNNNNNNNNNNN cccccccccccccc =

49 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 15: a.c. theory Learning outcome The learner will: 7. know the principles of a.c. theory. Assessment criteria The learner can: 7.1. state the effects of components in a.c. circuits. Range Effects: Resistance, inductance, capacitance, impedance, reactance. Resistance in an a.c. circuit Look at the circuit on the right. a.c. theory Here a resistor or load is connected to a d.c. supply via a switch. Using an ammeter and a voltmeter connected as shown, the resistance of the load can be determined using Ohm s law: RR = VV II There should be no problem carrying out calculations on this type of circuit. We will now consider the same component connected to an a.c. circuit. Here is the same diagram again. It is a similar diagram, but this time the load is connected to an a.c. supply. The first thing that should be noticed is that the symbol for the supply looks slightly different. It will be remembered from previous work that RMS value is used as far as a.c. is concerned and not the peak or peak to peak value. The RMS value is a relationship between a.c. and d.c. The RMS value is the equivalent d.c. heating effect. So if there is a 230V d.c. supply then this value can be related to a 230V RMS a.c. supply. So what is the difference between a d.c. and an a.c. supply in its effect on this circuit? Nothing! In a d.c. circuit the resistor assuming that it is a pure resistor (without inductance) acts to oppose the current flow. It acts a bit like a pinch on a hosepipe, which stops or slows the flow of water through the pipe. With an a.c. circuit the same pure resistor acts in exactly the same way and therefore there is no difference! However, the next step in thinking that you have to make is that of differentiating between what is happening to the current and what is happening to the voltage in this a.c. circuit. Since the a.c. supply is produced from a rotating alternator, the voltage and the current have both a quantity (how large is the value) and direction (in which sense is it acting). As they have quantity and direction, they can be said to be vector quantities.

50 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 15 This waveform shows the current and the voltage superimposed on the same diagram. It can be seen that, as time passes, the voltage and current get either bigger or smaller, depending on where they are in their cycle. Here the voltage and current depend on the size of the alternator, the number of windings and how they have been connected, and the speed of rotation of the rotor. The direction is continually changing because the rotor is continually turning. This is, of course, why we get an a.c. supply. Now that we know that a.c. is a vector quantity, we can draw any values that we have in the form of a vector diagram. In electrical terms, we call a vector diagram a phasor diagram. We can simplify the sine waves in the diagram above into a phasor diagram, which is an electrical vector diagram. Assuming the voltage is 100V and the current is 15A, the voltage is drawn to scale along the horizontal axis. The current, being in-phase with the voltage, is also drawn to scale along the same horizontal axis as the voltage. This may not appear to help us very much at the moment, but more will become apparent as we move on and look at other components. For now, you should know that the resistance in an a.c. circuit can be treated in exactly the same way as it was in a d.c. circuit. Summary Pure resistance causes the current to follow the voltage exactly and is therefore considered to be in phase. A good understanding of trigonometry is essential to an understanding of the effects of various components on an a.c. supply. Inductance in an a.c. circuit Inductance occurs when a length of wire is wrapped into a coil and the magnetic field around the coil increases as the current flows. The magnetic field that is produced changes for a short period while it builds up, and this change leads to an EMF being induced in the coil. This induced EMF opposes the supply voltage and this obeys Lenz s law. The induced, or back EMF, creates an induced current, which opposes the supply current. In this diagram it can be seen that during the positive half-cycle of the a.c. supply, the magnetic field flows towards the right (North). On the negative half-cycle, the magnetic field acts in the opposite direction and the North is now on the left. The level of inductance depends on the number of turns on the coil and the rate of change of flux (how quickly the magnetic field changes).

51 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 15 Rather than being bogged down in formulae, the important thing to note is that the induced EMF opposes the supply. This seems very strange, as a very real second voltage is produced. Now this inductive effect is compounded in an a.c. circuit. Since the supply is constantly changing (a.c. supply), the magnetic field surrounding the coil must be constantly changing. Also, as the magnetic field is constantly changing then the induced EMF will be continually changing. The supply current causes a magnetic field to be produced around the coil of the inductor. This coil then induces an EMF back into the coil, which we call a back EMF. The induced current, however, has a time delay in its effect on the supply current and we get what is called phase shift. The time delay is caused by the time taken for the magnetic field to build up. Effectively, the supply current is no longer in-phase with the supply voltage but has been shifted back 90. This diagram shows the supply voltage and the supply current if there is no inductive effect a pure resistor. Here with an inductor instead of a resistor it can be seen that the current waveform is lagging the voltage waveform by 90, or a quarter of a cycle. The thing to remember is that a resistor opposes current flow and an inductor opposes current change. The current flowing through a resistor is in-phase with the supply voltage, whilst the current through an inductor is lagging the voltage by 90, or a quarter of a cycle. Try to get into the habit of drawing the lines of a phasor to scale. As for the resistor, this effect can be drawn as a phasor diagram. Remember that the phasor diagram is a representation of the current and voltage in this particular instance. We have now looked at a pure resistor and a pure inductor. We know that it is almost impossible to have a pure anything. However, in order to give a clearer understanding of what is happening, we make certain assumptions; purity is one of these assumptions.

52 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 15 Capacitance in an a.c. circuit We are now going to look at what happens when a pure capacitor is connected in an a.c. circuit. By now it should be remembered that a capacitor is a device that is capable of storing an electric charge. The picture of the hydraulic pump shows how energy can be stored in a device even when there is no direct link between the two parts. As force is applied to the pump, the diaphragm moves. If the pump were then pulled in the opposite direction, the diaphragm would move in the opposite direction. If we now take this picture and apply it to our a.c. circuit, the capacitor is charging and discharging every time the alternating supply moves from its positive half-cycle to its negative half-cycle. In this circuit the capacitor is first charged by the positive halfcycle and then discharges when the negative half-cycle comes along. As you might expect, this charging and discharging of the capacitor have an effect on the supply current. We will look at what happens on the first half-cycle of the supply and you should get some idea of what effect there is on the current. You can see the a.c. waveform produced by the voltage supplied to the circuit. You now have to consider what happens to a capacitor when it is being charged up. When the supply is first switched on at that very instant there are zero volts applied to the capacitor. At that same point, however, the charging current is at a maximum. As the capacitor charges up, the voltage across it rises and the current begins to fall to the point where the voltage across the capacitor reaches a maximum and the current reaches 0A (the capacitor is fully charged). This effect causes the current in the capacitor to lead the voltage by 90, or a quarter of a cycle.

53 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 15 This relationship can be drawn using a phasor diagram, as we have done for the resistor and the inductor. This time, however, the current is shown leading the voltage. Notice that the phasor diagram for the capacitor is drawn exactly opposite to the phasor diagram of the inductor. In an a.c. circuit the capacitance does tend to counteract the effect of the inductor in the circuit, and this effect is used in power factor correction, which will be explained later on. Summary This session can be quite difficult to understand fully. Inductors and capacitors oppose current change, whilst resistors oppose current flow. In an a.c. circuit where the supply voltage is constantly changing, the effect of capacitors and inductors tends either to cause the current to lead or lag the supply voltage. This lead or lag effect is called a phase shift.

54 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 16: Inductive reactance Learning outcome The learner will: 7. know the principles of ac theory. Assessment criteria The learner can: 7.1. state the effects of components in a.c. circuits. Range Effects: Resistance, inductance, capacitance, impedance, reactance. Inductive reactance Determining the effects of inductance in an a.c. circuit In handout 15 we looked at what affects a resistor; an inductor and a capacitor have on the current in an a.c. circuit. However, we need to know how we can measure this effect. With our pure resistor, things are relatively simple and Ohm s law still applies, ie V = IR. With resistance, there is no difference in the value of the resistor, even when other things vary, such as frequency. It is solely affected by temperature change. With the inductor and capacitor, things are somewhat different. The term used for the inductor and capacitor when trying to express their effects on the circuit is called reactance. The label attached to reactance is X. If we are considering an inductor, we are looking at inductive reactance and the label is X L. If we happen to be considering a capacitor then we are dealing with capacitive reactance and the label is X C. Remember A resistor opposes or resists current flow, whilst the capacitor and inductor oppose or react against current change. The reactance or the opposition to change of both the inductor and the capacitor depends mainly on the frequency of the supply. There are two formulae that are attached to the inductor and capacitor. Inductive reactance XX LL = where: XX LL = IIIIIIIIIIIIIIIIII rrrrrrrrrrrrrrrrrr (ΩΩ) ff = FFFFFFFFFFFFFFFFFF (HHHH) LL = IIIIIIIIIIIIIIIIIIII (HH)

55 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 16 Example 1 A pure inductor has an inductance of 0.5H. Determine the inductive reactance if the frequency is: a) 50Hz b) 100Hz. a) X L = 2πfL = = b) X L = 2πfL NB = = The frequency doubles and so does the reactance. Reactance is still measured in ohms. Example 2 An inductor has a 60Hz supply and has an inductive reactance of 25Ω. Calculate the inductance of the inductor. X L = 2πfL Transpose to find L = X L 2πf = = henrys = 6666 mmmm The value of X L can be used in a variation on Ohm s law: VV = II. XX LL In this instance, X L has replaced the resistance value. Working out this type of problem is the same as it would be for a normal Ohm s law question.

56 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 16 Example 3 A 220V, 60Hz a.c. supply is connected to a pure inductor. The inductor has an inductance of 0.5H. Calculate the inductive reactance and the current flowing through it. X L = 2πfL = = V = I. X L I = V X L 220 = = In this example, we have first worked out the reactance. Remember that this is a measure of the opposition to current change in a circuit. After we know the reactance then we can determine the current flowing. It can be seen that when frequency changes, there will be a corresponding change in the current. Let s look at the last example with a reduction in frequency to 50Hz. X L = 2πfL = = V = I. X L I = V X L 220 = = Some quick maths tells us that for this reduction in frequency there has been nearly a 20% increase in current. This is why, of course, supply generators are only allowed to vary the frequency of the supply by 1% or 0.5Hz either side of 50Hz. We ll try another example before moving on to look at pure capacitance.

57 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 16 Example 4 A pure inductor of inductance 0.5H is connected across a 230V supply. Calculate the current flowing at a frequency of: a) 50Hz b) 25kHz. a) X L = 2πfL = = V = I. X L I = V X L 230 = = b) X L = 2πfL = = 7777, V = I. X L I = V X L = ,550 =

58 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 16 An increase in frequency leads to a reduction in the current. This effect has some very real practical benefits, particularly in the new, low power, mini-fluorescent all-in-one lamps, which run at a frequency of approximately 30kHz. Notice also that this allows cable sizes to be dramatically reduced for the same output. Have a look at the graph below. The vertical axis shows reactance and current, whilst the horizontal axis shows the frequency. As the frequency increases, the reactance increases in direct proportion. We can state this because we are getting a straight line. We can say that the inductive reactance is directly proportional to the frequency: f X L. As the frequency increases, the current falls, but not in a straight line. If the frequency were to fall to zero, which would effectively give us d.c., then the current would be infinite. This is nonsense. What would actually happen is that the supply would simply be looking at the resistance of the conductors and not the reactance, because we can never get a pure inductor.

59 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 17: Capacitive reactance Learning outcome The learner will: 7. know the principles of a.c. theory. Assessment criteria The learner can: 7.1. state the effects of components in a.c. circuits. Range Effects: Resistance, inductance, capacitance, impedance, reactance. Capacitive reactance The formula for capacitive reactance is as follows: XX CC = where: XX CC = CCCCCCCCCCCCCCCCCCCC rrrrrrrrrrrrrrrrrr (ΩΩ) ff = FFFFFFFFFFFFFFFFFF (HHHH) CC = CCCCCCCCCCCCCCCCCCCCCC (FF) Example 1 An 8.7µF capacitor is connected across an a.c. supply. Calculate the capacitive reactance if the frequency is 60Hz. X C = 1 2πfC = = Here the reactance is again measured in ohms, as it was with inductive reactance. Example 2 A 0.1µF capacitor is connected across an a.c. supply. Calculate the reactance at: a) 50Hz b) 15kHz. a) X C = 1 2πfC = = 3333,

60 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 17 b) X C = = 1 2πfC = Again, you should notice that frequency plays the largest part in the effect on the reactance in the circuit. Example 3 The capacitive reactance of a circuit is 250Ω. If the frequency of the supply is 50Hz, calculate the capacitance of the capacitor. X C = 1 2πfC Transpose to find C = 1 2πfX C = = F = µFF As with inductive reactance, there is a similarity to Ohm s law, when we are dealing with how the reactance relates to the current and voltage. The value of X C can be used in a variation on Ohm s law: VV = II. XX CC In this instance, X C has replaced the resistance value. Working out this type of problem is the same as it would be for a normal Ohm s law question. Example 4 A pure capacitor of capacitance 2.5µF is connected across a 220V supply. Calculate the current flowing at a frequency of: a) 50Hz b) 25kHz. a) X C = 1 2πfC = = 11,

61 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 17 b) X C = V = I. X C I = V = X C 220 1,273 = 0.173A = = 1 2πfC = V = I. X C I = V X C 220 = 2.55 = In the first part, the frequency of 50Hz creates a relatively small current, whereas at 25kHz, in the second part, the current has risen dramatically. A graph expresses this information, as with the inductive reactance. The vertical axis shows reactance and current, whilst the horizontal axis shows the frequency. As the frequency increases, the current increases in direct proportion. We can say that the current reactance is directly proportional to the frequency: I f. As the frequency increases, the reactance falls but not in a straight line. If the frequency were to fall to zero, which is in effect d.c., then the reactance would be infinite and current would be blocked. This is exactly what happens. In fact, in electronic circuits, capacitors are used to block direct current. If the frequency were to increase to infinity then the reactance would fall to zero and the current would correspondingly increase. This is exactly the opposite of what occurred with the inductive reactance.

62 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 17 Summary Resistance in an a.c. and d.c. circuit is the same and can be determined using Ohm s law. Resistance opposes current flow in a circuit. Phasor diagrams display vector quantities. In a purely resistive circuit, the voltage and current are in-phase with each other. A pure inductor opposes the current change in an a.c. circuit and the current lags the voltage by 90. A capacitor causes the current to lead the voltage by 90 and a capacitor has the opposite effect to an inductor.

63 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 18: Impedance Learning outcome The learner will: 7. know the principles of a.c. theory. Assessment criteria The learner can: 7.1. state the effects of components in a.c. circuits. Range Effects: Resistance, inductance, capacitance, impedance, reactance. Impedance In circuits where there is more than one component, we must find the total opposition (Ω) of the circuit before the amount of current flowing can be found. In an a.c. circuit the total opposition to current flow is called the impedance and the symbol for it is (Z). The total opposition is measured in ohms (Ω). Impedance is the name given to the combined effect of resistance (R) and inductive reactance (X L) or capacitive reactance (X C), and is measured in ohms (Ω). The impedance is calculated by adding the phasors of the resistance and reactance(s) together, ie the phasor sum of the reactances and resistances. For series circuits, impedance (Z) is calculated by one of the formulae given below: IIIIIIIIIIIIIIIIII cccccccccccccccc ZZ = RR 22 + XX LL 22 CCCCCCCCCCCCCCCCCCCC cccccccccccccccc ZZ = RR 22 + XX CC 22 CCCCCCCCCCCCCCCC cccccccccccccccc ZZ = RR 22 + (XX CC XX LL ) 22 With the last formula, whichever is the larger (X L or X C) is put first, ie (X L X C) or (X C X L) respectively. The current in any a.c. circuit may be found by first finding the impedance of the circuit (Z) and then by dividing V S by Z. II = VV SS ZZ A coil will always have both resistance and inductance, and although they cannot be physically separated, it is convenient for the purpose of calculating impedance, etc to show them as separate in circuit diagrams.

64 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 18 Example 1 A circuit supplied at 230V 50Hz consists of a resistor of 25Ω and an inductor of 0.05H connected in series. Calculate: a) the impedance of the circuit b) the current flowing in the circuit. a) X L = 2πfL b) = = 15.71Ω Z = R 2 + X L 2 = = I = V Z = = 7.7A Example 2 A circuit supplied at 100V 100Hz consists of a resistor of 100Ω and a capacitor of 10µF connected in series. Calculate: a) the impedance of the circuit b) the current flowing in the circuit. a) XC = = 1 2πfC = 159Ω Z = R 2 + X C 2 b) = = I = V Z = = 0.53A

65 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 18 Impedance triangles If we draw the phasor diagram for a circuit containing resistance and inductance, it will look like the phasor diagram shown immediately below. Since I is present in each term, if we divide each term by I then we get what is known as the impedance triangle. For a circuit containing R and X L If we draw the phasor diagram for a circuit containing resistance and capacitance, it will look like the phasor diagram shown immediately below.

66 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 18 For a circuit containing R and X C For a circuit containing R, X L and X C

67 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 19: a.c. power Learning outcome The learner will: 7. know the principles of a.c. theory. Assessment criteria The learner can: 7.2. state characteristics of power quantities for an a.c. circuit. Range Power quantities: Apparent power (KVA), Reactive power (KVAr), True power (KW), power triangles. a.c. power When we looked at the power consumed in a circuit connected to a d.c. supply, the supply voltage and current drawn from the supply were in-phase with each other, ie there was no difference in the starting point in time, between either the voltage or current. With a.c. this is not necessarily the case, as we have seen from the phasor diagrams in previous sections. If the voltage and current are not in-phase with each other then we cannot calculate power by the formula V x I, as this will give a false and much greater value for power than is actually being dissipated. The answer you get from multiplying V x I is the Volt Amperes or Apparent power referred to as VA. In order to obtain the True power (P) dissipated in a circuit, we have to take into account the phase angle between the supply voltage and supply current. The True power dissipated in a single phase a.c. circuit can be found by using either of the formulae shown below. TTTTTTTT pppppppppp, PP = II 22. RR or TTTTTTTT pppppppppp, PP = VV. II. cccccccc Cosφ is the phase angle and is the cosine of difference in degrees between the applied voltage and the supply current. The power factor (pf) is the cosine of the angle and is just a decimal number. For example, the cosine of 30 is The pf can never be greater than one (1), since the cosine wave maximum is one (1). The relationship between True and Apparent power is: pppp = PP VVVV where: pf = Power factor expressed as the cosine of the phase angle difference P = True power in watts VA = Apparent power in VA.

68 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 19 There is a further value that needs to be considered and that is power produced in the reactive components (inductors and capacitors). This does no useful work, as power is taken on one halfcycle and returned on the next half-cycle. This is referred to as VAr and is known as the Reactive power. Between this Reactive power and the true power there is a phase angle difference of 90. The relationship between these three values can be represented by the power triangle based on a right-angled triangle: Since the impedance triangle is a right angle triangle, if two of the values are known then the third can be found using Pythagoras Theorem. The following are the relevant formulae: AAAAAAAAAAAAAAAA pppppppppp = TTTTTTTT pppppppppp 22 + RRRRRRRRRRRRRRRR pppppppppp 22 TTTTTTTT pppppppppp = AAAAAAAAAAAAAAAA pppppppppp 22 RRRRRRRRRRRRRRRR pppppppppp 22 RRRRRRRRRRRRRRRR pppppppppp = AAAAAAAAAAAAAAAA pppppppppp 22 TTTTTTTT pppppppppp 22 Power factor can be determined by: pppp = TTTTTTTT pppppppppp AAAAAAAAAAAAAAAA pppppppppp

69 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 19 Example 1 When a voltage of 230 volts is applied to a circuit, a current of 5 amperes flows at a power factor of Calculate: a) the Apparent power b) the True power c) the Reactive power. a) VA = V I = = 11, b) P = V. I. cosϕ = = 977.5W c) VAr = VA 2 P 2 = 1, = 605.8VAr Example 2 When a voltage of 400 volts is applied to a circuit, a current of 25 amperes flows at a power factor of 0.6. Calculate: a) the Apparent power b) the True power c) the Reactive power. a) VA = V I = = 10,000VA = b) P = V. I. cosϕ = = 6,000W c) VAr = VA 2 P 2 = 10, ,000 2 = 8,000VAr = 8KVAr

70 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 20: Power factor Learning outcome The learner will: 7. know the principles of a.c. theory. Assessment criteria The learner can: 7.3. describe why power factor correction is required 7.4. describe how power factor correction may be achieved. Range Correction: Use of capacitors, load correction, bulk correction, synchronous motor. Power factor The power factor of an a.c. electrical power system is defined as the ratio of the True power flowing to the load to the Apparent power in the circuit, and is represented by a dimensionless number between 0 and 1. Power factor is a measure of how efficiently electrical power is consumed. In an ideal world, power factor would be unity (or 1). Unfortunately, in the real world power factor is reduced by highly inductive loads to 0.7 or less. This induction is caused by equipment, such as lightly loaded electric motors, luminaire transformers and fluorescent lighting ballasts, and welding sets. Drawback of low power factor In an electric power system, a load with a low power factor draws more current than a load with a high power factor for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system, and require larger wires and other equipment. Due to the costs of larger equipment and wasted energy, regional electricity companies will usually charge a higher rate to industrial or commercial customers where there is a low power factor. What causes low power factor? Inductive loads (which are sources of Reactive power) include: transformers induction motors fluorescent luminaires. These inductive loads constitute a major portion of the power consumed in industrial installations and cause a lagging power factor, ie the current lags the voltage. Power factor correction In order to correct a lagging power factor, we connect a device that creates a leading power factor. This usually takes the form of: capacitors synchronous motor. A synchronous motor can operate at leading or unity power factor and thereby provide powerfactor correction when connected to a circuit with a lagging power factor. Capacitors can be connected across the whole installation at the supply (bulk correction) or across individual pieces of equipment (load correction). For example, individual fluorescent light fittings have a capacitor connected across their incoming line and neutral terminals.

71 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 20 Consider an inductive load taking a current I L lagging by a phase angle ϕ from a supply of E volts. In order to improve the overall power factor (pf), a capacitor of C µf taking a current I C can be connected in parallel with the load, as shown below. The phasor sum of the capacitor current I C and the load current I L produces a resultant current I T which is at a smaller phase angle and, hence, a better power factor (pf) than the original load current I L. It can be seen from the above explanation that if the power factor of a circuit is low then the amount of current required to supply a given load can be large, but by adding a capacitor in parallel, the supply current is decreased and the power factor is increased. A LOW power factor has a LARGE angle in degrees. A HIGH power factor has a SMALL angle in degrees.

72 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 21: Basic mechanics Learning outcome The learner will: 4. know the principles of basic mechanics. Assessment criteria The learner can: 4.1 calculate quantities of mechanical loads 4.2 calculate the efficiency of a machine expressed as a percentage. Range Quantities: Force, mass, gravity, acceleration, work done, power, time. Levers: 1st order levers, 2nd order levers. Basic mechanics Mass Mass can be defined as the amount of material in an object. The SI unit of mass is the kilogram (kg). Acceleration When an aircraft takes off, it starts from rest and increases its velocity until it can fly. This change in velocity is called acceleration. Acceleration is the change in velocity with time. where: DDDDDDDDDDDDDDDD = mmmmmmmmmmmm TTTTTTTT = ssssssssssssss AAAAAAAAAAAAAAAAAAAAAAAA = DDDDDDDDDDDDDDDD TTTTTTTT (mm ss22 ) The SI unit for acceleration is the metre per second per second or m/s 2. Example 1 If a car accelerates from a velocity of 3 m/s to 15 m/s in 4 seconds, calculate its average acceleration. Average velocity = 15 m s 3 m s = 12 m s Average acceleration = Velocity Time = 12 4 = 33 mm ss 22 Thus, the average acceleration is 3 metres per second, every second.

73 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 21 Force The SI unit of force is the Newton (N). A mass of 1 kilogram experiences a force of 9.81 Newtons as a result of gravity. In other words, it would require a force of 9.81 Newtons to raise a mass of 1kg against the pull of gravity (on Earth). Example 2 A bundle of conduit has a mass of 800kg. What force will be required to lift it? Force required = Mass Newtons = = 77, NNNNNNNNNNNNNN A force can be described as a push or a pull. When a force acts on a body, it may: a) accelerate the body b) decelerate the body c) deform the body, ie change its shape d) be exactly resisted by other forces (equilibrium). Work done If a force is applied to a body and movement takes place then it is said that work has been done. For example, when a weight is lifted, work is done. Work is measured in terms of the distance moved by the body and the force that caused its movement. When the movement is in the same direction as the force, the work done is equal to the distance moved, multiplied by the force exerted. WWWWWWWW dddddddd = FFFFFFFFFF DDDDDDDDDDDDDDDD (NNNN) Force is measured in Newtons. In order to convert weight (kg) into force, we must multiply the weight by 9.81, ie force = kg x 9.81 Newtons FFFFFFFFFF = MMMMMMMM NNNNNNNNNNNNNN Example 3 What work must be done to lift a 200kg bundle of conduit from the floor on to a rack 2 metres high? WWWWWWWW dddddddd = FFFFFFFFFF DDDDDDDDDDDDDDDD = = 33, NNNN

74 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 21 Energy Energy is the capacity to do work. It may take many forms, such as nuclear, chemical, heat, mechanical or electrical energy. The units of energy are the same as those for work, ie the Newton-metre or Joule. Power Efficiency FFFFFFFFFF = JJJJJJJJJJJJ = WWWWWWWWWW ssssssssssss WWWWWWWW oooo wwwwwwww dddddddd = EEEEEEEEEEEE = FFFFFFFFFF DDDDDDDDDDDDDDDD EEEEEEEEEEEE = JJJJJJJJJJJJ = WWWWWWWWWW ssssssssssssss PPPPPPPPPP iiii dddddddddddddd aaaa TTTTTT RRRRRRRR OOOO DDDDDDDDDD WWWWWWWW The unit of Power is the watt. Since the watt is a small unit, the kilowatt is often used. 1kW = 1,000 watts 1 hour = seconds = 3,600 seconds 1kWh = 1,000 3,600 Average power = = 3,600,000 joules amount of work done time taken to do it The efficiency of a system can be defined as the ratio of output power to input power and is usually expressed as a percentage. EEEEEEEEEEEEEEEEEEEE = OOOOOOOOOOOO eeeeeeeeeeee oooo pppppppppp IIIIIIIIII eeeeeeeeeeee oooo pppppppppp % The difference between input power and output power is the wasted energy or losses. Therefore: Output = Input Losses or Input = Output + Losses

75 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 21 Example 4 A motor produces 100 watts output power and is: a) 50% efficient b) 70% efficient. Calculate its input power in each case. Efficiency = Output power Input power 100 Input power = Output power Efficiency a) Input power = = wwwwwwwwww b) Input power = = wwwwwwwwww 100 Example 5 An electric motor drives a pump that lifts 1,000 litres of water each minute to a tank 20 metres above ground level. Calculate the power that the motor must provide if the pump is only: a) 50% efficient b) 80% efficient. NB: One litre of water weighs 1kg (1kg = 9.81N) Work = Force Distance Force = Mass in kg 9.81 = 1, = 196,200 Nm minute or Joules Joules = Watts seconds Watts = Joules seconds

76 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 21 a) b) Power (Watts) = Efficiency = Input power = 196, (because time taken is 1 minute) = 3,270 watts (output power) Output power Input power 100 Output power Efficiency Input power = 3, = 66, wwwwwwwwww Input power = 3, = 44, wwwwwwwwww 100

77 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 22: Levers Learning outcome The learner will: 4. know the principles of basic mechanics. Assessment criteria The learner can: 4.3 calculate mechanical advantage gained by use of levers. Range Quantities: Force, mass, gravity, acceleration, work done, power, time. Levers: 1 st order levers, 2 nd order levers. Levers A lever is a machine consisting of a beam or rigid rod pivoted at a fixed hinge, or fulcrum. A lever amplifies an input force to provide a greater output force, which is said to provide leverage. Classes of lever Levers are classified by the relative positions of the fulcrum and the input and output forces. It is common to call the input force the effort and the output force the load. This allows the identification of three classes of levers (or order) by the relative locations of the fulcrum, the load and the effort. Class 1. Fulcrum in the middle: the effort is applied on one side of the fulcrum and the resistance on the other side; for example, a crowbar or a pair of scissors. Class 2. Resistance in the middle: the effort is applied on one side of the resistance and the fulcrum is located on the other side; for example, a wheelbarrow or a nutcracker or a bottle opener.

78 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 22 Class 3. Effort in the middle: the resistance is on one side of the effort and the fulcrum is located on the other side; for example, a pair of tweezers or the human mandible. Lever calculations The effort to be applied to a lever will depend on the weight of the load, how far from the fulcrum the load is and how far from the fulcrum the effort is applied. This can be summarised as follows: EEEEEEEEEEEE = LLLLLLLL LLLLLLLL tttt ffffffffffffff dddddddddddddddd EEEEEEEEEEEE tttt ffffffffffffff dddddddddddddddd Example 1 A crowbar is used to lift a packing case. The load exerted by the packing case is 1,200N. Determine the effort needed to lift the packing case if the load is 10cm from the pivot and the effort is 1.0m from the pivot. Effort = Load Load to fulcrum distance Effort to fulcrum distance = 1, = Example 2 A horizontal bar 1.75m in length is pivoted at a point 0.75m from one end and a downward force of 150N is applied at right angles to this end of the bar. Calculate the force that must be applied to the other end of the bar to maintain it in a horizontal position. Ignore the weight of the bar. Effort = Load Load to fulcrum distance Effort to fulcrum distance = =

79 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 23: Force on current-carrying conductor Learning outcome The learner will: 6. know the operating principle of a range of electrical equipment. Assessment criteria The learner can: 6.1 describe the operating principle of electrical equipment 6.2 apply Fleming s left hand rule to determine the direction of rotation of a motor. Range Electrical equipment: Basic a.c. motors, basic d.c. machines, fluorescent luminaires, relays. Force on current-carrying conductor When a conductor is situated in a magnetic field at right angles to it and a current is then passed through the conductor, the latter will experience a mechanical force, causing it to move due to the interaction of the two magnetic fields. If the direction of the main magnetic field or the current in the conductor were reversed then the force would be in the opposite direction, as shown. Changing direction of current

80 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 23 Changing direction of flux The magnitude of the force on a current flowing at right angles to a magnetic field is proportional to: 1. the flux density of the magnetic field 2. the current flowing in the conductor 3. the length of the conductor in the magnetic field. The strength of the mechanical force is given by the formula: FF = BB. II. LL where: FF = MMMMMMMMMMMMMMMMMMMM ffffffffff eeeeeeeeeeeeee oooo tttttt cccccccccccccccccc mmmmmmmmmmmmmmmm iiii NNNNNNNNNNNNNN (NN) BB = FFFFFFFF dddddddddddddd oooo mmmmmmmm mmmmmmmmmmmmmmmm ffffffffff mmmmmmmmmmmmmmmm iiii WWWW mm 22 tttttttttttt (TT) II = CCCCCCCCCCCCCC ffffffffffffff iiii cccccccccccccccccc mmmmmmmmmmmmmmmm iiii aaaaaaaaaaaaaa (II) LL = LLLLLLLLLLLL oooo cccccccccccccccccc iiii tttttt mmmmmmmmmmmmmmmm ffffffffff mmmmmmmmmmmmmmmm iiii mmmmmmmmmmmm (mm) This principle of motion is how a motor works, ie the conversion of electrical energy into mechanical motion. Example 1 A conductor of 20cm in length is situated perpendicularly in a magnetic field of flux density 5 teslas and has a current of 10A flowing through it. Calculate the force on the conductor. F = B. I. L = = 1111 NNNNNNNNNNNNNN Example 2 A conductor of 0.5m in length is situated perpendicularly in a magnetic field of flux density of 10 teslas and has a current of 15A flowing through it. Calculate the force on the conductor. F = B. I. L = = 7777 NNNNNNNNNNNNNN

81 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 23 The direction of this force can be found by using: Fleming s left hand motor rule

82 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 24: Electrical equipment three phase a.c. motors Learning outcome The learner will: 6. know the operating principle of a range of electrical equipment. Assessment criteria The learner can: 6.1 describe the operating principle of electrical equipment. Range Electrical equipment: Basic a.c. motors, basic d.c. machines, fluorescent luminaires, relays. Principle of motor operation Electrical equipment three phase a.c. motors The basic principle of all motors can easily be shown using two electromagnets and a permanent magnet, as shown below. Current is passed through one coil in such a direction that a North Pole is established, and through the second coil in such a direction that a South Pole is established. A permanent magnet with a North and South Pole is the moving part of this simple motor. In picture (a) the North Pole of the permanent magnet is opposite the North Pole of the electromagnet. Similarly, the South Poles are opposite each other. Like magnetic poles will repel each other, causing the movable permanent magnet to begin to turn. After it turns part way around, the force of attraction between the unlike Poles becomes strong enough to keep the permanent magnet rotating. The rotating magnet continues to turn until the unlike Poles are lined up. At this point the rotor would normally stop because of the attraction between the unlike Poles see picture (b). If, however, the direction of currents in the electromagnetic coils were reversed, thereby reversing the polarity of the two coils, then adjacent poles would again be the same and repel each other see picture (c). The movable permanent magnet would then continue to rotate. If the current direction in the electromagnetic coils was changed every time the magnet turned 180 degrees, or halfway around, then the magnet would continue to rotate. This simple device is a motor in its simplest form. An actual motor is more complex than the simple device described above, but the principle is the same.

83 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 24 In order to reverse the current with dc. machines, the shaft of the motor is fitted with a commutator and brushes connect the current to the commutator; this arrangement periodically reverses the supply to gain continuous rotation. With a.c. machines, because of the nature of a.c., the supply will naturally reverse itself periodically so no further action is required to do this. Basic a.c. motor operation An a.c. motor has two basic electrical parts: a stator and a rotor, as shown in the diagram below. The stator is in the stationary electrical component. It consists of a group of individual electromagnets arranged in such a way that they form a hollow cylinder, with one pole of each magnet facing toward the centre of the group. The rotor is the rotating electrical component. It also consists of a group of electromagnets arranged around a cylinder, with the poles facing toward the stator poles. The rotor, obviously, is located inside the stator and is mounted on the motor s shaft. The most common form of a.c. machine available is the induction motor. This type of machine produces a rotating magnetic field in the stator that induces an EMF into the rotor. The interaction between the rotating magnetic field in the stator and the magnetic field produced in the rotor by the induced EMF will cause torque (tendency of a force to rotate an object about an axis, fulcrum or pivot) to be produced in the rotor. It is easier to discuss torque production with three phase machines. Production of a rotating magnetic field In order for an a.c. machine to work, it must produce a rotating magnetic field in the stator. This is achieved in a three phase machine by distributing three windings, one for each phase, around the stator. Since the three phases are 120 out of phase from each other, the magnetic fields produced by each phase will interact to produce a rotating magnetic field.

84 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 24 The diagrams on this page show how this rotating magnetic field is produced in three sets of windings distributed around the stator and fed with a three phase supply. The diagram shows the magnetic field produced by currents flowing in the three phase stator windings when supplied with an alternating current. The positive half of the cycle produces currents shown in the start of the coils (R, Y and B) and in the end of the coils (R, Y and b ). The negative half of the cycles produces opposite polarity, ie in the start of the coils (R, Y and B) and in the end of the coils (R, Y and B ).

85 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 24 Synchronous speed The speed of the rotating magnetic field is reliant on two factors: the supply frequency and the number of pairs of poles in the stator. This can be summarised by the following formula: NN (rrrrrr) = FF PP NN (rrrrrr) = FF 6666 PP where: NN = RRRRRRRRRRRRRRRRRRRRRR rrrrrr iiii rrrrrrrrrrrrrrrrrrrrrr pppppp ssssssssssss rrrrrr iiii rrrrrrrrrrrrrrrrrrrrrr pppppp mmmmmmmmmmmm FF = FFFFFFFFFFFFFFFFFF iiii HHHHHHHHHH (HHHH) PP = TTTTTT nnnnnnnnnnnn oooo pppppppppp oooo pppppppppp Example 1 A 4 pole a.c. three phase machine is fed with a supply at a frequency of 50Hz. Calculate synchronous speed in: a) rps b) rpm. NB: the machine has 4 poles so that is 2 pairs of poles. a) rps = F P = 50 2 = 2222 rrrrrrrr pppppp ssssssssssss b) rpm = F 60 P = = 11, rrrrrrrr pppppp mmmmmmmmmmmm

86 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 24 Production of torque in an induction motor The most common type of induction motor uses a cage rotor sometimes referred to as a squirrel cage rotor. The basic arrangement is shown below. When the supply is connected to the stator, a rotating magnetic field is produced, as described previously. The lines of flux will cut the bars of the cage rotor, which will induce an EMF into these bars. Due to the shorting rings at either end of the cage rotor, the induced EMF will produce a current flowing through the bars and the shorting rings. This current will produce its own magnetic field that will interact with the stator magnetic field and produce torque. Slip As the rotor speed increases, the rate at which the rotating magnetic field flux cuts the cage rotor will reduce, thus reducing the induced EMF; the resulting cage current and rotor magnetic field and torque will reduce, as rotor speed approaches synchronous speed. The rotor speed can never reach synchronous speed because in this situation no EMF will be induced in the rotor and no torque produced. Therefore, rotor speed will always be less than synchronous speed and the difference between these two values is referred to as slip. Slip is defined in two ways: per unit slip percentage slip and can be calculated as follows: PPPPPP uuuuuuuu ssssssss = (nn ss nn rr ) % ssssssss = (nn ss nn rr ) nn ss nn ss where: nn ss = SSSSSSSSSSSSSSSSSSSSSS ssssssssss nn rr = RRRRRRRRRR ssssssssss The calculation can be carried out using either revs per second or revs per minute, as long as the same units are used for both synchronous and rotor speeds.

87 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 24 Example 2 A two-pole induction motor runs at 2,880 rpm when connected to the 50Hz mains supply. Calculate the: a) per unit slip b) percentage slip. Sync speed, n s = F 60 P = = 3,000 rpm a) Per unit slip, S = (n s n r ) n s = (3,000 2,880) 3,000 = b) % slip = (n s n r ) n s 100 = (3,000 2,880) 3, = 44% Example 3 A 4 pole 50Hz induction motor has a per-unit slip of 0.03 on full load. Calculate the full load speed. Sync speed, n s = F 60 P = = 1,500 rpm Per unit slip, S = (n s n r ) n s S n s = n s n r S n s + n r = n s n r = n s S n s = 1,500 (0.03 1,500) = 1, = 11, rrrrrr

88 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 25: Electrical equipment single-phase a.c. motors Learning outcome The learner will: 6. know the operating principle of a range of electrical equipment. Assessment criteria The learner can: 6.1 describe the operating principle of electrical equipment. Range Electrical equipment: Basic a.c. motors, basic d.c. machines, fluorescent luminaires, relays. Electrical equipment single-phase a.c. motors With three-phase machines the interaction of three out of phase windings gives us the rotating magnetic field necessary to produce induction in the rotor and therefore torque. This rotating magnetic field cannot be produced in the same way with single phase. In order to start a single-phase induction motor, we must artificially create a rotating magnetic field. In order to achieve this artificial rotating field, two windings are provided: one a start or auxiliary winding and the other a run winding. By various means, the start or auxiliary winding is slightly out of phase with the run winding. Once the machine has started rotating, we no longer need this artificial rotating magnetic field and in most cases the start or auxiliary winding is disconnected. The names of the various types of single-phase machine describe the means of achieving the phase difference. Split-phase (resistance start) motors Below is a basic one-line diagram of the split-phase motor. It shows the run and start winding of the starter, as well as the centrifugal switch (CS). The resistance of the start or auxiliary winding will be different to the run winding and this will cause it to have a different impedance to the run winding and, hence, a different phase angle between the two windings. This provides sufficient phase difference to provide a starting torque, albeit very small. Reversal of direction of rotation: The rotor will always turn from the start winding to the adjacent run winding of the same polarity. Therefore, the relationship between the start and run windings must be changed. In order to change the relationship and the direction of rotation, the polarity of only one of the fields must be reversed. Split-phase motor applications: Split-phase motors are generally limited to the ⅓ horsepower size. They are simple to manufacture and inexpensive. The starting torque is very low and can be used for starting small loads only.

89 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 25 Capacitor-start motors In order to increase the phase angle difference and therefore increase the torque, a capacitor is placed in series with the start (or auxiliary) winding, as shown in the diagram below. Once the motor has attained 75% of its rated speed, the start capacitor and the start winding can be eliminated by the centrifugal switch. It is not necessary for this motor to operate on both windings continuously. The capacitor of the capacitor-start motor improves the power factor of the electrical system only on starting. Letting a capacitor remain in the circuit will improve the electrical power factor that was modified initially by the use of a motor. The permanent capacitor is placed in series with one of the windings. The two windings are now called the main and auxiliary windings. They are constructed exactly alike and both are left in the circuit during the operation of the motor. A centrifugal switch is no longer needed. Capacitor-start/capacitor-run motor When additional torque is required to start and keep a motor operating, additional capacitors can be added. The diagram below shows the two-capacitor motor, commonly referred to as the capacitorstart/capacitor-run motor. Notice that the start capacitor is in series with the auxiliary winding. The centrifugal switch is used to control the start capacitor in the same manner as it did in the capacitor-start motor. This capacitor is used only to develop enough torque to start the motor turning. The run capacitor is connected in parallel with the start capacitor. In this manner, both capacitor capacitances add together to increase the total phase angle displacement when the motor is started. Also, the run capacitor is connected in series with the auxiliary winding. And therefore the motor always has the auxiliary winding operating, and increased torque is available.

90 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 26: Electrical equipment d.c. motors Learning outcome The learner will: 6. know the operating principle of a range of electrical equipment. Assessment criteria The learner can: 6.1 describe the operating principle of electrical equipment. Range Electrical equipment: Basic a.c. motors, basic d.c. machines, fluorescent luminaires, relays. Electrical equipment d.c. motors Generally, d.c. motors contain two windings, as follows: field winding armature winding. The rotating part of the machine is referred to as the armature. Mounted on this is the armature winding which is connected to the outside world by brushes bearing on a commutator. The field winding is mounted in the machine body and surrounds the armature. Both of these windings will produce a magnetic field, and reaction between the two windings will produce the torque causing the armature to rotate. The type of d.c. motor is classified by how the two windings are connected together, with each type having advantages and disadvantages over the others. Three main types are: series wound motor parallel (or shunt) wound motor compound wound motor. Series wound motor The diagram on the right shows the arrangement for the series wound motor. Since the series and armature windings are connected in series, when the machine is loaded the armature draws more current; this will increase the strength of the magnetic field generated by the field winding. Main characteristics High starting torque. Speed falls off as load increases (and vice versa). Must be started and run with load permanently connected, as motor will speed up to dangerous levels off load. Reversal of motor can be achieved by reversing either the field or armature not both.

91 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 26 Parallel (or shunt) wound motor The diagram on the right shows the arrangement for the parallel wound motor, also known as a shunt wound motor. Since the field winding is connected in parallel across the supply, it will always draw the same current irrespective of the current drawn by the armature. This means that regardless of the load on the shaft, and hence the armature current, the speed of the armature will be relatively constant, even when the machine is not mechanically loaded. The parallel wound motor has low starting torque so is only suitable for small loads. Main characteristics Low starting torque. Speed falls off by only about 5% as load is increased and therefore has constant speed once started. Start with load disconnected. Reversal of motor can be achieved by reversing either the field or armature not both. Compound wound motor The diagram on the right shows the arrangement for the compound wound motor. It has two field windings: one connected in series and one in parallel (shunt). This machine combines the best characteristics of the series and shunt machines: good starting torque and relatively constant speed. Main characteristics Good starting torque. Speed is almost constant under varying load conditions. Reversal same as for series and shunt.

92 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 27: Electrical equipment fluorescent luminaires and relays Learning outcome The learner will: 6. know the operating principle of a range of electrical equipment. Assessment criteria The learner can: 6.1 describe the operating principle of electrical equipment. Range Electrical equipment: Basic a.c. motors, basic d.c. machines, fluorescent luminaires, relays. Electrical equipment fluorescent luminaires and relays Fluorescent lamps are correctly referred to as low pressure mercury vapour lamps. They are comprise of a glass tube with cathode filament at either end. The sealed glass tube contains argon and a small quantity of liquid mercury. If a large voltage is applied to either end of the tube, the gas in the tube ionises and becomes conductive. This results in a flow of current from one end of the tube to the other through the ionised gas, causing the emission of ultraviolet light. The inside of the tube is coated with a phosphor coating which converts the ultraviolet light into useable light of a shade determined by the nature of the phosphor coating. The circuitry associated with the fluorescent tube is required to provide the large voltage to initiate ionisation and, hence, current flow. The circuit of the traditional switch-start fluorescent circuit is shown below: When the supply is applied, current will flow through the choke, then the starter switch and back to the supply; this will cause energy to be stored in the choke in the form of a magnetic field. Although the glow starter switch is open circuit at switch-on, conduction occurs through the small gap due to ionisation. This ionisation produces heat that causes the bimetal strip forming the contacts to bend together. When the contacts touch, the ionisation stops and the bimetal strip starts to cool and the contacts open.

93 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 27 This interrupts the supply current, causing the choke to discharge its energy across the ends of the tube in the form of a large voltage. This voltage is large enough to cause ionisation and, hence, conduction along the tube and it then emits light. If the tube fails to strike, the process repeats itself until the tube strikes. Once the tube has struck, the high voltage is not required, as the effective resistance from one end of the tube to the other is greatly reduced. The choke then limits the current flow in the circuit. High frequency (HF) High frequency electronic control gear consists of a single ballast unit that performs the functions of all of the major components in a standard switch-start circuit. Through the use of solid state electronics, the HF ballast creates a discharge frequency of 25 40kHz. This is far higher than standard switch-start circuits and results in a vast improvement in the quality of the light produced. Lamp flicker is eliminated and as HF operation is more efficient than normal operation, significant savings can be made in energy costs. Relays A relay is an electro-magnetic device that has many uses but mainly in control circuits where small control currents are used to switch large load currents. A typical relay is shown in the diagram right. The basic relay consists of a coil wound around a soft iron core which is connected to a soft iron yoke. Pivoted on the end of the yoke is an armature, also made of soft iron. When the relay coil is energised, it magnetises and the armature is then attracted to the pole face; the insulated push rod at the top of the armature will then operate contacts on the contact set. When the supply to coil is removed, the coil will demagnetise, the armature will return to its original position and the contacts will also return to their original position. The magnetic circuit is made of soft iron, as this magnetises easily when the magnetic field is established but loses its magnetism when the coil is de-energised. More than one set of contacts can be operated simultaneously from the armature, controlling a number of different circuits at the same time. A normally open contact is shown in the diagram above but many combinations can be incorporated on the same relay, including normally open, normally closed, changeover and make before break, just to mention the more common arrangements. Some of these are shown below: A contactor is basically a heavy-duty relay and works to the same general principle, ie an electromagnet energising and causing an armature to be attracted, thus operating contacts and, when de-energised, operating the contacts back to their normal position.

94 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 27 In most applications, relay contacts shown on diagrams are normally shown in the de-energised position. The diagram below shows a circuit diagram for a typical direct-on-line motor starter employing a contactor.

95 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 28: Transformers Learning outcome The learner will: 6. know the operating principle of a range of electrical equipment. Assessment criteria The learner can: 6.3 describe the operating principle of transformers. Range Transformers: Self-inductance, mutual inductance, turns ratio, potential transformer, current transformer, isolating transformer. Electro-magnetic induction Transformers When a conductor cuts or is cut by magnetic lines of force (magnetic flux), a voltage is induced in the conductor. This is known as electromagnetic induction. If a conductor is held steady in a magnetic field, or is moved parallel to a magnetic field, then no lines of force will be cut and therefore no voltage is induced in the conductor. In general, electro-magnetic induction is the production of an EMF in a circuit due to a change of flux linkage within the circuit which can be shown in two ways. Self-inductance This is the changing flux linkage within the same circuit. Any circuit in which a change of current produces a change of flux and therefore produces an induced EMF is said to possess self-inductance. The unit of self-inductance, called the Henry (H), is the inductance of a closed circuit in which an EMF of 1 volt is produced when an electric current in the circuit is varied uniformly at the rate of 1 ampere in 1 second. IIIIIIIIIIIIII EEEEEE = LL. (II 22 II 11 ) vvvvvvvvvv tt Where: L = The value of inductance in Henrys (H) t = Time in seconds (s) I 1 and I 2 = are values of current NB: The negative sign indicates that the voltage opposes the current change.

96 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 28 Example 1 If the current flowing through a coil of inductance of 0.7H increases from 2A to 10A in 40 milliseconds, calculate the average value of an induced EMF. E = L. (I 2 I 1 ) t 0.7 (10 2) = = = = vvvvvvvvvv Mutual inductance This is when the induced EMF in a circuit is due to a changing current in another circuit, ie a transformer. Practical transformer When a.c. voltage is applied to the primary winding, it produces an a.c. current flowing through the primary winding which produces a flux in the iron core that is also alternating. This alternating flux cuts the conductors in secondary winding and induces an EMF into this secondary winding.

97 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 28 The EMF induced in each turn of the secondary will be the same as the voltage across each turn in the primary. If N 1 and N 2 are the number of turns on primary and secondary, respectively, then since both coils of the transformer are linked by the same flux, their induced EMFs will be proportional to the number of turns in each coil. Therefore: V P (primary EMF) V S (secondary EMF) = N P(primary turns) N S (secondary turns) VV PP VV SS = NN PP NN SS oooo VV 11 VV 22 = NN 11 NN 22 Since the full load efficiency of a transformer is nearly 100%: V 1. I 1 = V 2. I 2 then V 1 V 2 = I 2 I 1 Hence VV 11 VV 22 = II 22 II 11 = NN 11 NN 22 The above is known as the transformer equation. Example 2 A transformer connected to a 230 volt 50Hz supply has a primary winding of 1,200 turns. Calculate: a) the EMF induced in a secondary winding of 300 turns b) the number of turns that would be required in the secondary winding if it is to produce an EMF of 100 volts c) the volts per turn there are on the primary and secondary windings in part (a) above. a) V 1 V 2 = N 1 N 2 V 2 = = N 2 N 1 V , = b) V 1 V 2 = N 1 N 2 N 2 = V 2 V 1 N = 230 1,200 = tttttttttt

98 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 28 c) Primary volts turn = V 1 N 1 = 230 1,200 = vvvvvvvvvv Secondary volts turn = V 2 N 2 = = vvvvvvvvvv It can be seen from the above calculations that the volts per turn on a transformer are the same on the secondary winding as on the primary winding. Turns ratio It is not necessary to know the exact number of turns on the primary and secondary of the transformer. We can carry out calculations as long as we know the ratio between the primary and secondary turns. Example 3 A transformer connected to a 240 volt 50Hz supply has a primary/secondary turns ratio of 20:1. Calculate the EMF induced in a secondary winding. V 1 V 2 = N 1 N 2 N 2 V 2 = V N 1 1 = = Transformers are rated by their output capabilities The output of a transformer is measured in kva or MVA. Therefore, if a transformer is rated at 20.7kVA at 230 volts, it is capable of delivering a load current of 90 amps. This is obtained by dividing the kva rating by the rated voltage. Since kva = Volts amps then kva amps = Volts eg 20.7kVA I = 230V = 9999 aaaaaaaaaaaaaa

99 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 28 Example 4 A 50kVA single-phase transformer has a turns ratio of 440/20. The primary winding is connected to a 220 volt 50Hz supply. Calculate: a) the secondary voltage on no load b) the approximate values of primary and secondary currents on full load. a) V 1 V 2 = N 1 N 2 b) V 2 = N 2 N 1 V 1 20 = = I 1 = kva V 1 = 50kVA 220V = I 2 = kva V 2 = 50kVA 10V = 55,

100 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout : Principles of electrical science Handout 29: Special transformers Learning outcome The learner will: 6. know the operating principle of a range of electrical equipment. Assessment criteria The learner can: 6.3 describe the operating principle of transformers. Range Transformers: Self-inductance, mutual inductance, turns ratio, potential transformer, current transformer, isolating transformer. Special transformers Voltage and current (instrument) transformers Most test instruments have a limited range of operation but this can be extended by using a voltage transformer (VT) connected to a voltmeter to measure high voltage values and a current transformer (CT) connected to an ammeter to measure high current values. These are collectively referred to as instrument transformers and will only work on a.c. circuits. The high current or voltage to be measured is connected to the primary of the instrument transformer, whilst the instrument itself is connected to the secondary. Since the secondary value is proportional to the primary value, the instrument measures the small secondary value that represents the value on the secondary. Advantages of using instrument transformers are: the secondary side of the instrument transformer is wound for low voltage, which simplifies the insulation of the measuring instrument and makes it safer to handle the transformer isolates the instrument from the main circuit the measuring instrument can be read in a remote, convenient position connected by long leads to the instrument transformer the secondary voltage or current can be standardised (usually 110V and 5A), which simplifies instrument changes. Voltage transformers (VT) Voltage transformers are operated as a step-down transformer and the construction of a VT is similar to a basic power transformer. The secondary winding of the VT is connected to a voltmeter or the voltage coil of a wattmeter, as shown in the diagram on the right. A large number of turns are wound on the primary and a few on the secondary. Since: VV PP VV SS = NN PP NN SS the voltmeter reading must be multiplied by the turns ratio to determine the load voltage.

101 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 29 Example 1 A voltmeter is connected to 50 turns on the secondary winding of a VT. The primary winding of 250 turns is connected to the main supply. Calculate the supply voltage if the voltmeter reading was 80V. Primary voltage, V P = N P N S V S 250 = = vvvvvvvvvv As an alternative solution, we could say that the turns ratio is 250:50, ie 5:1, and therefore the supply voltage is 5 80 = 400V. Example 2 An electrical contractor wishes to monitor a 660V supply with a standard 110V voltmeter. Determine the turns ratio of the VT to perform this task. V P N P = V S = = N S N P N S N P N S = The turns ratio is 6:1. This means that the number of turns on the primary side of the VT must be six times greater than the number of turns on the secondary, which is connected to the 110V voltmeter. Current transformers (CT) Although the basic transformer principles remain the same, the operation of a CT is different to a basic power transformer. Many turns make up the secondary winding, whilst the primary is made up of very few turns. In some cases the secondary windings are wound on to the busbar itself. The secondary is connected to an ammeter or the current coil of a wattmeter, as shown in the diagram on the right. VV PP VV SS = II SS II PP

102 Level 2 Diploma in Electrical Installations (Buildings and Structures) Unit 202 Handout 29 Example 3 An ammeter having a full scale deflection of 5A is used to measure a line current of 200A. If the primary is wound with two turns, calculate the number of secondary turns required to give full scale deflection. I S I P = N P N S N S = I P N I P S N S = = 8888 tttttttttt In order to cause a primary current to flow in a power transformer, it is necessary to have current flow in the secondary. However, with a CT, the current in the primary is also the main circuit current which will flow even without the secondary being connected. Since the secondary current is necessary to stabilise the core magnetic flux, if the ammeter is disconnected then the voltage on the secondary terminals could reach an unacceptably high level, which would be extremely dangerous. This could cause overheating of the core and possible breakdown of the insulation. A CT must never be operated with the secondary open-circuited. When it is necessary to, for example, change the ammeter, the secondary terminals must be short-circuited. This will prevent a dangerous situation from occurring and will not damage the CT. Like other transformers, instrument transformers are rated in volt amperes and are referred to as the burden. Isolating transformer Aside from the ability to easily convert between different levels of voltage and current in a.c. circuits, transformers also provide an extremely useful feature called isolation, which is the ability to couple one circuit to another without the use of direct wire connections. Isolating transformers do not usually step up or step down voltages and the secondary voltage is the same as the primary voltage. The supply system neutral in the UK is connected to earth at the sub-station transformer so touching line and earth will result in an electric shock of the same magnitude as touching line and neutral. However, if the secondary of the isolating transformer is not connected to earth then touching both pole and earth should not result in an electric shock. Of course, touching both poles simultaneously will result in an electric shock. This is the arrangement normally employed in shaver sockets providing 230V for shavers but if a person touches both a pole and earth they should not get an electric shock.