Smart Electrical Technology Textbook PDF 2023 1-30

Smart Electrical Technology UNIT 1 ELECTRICAL PRINCIPLES AND CIRCUITS At the end of the unit, students should be able to:  understand the basic principles of electricity  connect simple electrical lighting circuits for residential premises  use a multimeter to test for electrical continuity The section is organised into 11 sub-units 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.1 Electrical Safety Electric Circuits Electric Circuit Laws Electric Circuit Connections Power and Energy in an Electric Circuit Electric Power Sources Electrical Hazards and Protection Electrical Cables Electrical Test Instruments Conventional Lighting Circuits Electrical Supply Systems Electrical Safety Learning Objectives  Explain the two types of electric shock: • Direct contact • Indirect contact  Explain the potential dangers in electrical work  Understand the danger of hazardous work practices  Explain the precautions and procedures for safe electrical work  Explain the benefits of good housekeeping in electrical work  Recommend measures to protect against electrical hazards 1 Smart Electrical Technology 1. How an Electric Shock Happens An electric shock occurs when an electrical current flows through the human body (Figs. 1.1-1 and 1.1-2). From L supply N transformer E y 2. Fig. 1.1-1: A person who has received an electric shock y Fig. 1.1-2: How an electric shock occurs Types of Electric Shock There are two types of electric shock: • direct contact • indirect contact 2.1 Electricity A Direct Contact (current) normal use is flowing through during -> Live wire/Phase wire . Direct contact occurs when a person comes into contact with a conductor that is live under normal conditions (Fig. 1.1-3). Supply transformer P Earth metal N Protective conductor y Fig. 1.1-3: Electric shock due to direct contact 2 Smart Electrical Technology 2.2 Indirect Contact Indirect contact occurs when a person comes into contact with exposed conductive parts that have become live under fault conditions, i.e., parts which are not normally live but have become live due to faults such as insulation failure (Fig. 1.1-4). Supply transformer Fault P Earth metal N Protective conductor y Fig. 1.1-4: Electric shock due to indirect contact The severity of injury from an electric shock depends on: • the amount of current flowing through the body; and • the length of time the current flows through the body. • the pathway the current flows through the body A shock current of 50 mA can be fatal. Figure 1.1-5 shows a hand burned by electricity. I 50 milliamperes . al ↳ convert m alternating current . A to A 1000 mA 1 mA 50mA . = = = 1 A 0 . 001A 50x0 001 . =0 . 05A - in A A = - A mA : : = x # y Fig. 1.1-5: Electrical burn 1000 1000 3 Smart Electrical Technology 3. What to Do When Someone Is Shocked or Burned by Electricity • Switch off the electrical supply if the person is still in contact with the live circuit. At the same time, get someone else to call for help. • If you cannot find the electrical supply source, do not take hold of the victim, otherwise the current may pass through you too. Use a dry towel or scarf to free the victim, or use a piece of wood to knock the victim’s hand free of the electrical equipment. • As a last resort, take hold of the victim’s clothing – without touching the body – to pull the victim free. • Do not try to move a victim who has fallen due to electric shock, except to shift the body into the recovery position, as the victim may have sustained other injuries. 4. How to Work Safely In addition to maintaining a safe work environment, we must also work safely. Safe work practices help reduce the risk of injury or death from workplace hazards. Here are some examples of safe work practices: • Use and maintain tools properly. • Inspect tools before using them. • Switch off the power before working on a circuit. 5. Good Housekeeping In order to work safely, we should keep our workspaces tidy and well-arranged. Good housekeeping: • lowers the risk of accidents and fire; • improves productivity; • makes better use of space; and • reflects a well-managed operation. 6. General Safety Rules in the Electrical Laboratory 6.1 Dos • • • • • • Use only insulated tools. All electrical work must be completed by a suitably qualified person. Always switch off the power and remove the plug before any electrical work. Ensure that your electrical equipment is properly inspected and maintained. Disconnect broken appliances and replace frayed cords or broken power points. Use test equipment correctly. Read the instruction booklet and follow all instructions. 4 Smart Electrical Technology • Keep electrical cords off the floor to prevent them from being damaged from dragging or contact with sharp objects. A damaged electrical cord can cause a fatal electric shock. • Know the location of the main electricity supply. • Keep electrical equipment away from water and wet areas to lower the risk of electric shock. 6.2 Don'ts • Never take risks. • Do not re-close a tripped circuit breaker or replace a blown fuse until the cause has been found and rectified. • Do not misuse electrical equipment and appliances. Keep them dry. • Do not use flammable solvents near an electrical apparatus unless the apparatus is labelled “flameproof”. • Do not use a fire extinguisher on electrical fires unless it is an appropriate type, such as a carbon-dioxide or dry-powder extinguisher. Switch off the power as soon as possible. • Do not overload circuits and fuses by plugging in too many appliances into one power point. • Use a power board with individual switches instead of double adapters. 5 Smart Electrical Technology 1.2 Electric Circuits Learning Objectives  Explain how an electric circuit works  State the three basic electrical quantities: voltage, current and resistance  State the units of measurement for voltage, current and resistance  Describe the use of a voltmeter for measuring voltage  Describe the use of an ammeter for measuring current  Describe the use of an ohmmeter for measuring resistance  State the different uses of a multimeter  Use the multimeter to measure current, voltage and resistance, and to check for continuity in an electrical installation  Exercise safety precautions when handling and using measuring instruments 1. Electric Circuits An electric circuit is the physical pathway for a current to flow. A simple electric circuit consists of: #Voltage • a source of electromotive force (emf), such as a battery or generator; • a load (which has resistance), such as a lamp; • conducting/connecting wires to connect the various parts of the circuit; and • additional components such as a switch, fuse or circuit breaker, and a measuring instrument. ↑ Switch A Battery Height potential = y . Connecting wires Fig. 1.2-1: Simple electric circuit Eg 1 . IV I => I I Lo - - - im↑ -1 ↓ Die Lamp ~ from high potential to low potential flows from high potential (voltage) to , 5V ④ · S zvo-8 q 1 Eg 5~zV . . . - - ! 1 3V water flows current - low potential (voltage) . 6 Smart Electrical Technology 2. Voltage, Current and Resistance The three basic electrical quantities in a basic electric circuit are: • supply voltage or electromotive force (EMF) circuit • current • resistance - - 2.1 - C C . Supply Voltage / Electromotive Force 0 C . . 1 This is the force required to cause the current to flow in a closed circuit. It is represented by V (supply voltage) or emf (electromotive force) and is measured in volts (V). 2.2 nit for voltage r Current is flowing charges that allows the load to function . It is the electrical quantity that is needed for the electrical load to function, i.e., the flow of current allows the lamp in Fig. 1.2-1 to light up. It is represented by I and is measured in amperes (A). 2.3 Fans Resistance , Lights air . con etc . Resistance helps limit the current flow in the circuit. When there is no or very little resistance in the circuit, such as when a short circuit occurs, the current will be very high and will damage the equipment if the circuit is not protected. It is represented by R and is measured in ohms (Ω). 3. Ammeter An ammeter is used for measuring current flowing in the circuit. It is always connected in series to the load in the circuit. A e The symbol for an ammeter is A I Figure 1.2-2 shows how an ammeter should be connected to measure the current in an electric circuit. - connected in es A ② V y Lamp Fig. 1.2-2: Ammeter connection -voltmeter Lov in connected parallel . 7 Smart Electrical Technology 4. Voltmeter Electromotive Force V - = Voltage A voltmeter is used for measuring the emf of the battery, supply voltage and voltage across the load in the circuit. It is always connected in parallel to the load or supply source where the voltage is to be measured. The symbol for a voltmeter is V · Figure 1.2-3 shows how a voltmeter should be connected to measure the voltage in an electric circuit. V V Lamp overGre ne i measure voltage across the - battery . y = 5. Fig. 1.2-3: Voltmeter connection Ohmmeter An ohmmeter is used for measuring the resistance of an electrical load in the circuit. It is always connected in parallel to the load where the resistance is to be measured. The symbol for an ohmmeter is Ω -- Figure 1.2-4 shows how an ohmmeter should be connected to measure the resistance of an electrical load. NOTE: When using an ohmmeter to measure resistance, make sure that the power is switched off. Otherwise, the user may receive an electric shock and the ohmmeter may be damaged. Switch measuring - F when R Fo· Ω · t - - Electrical load ⑲ -↳ y Fig. 1.2-4: Ohmmeter connection 8 Smart Electrical Technology 6. Multimeter A multimeter is a device that can measure voltage, current and resistance. It can also be used to diagnose electrical problems. test -> connectivity . There are two kinds of multimeter: • the analogue multimeter (Fig. 1.2-5), which uses an indicator needle with a measurement scale; and • the digital multimeter (Fig. 1.2-6), which uses a numeric LCD display. The multimeter uses a rotating switch to select the quantity to be measured. It has two metal-tipped wires called probes, one red (+) and one black (-). y Fig. 1.2-5: Analogue multimeter y Fig. 1.2-6: Digital multimeter When using a multimeter to measure resistance, make sure that the power is switched off. However, the measurement of voltage can only be carried out with the power on. Due to the risk of electric shock, only trained individuals should conduct voltage tests. 9 Smart Electrical Technology 1.3 Electric Circuit Laws Learning Objectives  Understand the relationship between current, resistance and voltage (Ohm’s law): V = I × R  Apply Ohm's law to determine current, resistance or voltage in an electric circuit  Connect a simple electric circuit comprising an ammeter, a voltmeter, a load and a power supply to verify Ohm’s law - Resistance = Current 1. is doesn't change directly proportional to voltage when R Ohm's Discovery ↳ when Voltage increase current also , . increase . The relationship between V, I and R was discovered by scientist Georg Ohm. This discovery, known as Ohm’s law, is the basic formula used in all electric circuits. - Ohm’s law states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) applied across its ends, provided the temperature remains constant. R I V y Fig. 1.3-1: Simple electric circuit V=I×R The following formula is derived from Ohm’s law: Ohm's Law Triangle V . - I - V X - I R Y I R = x **! R = I where V is in volts (V) I is in amperes (A) R is in ohms (Ω) - -> - =k I* I X IP - x = E 10 Smart Electrical Technology 2. Worked Example R I R V y (A) Fig. 1.3-2: Simple electric circuit Refer to Fig. 1.3-2. Determine V if I = 0.5 A and R = 20 Ω. V = I × R = 0.5 A × 20 Ω = 10 V 0 51 V IxR = = (B) . + 20 = x A load of resistance 500 Ω is connected to a supply of 230 V. What is the current drawn? V 230V R 500 2 V 230 V = I= = 0.46 A 250V R 500 Ω = = , I 3. = E = = 5002 A - x Tutorial R V I R I (A) V y Fig. 1.3-3: Simple electric circuit Refer to Fig. 1.3-3 for the following questions. Determine: (i) V if I = 0.5 A and R = 100 Ω V = Ix R = 0 . 5A x 100 = 5Vx (50 V) (ii) I if V = 110 V and R = 550 Ω I = 4 (0.2 A) = = A 11 Smart Electrical Technology (iii) R if V = 230 V and I = 1.2 A R (191.67 Ω) " I = = 02ex(35 f) = . (iv) I if V = 24 V and R = 10 kilo-ohms (kΩ) I (B) = I toooooo (0.0024 A) e A load of resistance 200 Ω is connected to a supply of 110 V. What is the current drawn? I (C) = 4 40X = (0.55 A) 55Ax = The current flowing in a circuit of 50 Ω is 0.2 A. What is the supply voltage? V IxR = = Convert in mA divide 5 to illiampere = ampere 1000 . . 2A 50 - (10 V) = x Convert kilo-ohms to otms . A = = by 0 x k = 0 005 . 201 = = M l multiply by 1000 . $1000 > - 20x1000 20000E -micro M-MEO 12 Smart Electrical Technology 1.4 Electric Circuit Connections Learning Objectives  Identify the three methods of connecting electrical loads: • Series • Parallel • Series-parallel  State the characteristics of a series circuit: • One path for current flow • Supply voltage is equal to the sum of all individual voltages • Total resistance is equal to the sum of all individual resistances  State the characteristics of a parallel circuit: • Supply voltage is the same as all branch voltages • Supply or total current is equal to the sum of all individual branch currents • Total resistance is lower than the lowest individual resistance  Determine the total resistance of a series circuit comprising two resistors using the formula: RT = R1 + R2  Determine the total resistance of a parallel circuit comprising two 1 1 1 resistors using the formula: = + RT R1 R2  Connect a series circuit comprising two resistors, a voltmeter and an ammeter for the purpose of verifying the characteristics of a series circuit  Connect a parallel circuit comprising two resistors, a voltmeter and an ammeter for the purpose of verifying the characteristics of a parallel circuit  Recognise that a series-parallel circuit has the characteristics of both series and parallel circuits 1. Types of Circuit There are basically three types of circuit, which are classified in terms of how their components (or loads) are connected to each other. They are: • series circuit • parallel circuit • series-parallel circuit Rel s RI * I +11- Saralle R2 R3 ⑳ . + /113 Smart Electrical Technology 2. 7 Voltage Characteristics of a Series Circuit W /Current flowing across R1 into R1 R1 I1 . R2 I2 V1 Supply Current It : Total V2 IS Current VS -> What comes out from the battery is always Vs OR Is y Supply Voltage Fig. 1.4-1: Series circuit . (A) ↑ Total Voltage There is only one path for the current to flow, that is: SERIES: (B) Current is IS or IT = I1 = I2 Equal/the same The supply voltage, VS or VT , is the sum of all the individual voltages: (C) Voltage VS or VT = V1 + V2 SERIES : adds up . The total resistance, RT , is the sum of all the individual resistances: SERIES Resistance RT = R1 + R2 ⑧ B adds up . Note that the total resistance is higher than the highest individual resistance in the circuit. 3. Applying Ohm's Law to a Series Circuit VT = IT × RT 20 IOR V1 = I1 × R1 5 5a T i --T - - II "sor T] I 52 M - > V2 = I2 × R2 50 -7 ,- ↳ Li -- i - h 902 I -I T > 14 Smart Electrical Technology 4. Worked Examples R1 I1 I2 V1 R2 V2 IS VS y (A) Fig. 1.4-2: Series circuit Refer to Fig. 1.4-2. Let R1 = 10 Ω, R2 = 20 Ω and VS = 120 V. Determine: (i) the total resistance, RT RT = R1+ R2 = 10 Ω + 20 Ω = 30 Ω (ii) the supply current, IS IS = VS RT = 120 V 30 Ω =4A (iii) the voltage across R1 VS = IS × R1= 4 A × 10 Ω = 40 V (B) Refer to Fig. 1.4-2. Let R1 = 40 Ω, IS = 0.5 A and VT = 60 V. Determine: (i) the total resistance, RT VT 60 V = RT = = 120 Ω 0.5 A IS (ii) the resistance, R2 R2 = RT − R1= 120 Ω − 40 Ω = 80 Ω (iii) the voltage across R1 V1 = IS × R1= 0.5 A × 40 Ω = 20 V 15 Smart Electrical Technology 5. Characteristics of a Parallel Circuit I1 R1 V1 I2 R2 V2 IS 2 VS y (A) I -I ↑~47 07 Fig. 1.4-3: Parallel circuit - I The supply voltage is equal to all branch voltages, that is: Tovs L- VS or VT = V1 = V2 (B) - . The supply current or total current is equal to the sum of all individual branch currents: IS or IT = I1 + I2 (C) (i) The total or equivalent resistance is equal to the reciprocal of the sum of the reciprocal of individual resistances: 1 RT = 1 R1 + 1 R2 (ii) When there are only two resistances, the following formula can be used: RT = R1 × R2 R1 + R2 Note that the total or equivalent resistance is lower than the lowest individual resistance in the circuit. 3 resistors Find Ri · resistors : For - 1OR R = + + . ⑧ x = 10 - 10 = +3 tot - 0 + 10 10h - 10 R2 = E + E R Rix Ra R = . 10r =0 3 . low - . 8 ⑧ R + = 3 . 33x I e 16 Smart Electrical Technology 6. Applying Ohm’s Law to a Parallel Circuit VT = IT × RT 7. V1 = I1 × R1 V2 = I2 × R2 Worked Examples I1 R1 V1 I2 R2 V2 IS VS y (A) Fig. 1.4-4: Parallel circuit Refer to Fig. 1.4-4. Let R1 = 30 Ω, R2 = 20 Ω and VS = 120 V. Determine: (i) the total resistance, RT RT = 1 1 1 + R1 R2 = 1 1 1 + 30 Ω 20 Ω = 12 Ω (ii) the supply current, IS VS 120 V = IS = = 10 A RT 12 Ω (iii) the current flowing in R1 V1 = VT = 120 A I1 = V1 120 V = =4A 30 Ω R1 (iv) the current flowing in R2 I2 = V2 120 V = =6A 20 Ω R2 17 Smart Electrical Technology (B) Two resistors of 80 Ω and 60 Ω respectively are connected in parallel across a 100 V supply. Determine the total resistance of the circuit and the current flowing in each of the branches. RT = 8. 1 = 1 1 + R1 R2 1 1 1 + 80 Ω 60 Ω I1 = V1 100 V = = 1.25 A 80 Ω R1 I2 = V2 100 V = = 1.67 A 60 Ω R2 = 34.30 Ω Pl 18 Tutorial . R1 I1 ⑮ries 80 R2 I2 e4o 1602 V1 . V2 R Is IS - 0 2A Vs . 0 2A VS . y V 1 + = Fig. 1.4-5: Series circuit = 02 . (A) . Refer to Fig. 1.4-5. Let R1 = 80 Ω, R2 = 160 Ω and IS = 0.2 A. Determine: (i) the total resistance, RT R + = = (ii) the supply voltage, VS V V = = R + L 240 * (240 Ω) 48 (120 V) + , Vex U (iii) the voltage across R1 = I xR 0 2 x 240= . (40 V) - - - V I R = = , , 0 2Ax 802 . = V 18 Smart Electrical Technology (iv) the voltage across R2 (80 V) (B) Two resistors of 50 Ω and 70 Ω respectively are connected in series across a 100 V supply. Determine the total resistance of the circuit and the supply current. (120 Ω; 0.83 A) (C) Two resistors of 100 Ω and R Ω respectively are connected in series across a 110 V supply. If the current drawn is 0.5 A, determine the value of R. (120 Ω) (D) Refer to Fig. 1.4-6. Let R1 = 60 Ω, R2 = 80 Ω and IT = 1 A. I1 R1 V1 I2 R2 V2 IS VS y Fig. 1.4-6: Parallel circuit Determine: (i) the total resistance, RT (ii) the supply voltage, VS (34.29 Ω) (34.29 V) 19 Smart Electrical Technology (iii) the current flowing in R1 (iv) the current flowing in R2 (E) (0.43 A) Two resistors of 60 Ω and 90 Ω respectively are connected in parallel across a 48 V supply. Determine: (i) the total resistance of the circuit (ii) the supply current (iii) the current flowing in each of the two branches (F) (0.57 A) (36 Ω) (1.33 A) (0.8 A; 0.53 A) The total resistance of a parallel circuit with two resistors is 300 Ω. The resistance of one of the resistors is 500 Ω. If the supply voltage is 100 V, determine: (i) the supply current (ii) the current flowing in each of the two branches (iii) the resistance of the other resistor (0.33 A) (0.2A; 0.13 A) (769.23 Ω) 20 P= I (ER) ETYI at p -TV Smart Electrical Technology = 1.5 Power and Energy in an Electric Circuit Learning Objectives  Define the terms “power” and “energy” in electric circuits  State the units of measurement for power (watt) and energy (joule)  State and apply the formula to determine the power in an electric circuit comprising one electrical load: P = V × I  State and apply the formula to determine the energy in an electric circuit comprising one electrical load: E = P × t  State the practical unit of energy consumption in a household (kilowatt-hour or kWh)  Calculate the energy consumption of an electrical load in kilowatt-hours 1. Power VFIB V R I Ohn's Law Triangle Power is the rate of doing work. The faster the work is done, the more power is required to do it. Power is represented by P and is measured in watts (W) or joules per second (J/s). In a direct current circuit, power can be determined by the following formulae: P = VI P = I 2R V2 P= R 2. * P I V P-V-I Relationship Triangle y Fig. 1.5-1: Formulae for determining power Power Rating of Electrical Load Most electrical loads have ratings that include both voltage and power. The voltage rating of the load is an indication of the voltage the load is designed for. The power rating will determine how much work the load can do. Here are some common ratings of home appliances: • Light bulb (60 W, 230 V; 40 W, 230 V) • Fluorescent lamp (32 W, 230 V) • Water heater (3,000 W, 230 V) 60W W lightbulb is the will unit be brighter than a for power Watt 40W . . . A 60 W bulb can do more work, and would thus be brighter, than a 40 W bulb. 21 V IR = P Power Triangle Ohi's Law = Triangle P V I IV R I V P IR = p = 100 Example) ① E Qu Power of the -> bulb? I'll LOV V 20V = R Method 1 : - P 12 = E 100 = Watts . = 100R I Method 2 : I= = 20V - 100R I 0 2A . P IV = =0 2x 20 . =N Smart Electrical Technology 3. Worked Examples (A) Determine the rated current and resistance of a 60 W, 230 V bulb. W L P 60 W = 0.26 A I= V = 230 V V 230 V = R= = 884.62 Ω I 0.26 A (B) I (0.26 A, 884.62 Ω) P IV I I = Ne = = = - = 884 62e . - A resistance of 10 Ω is connected to a voltage of 220 V. Determine the power dissipated. V 2 (220 V)2 = P= = 4840 W = 4.84 kW R 10 Ω = 10 (4.84kW) . V = 220V use y2 P= E . The power dissipated in a 1,000 Ω resistor is 50 W. Determine the supply voltage and current. (223.61 V; 0.22 A) Given P= V R : 2 R = p = 1000 - BOW . V2 = P × R V= I= 4. R 6A = Given · R (C) Resistance of the bulb? = P×R = p= I PxR 50 W × 1000 Ω = P 50 W = = 0.22 A V 223.61 V 50000 = 223.61 V = V V V = = > PXR NFR Energy Energy is defined as the capacity to perform work. Energy exists in many forms and may be converted from one form into another. For example, a lead-acid battery converts chemical energy into electrical energy when it discharges, and vice-versa when it is being charged; a generator converts mechanical energy into electrical energy; and an electric radiator converts electrical energy into heat energy. Energy must be consumed in order to do work, and the amount of work done is directly proportional to the energy used. 22 Smart Electrical Technology The electrical energy taken from a source depends on the electric power of the appliance and the length of time used. Energy is represented by W and is measured in joules (J) or watt-seconds (Ws). 1 joule 1 watt 100 watts = 1 watt-second = 1 joule per second = 100 joules per second 1kW 1000N Seed = ↳ KIE multipl However, the joule or watt-second is too small to be used for computing the electrical energy consumed by households or industries. The practical unit used for this purpose is kilowatt-hour (kWh). & Electrical energy in kWh can be determined by the following formula: Energy (W) = Power (P) × Time (t) where power is in kilowatts (kW) and time is in hours (h). 5. Cost of Energy (Energy Bill) The cost of electrical energy is calculated by the product of energy consumed in kWh and the rate per unit. 1 unit = 1 kWh Cost of Energy = Energy (kWh) × Rate ($) Energy SUMMARY 15 - * L soules [51 kW = 1Ws Slice as Waitt kilowatt hour practical unit IkWL) second AsI F birth - . Energy - - - Power IkW] + + time I] 23 Note Jouks IWs] 1kW= 1000 jox IsWhIkWLI = 2 kWh = . ? Joues x - = 1kWh 106j = 1 = 3600000 million J . . = - 3600000 J 36 . 1kWh 3 6 3600000 Is 15 units - - = = 2 Conversion between the : kNG kWK 20-7 Smart Electrical Technology . 3 6 x10 % . . 6. Worked Examples (A) A 1,000 W, 230 V heater is turned on for 10 hours. Determine the electrical energy used in: Energy = Powerx time . (i) joules IWsI W = P × t = 1000 W × (10 × 60 × 60) s = 36000000 J = 3.6 × 107 J me seconds Power 36 million J (ii) kilowatt-hours IkWh] W = P × t = 1 kW × 10 h = 10 kWh Tier urs (B) A 2 kW refrigerator is turned on 24 hours per day. Determine: C Power Rating . (i) the energy consumed (kWh) in one month (30 days) I day 247 30 : ↓ X days W = 2 kW × (24 × 30) h = 1440 kWh nee e k I unit (ii) the cost of using the refrigerator at $0.25 per unit = =1 kNG $0 25 per . kWh Cost = 1440 kWh × $0.25 = $360 trogy htper unit used (C) What is the total energy consumed by a 1,000 W heater for 30 days if it is turned on for an average of 10 hours per day? If the rate of energy is $0.25 per unit, what is the bill for using this heater? Total estBods e Energy W = 1 kW × (10 × 30) h = 300 kWh ~ power my Cost = 300 kWh × $0.25 = $75 7. How to Interpret an Energy Bill Most households in Singapore pay their utilities bill to SP Services. The bill has three main components: • electricity services • gas services • water services 24 y Fig. 1.5-2: Utilities bill of a residential premise Smart Electrical Technology 25 Smart Electrical Technology 8. High-Energy Consumption Home Appliances The higher the power rating, the more energy an appliance consumes. Home appliances that consume a lot of energy include the: • refrigerator • air conditioner • water heater • electric iron 9. Strategies to Reduce Electrical Energy Consumption It is important for us to reduce our energy consumption. Not only does this reduce our energy bills, it also helps lessen the effects of air pollution and global warming. Making energy-efficient choices allows us to save electricity and money without compromising on our quality of life. Here are some easy energy-saving habits: • Unplug Unplug chargers that are not in use. Switch off the power supply to televisions, home theatre equipment and stereos when not in use. If you switch them off using only the on/off button, their combined stand-by consumption may be equivalent to that of a 75 or 100 W light bulb. • Set Computers to “Sleep” and “Hibernate” Enable the “sleep” mode on the computer to reduce power consumption during periods of inactivity. Set the computer to automatically “hibernate” after 30 minutes of inactivity. When you are done working for the day, shut down the computer. • Take Control of the Temperature Set the air conditioner to about 25 °C. • Use Appliances Efficiently Check that the seal on the oven door is undamaged and tight. When cooking with the oven, do not open the door unnecessarily. Use a microwave oven to reheat small amounts of food. When using the washing machine, set the water level to match the size of each load. Use cold water to wash and rinse clothes. • Use Energy-Saving Light Bulbs Replace incandescent light bulbs with energy-saving compact fluorescent bulbs or LED bulbs. • Flick a Switch Switch off the lights when you leave a room. 26 P IV V IR = = Smart Electrical Technology 10. Tutorial (A) A resistor of 500 Ω is connected to a supply voltage of 230 V. Determine the power dissipated. (105.8 W) 1 P /V R I = = R =0 H6A . - + 230V =8 230V -500R (B) V 06 The power dissipated in a 200 Ω resistor is 1,000 W. Determine the supply voltage and current. (2.24 A; 448 V) IR P 1000 = = = I = I 200 = I I N5 = 100 (C) = a =24A A 100 W, 230 V heater is turned on for 10 hours. Determine the energy used in: (i) joules (ii) kilowatt-hours (i) . Energy (Joules) = Time in seconds) 100N-(10/60 60s) + = 100 + 36000s . -0005 - (D) I (ii) YOOW 11 kW = Energy (kWh) =0 1kW . = (3,600,000 J) (1 kWh) 10 hours + 1 - - A 1 kW refrigerator is turned on for 24 hours per day. What is the energy consumed in one month (30 days)? (720 kWh) Total hours 30 x 24 Feohours : = . Energy (E) = 1kW + 7207 G = * I A 1,000 W air conditioner in a residential premise is turned on for six hours per day. Determine: (i) the energy consumed in 30 days (ii) the cost of using the air conditioner for 30 days if the rate of energy is $0.25 per unit (i) Total kours = 30 days - 64 =180 hours Energy = 1kW = x (ii) Cost $0 = . 25 x 180 (180 kWh) ($45) = 180 hours N 27

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