SET Chapter 5 PDF

Learning Outcomes Define the terms “power” and “energy” in electric circuits State the units of measurement for power (watt) and energy (joule) Or (kilowatthour) State and apply the formula to determine the power in an electric circuit comprising one electrical load: P =...

Learning Outcomes Define the terms “power” and “energy” in electric circuits State the units of measurement for power (watt) and energy (joule) Or (kilowatthour) 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 Electrical how much energy per second t 1.5.1 Power 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: V2 P = VI P=I R 2 P= R where P is power in watts (W) V is voltage in volts (V) I is current in amperes (A) 0 InA E.g.. -- 0 1 + 1000. 00 = Unit 1.5 | Power and Energy in an Electric Circuit 29  1.5.2 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 the amount of work the load can do within a unit of time. Here are some common ratings of home appliances: Voltage Rating How much voltage the : Light bulb (60 W, 230 V; 40 W, 230 V) appliance is designed for. Fluorescent lamp (32 W, 230 V) Water heater (3000 W, 230 V) Power Rating How much : energy the appliance do second. can per A 60 W bulb can do more work than a 40 W bulb within the same unit of time, and thus would be brighter. 1.5.3 Worked Examples (a) For a 60 W, 230 V bulb, determine: P (i) the rated current. I V P P I = I = v V 60 W GOW = I 230 V 230V = 0.26 A =A (ii) the resistance. V R= V R = I V I 230 V R 230 = I = 0.26 A 0. 26A = 884.62 Ω = 8852 - ⑭ O (b) A resistor of resistance 10 Ω is connected to a supply voltage of 220 V. Determine the power dissipated. W P = V2 # formula to calculate Power 3. R (220 V)2 OP = IV = 10 Ω = 4840 W = 4.84 kW ②P = TPR 1 kW = 1000 W its 1000 w = 1kW ③ P = Unit 1.5 | Power and Energy in an Electric Circuit 30 (c) The power dissipated by a 1000 Ω resistor is 50 W. Determine: (i) the supply voltage. P= V2 P = R V2 50 = V =P×R 2 1000 V=√P×R 50 x 1000 = 12 = √ 50 W × 1000 Ω V2 = 50000 = √ 50 000 N = 50000 = 223.61 V V (ii) the current. P I = V 50 W = 223.61 V = 0.22 A Unit 1.5 | Power and Energy in an Electric Circuit 31  Power is the rate of doing work. [W]: watts Energy is the capacity to perform work. [kWh]: kilowatt-hour 1.5.4 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 is needed to do work, and the amount of work done is directly proportional to the energy used. The electrical energy transferred from a source to an appliance depends on the electric power of the appliance and the length of time used. Energy is represented by E and is measured in either joules (J) or watt-seconds (Ws). However, a joule or a watt-second is a small amount of energy, making it inappropriate for use to describe or compute large amount of electrical energy which households or industries tend to use most of the time. For example, a typical household refrigerator can use up to 3,600,000 J of electrical energy in a day. Thus, we use kilowatt-hour (kWh) as the unit for electrical energy usage or consumption to reduce the number of numerical digits involved in presentation and calculations. To put it simply, using kilowatt-hour as the unit for electrical energy is more practical. For example, 3,600,000 J of electrical energy is equivalent to only 1 kWh. 36000005 = 1kWh. Electrical energy in kWh can be determined by the following formula: E=P×t where E is energy in watts (kWh) P is power in watts (kW) I is time in hours (h) 1J 1Ws Notes! for Energy = Unit Energy Power = time 1h = 60min x60s. Joules [J] IWstJ [J] IWIx Is] 600sec = - =. - Watt-second - kilowatt-hour IkWh] IkWh] = IkW] x Ih/ Unit 1.5 | Power and Energy in an Electric Circuit 32  1.5.5 Cost of Electrical Energy (Electrical Bill) The cost of electrical energy is calculated by the product of energy consumed (used) in kWh and the rate of per unit of electrical energy. 1 unit = 1 kWh Cost of Electrical Energy ($) = Energy Consumed (kWh) x Rate ($/kWh) 1.5.6 Worked Examples Power Voltage time (a) A 1000 W, 230 V heater is turned on for 10 hours. Determine the electrical energy used in: watts Time seconds (i) joules. in ~ Power in. - E=P×t to seconds = 1000 W × 10 h ↳ change. = 1000 W × (10 × 60 × 60) s = 1000 W × 36 000 s 1 min = 60 s = 36 000 000 J 1 h = 60 mins = 3600 s = 3.6 × 107 J (ii) kilowatt-hours. IW E=P×t = 1000 W × 10 h = 1 kW × 10 h 1 kW = 1000 W = 10 kWh (b) A 2 kW refrigerator is turned on 24 hours per day. Determine: (i) the energy consumed (kWh) in one month (30 days). E=P×t = 2 kW × 24 h × 30 days = 2 kW × 720 h 1 unit 1kWh. = = 1440 kWh Y (ii) the cost of using the refrigerator at $0.25 per unit. Cost = 1440 kWh × $0.25/kWh = $360 Unit 1.5 | Power and Energy in an Electric Circuit 33 (c) A 900 W heater is turned on for an average of 30 minutes per day for a period of 1 week. If the rate of energy is $0.25 per unit, what is the total cost of energy for using this heater? E=P×t = 900 W × 30 mins × 7 days = 0.9 kW × 0.5 h × 7 days = 0.9 kW × 3.5 h = 3.15 kWh Cost = 3.15 kWh × $0.25/kWh = $0.79 Unit 1.5 | Power and Energy in an Electric Circuit 34  1.5.7 How to Interpret a Utilities Bill Households in Singapore pay their utilities bill to electrical suppliers (e.g., SP Services). The bill has three main components: electricity services gas services water services 229 kWh of electrical energy used within the billing period Fig. 1.5-1a: Utilities bill of a residential premise (Part 1) Unit 1.5 | Power and Energy in an Electric Circuit 35  cost of electrical energy = 229 kWh × $0.2241 = $51.32 electrical energy used within the billing period rate of per unit of electrical energy Fig. 1.5-1b: Utilities bill of a residential premise (Part 2) 1.5.8 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: refrigerator; air conditioner; water heater; and electric iron. Unit 1.5 | Power and Energy in an Electric Circuit 36  1.5.9 Strategies to Reduce Electrical Energy Consumption It is important for us to reduce our energy consumption by conserving electricity. Conserving electricity can reduce our expenses on electricity bills, contribute to our sustainability efforts and help to mitigate climate change. Making energy-efficient choices allows us to conserve electricity and save money without compromising on our quality of life. Here are some easy energy-saving habits: (a) Unplug Unplug chargers that are not in use. Switch off the power supply to electrical appliances when they are not in use. For example, when we switch the television, home theatre equipment and stereos to stand-by mode instead of switching them off completely, their combined electricity consumption may be equivalent to that of a 75 or 100 W light bulb. (b) 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 a short period (e.g., 15 minutes) of inactivity. When you are done working for the day, shut down the computer. (c) Take Control of the Temperature Set the air conditioner to about 25 °C. (d) Use Appliances Efficiently Check that the seal on the oven door is undamaged and the oven can be tightly closed. When cooking with the oven, do not open the door unnecessarily. When using the washing machine, set the water level to match the size of each load. Use cold water to wash and rinse clothes. (e) Consider Purchasing Appliances with More Ticks on the Energy Labels Consider purchasing electrical appliances with more ticks on the Energy Labels as more ticks means these appliances have better energy-saving capability. 6 Fig. 1.5-2: Sample Energy Label for a certain air-conditioner Unit 1.5 | Power and Energy in an Electric Circuit 37 (f) Use Energy-Saving Light Bulbs Replace incandescent light bulbs with energy-saving compact fluorescent bulbs or LED bulbs. (g) Flick a Switch Switch off the lights when they are not in use. Unit 1.5 | Power and Energy in an Electric Circuit 38 Pg3R IV P E Pt 1.5.10 Practice Questions R V P? P = 11/ (a) A 500 Ω resistor is connected to a supply voltage of 230 V. Determine the power dissipated. - [105.8 W] P F2R = P-230 500 = 105 8W. P= given out a (b) The power dissipated by a 200 Ω resistor is 1000 W. Determine: R P (i) the supply voltage. [447.2 V] ya 1000 (200) = Va P = R V=72V v2 v2 = 200. 000 1000 = 200 V = 200 , 000 (ii) the supply current. [2.24 A] 1 = = (c) A 100 W, 230 V heater is turned on for 10 hours. Determine the energy used in: W (i) joules. ↑ I [3 600 000 J] Energy P = + t = low (0 + 60x60)s. = looW 36000s - = 36000005. = 3 &. 6 x106] - ki h. (ii) kilowatt-hours. ↑ ↑ [1 kWh] Energy Pxt = & 100W = 0 1kWx10h. 1 kW Ih = 0. = Unit 1.5 | Power and Energy in an Electric Circuit 39 2024 T3WG-7 : WA3 Practical Sep/Oct : EYE MCQ (30Qns) 205 Fo4 WAL T1 SET P2 PatA Part B > - WAZ T2 - > Part C Prelims SET 93 : Home Automat JulAus. SET P1 30 MCQ :.  Current : Amperes [A] Power KilowattIkW] : Power Volts IVI Energy : Kilowatt-hour[kWh] Voltage : time : hours [hI Resistance Ohusie]: (d) A 1 kW refrigerator is turned on for 24 hours per day. What is the energy consumed in one month (30 days)? [720 kWh] Energy Power= time = 1kW (24h +30 days) = 20kWh (e) A 1000 W air conditioner in a residential premise is turned on for 6 hours per day. Determine: (i) the energy consumed in 30 days in kWh. [180 kWh] 1000W Energy-power time = 1kW x(6h x 30 days) =Wh. (ii) the cost of using the air conditioner for 30 days if the rate of energy is $0.25 per unit. 25 cents [$45] per unit $0 25 per kWh. Cost Energy Rate. = x. 1kWh cost $0 25.. = 180kwh x $0 25. = Unit 1.5 | Power and Energy in an Electric Circuit 40