22.2.5 Power (Prior Learning) When we talk about Power what we mean is “the amount of energy delivered per second”1 Joule / 1 Second = 1 WattIt then makes sense that the Power used by a component can be found from the product of current throughand voltage across the component;Power = Voltage x CurrentP = V x I
3AnalogyAnother way of thinking about it is saying that the current carries the energy;4V1A1J1J1J1J1J1J1J1JCCCCCCCCCCC= 1 Coulombof charge1J= 1 Joule of energyAs the Coulombs of Charge move they release their energy as heat and light (through the bulb)= 1 Second of time
4Analogy 2If the voltage increases, more energy is delivered so the power increases;5V1A1J1J1J1J1J1J1J1J1J1J1J1JCCCCCCCCCCC= 1 Coulombof charge1J= 1 Joule of energyPower = 5V x 1A = 5J/s = 5W= 1 Second of time
5Analogy 3If the current increases, more energy is delivered so the power increases;4V2A1JCCCCCCCCC= 1 Coulombof chargeCC1J1J1J1JC1J1J1J1J= 1 Joule of energy1J= 1 Second of timePower = 4V x 2A = 8J/s = 8W
6Flow of ChargeWhen an electrical appliance is on, electrons are pushed through the appliance by the potential differenceThe potential difference causes a flow of charge particles (free electrons). The rate of flow of charge is the electric current through the appliance. 1A = 1C/sThe unit of charge, the coulomb (C).The charge passing along a wire or through a component in a certain time depends on: the current, and the time.We can calculate the charge using the equation:Q=It
7Energy and Potential Difference E = VQQ=ItE = VItWhen a resistor is connected to a battery, electrons are made to pass through the resistor by the battery.Each electron repeatedly collides with the vibrating atoms of the resistor, transferring energy to them.The atoms of the resistor therefore gain kinetic energy and vibrate even more. The resistor becomes hotter.When charge flows through a resistor, electrical energy is transformed into heat energy.The energy transformed in a certain time in a resistor depends on: the amount of charge that passes through it, and the potential difference across the resistor.
82.2.5 Power Assessable learning outcomes describe power as the rate of energy transfer;select and use power equations P = VI, P = I2R, P = V2/Rexplain how a fuse works as a safety device (HSW 6a);determine the correct fuse for an electrical device;select and use the equation W = IVt;define the kilowatt-hour (kW h) as a unit of energy;calculate energy in kW h and the cost of this energy when solving problems (HSW 6a).A Joulemeter (V, I and t) or a data-logger may be used to determine energy transfer.A utilities statement can be used to illustrate the use of the kW h by electricity companies. (Homework)
9a) describe power as the rate of energy transfer;
10a) HSW - PC Energy Use Question Here is an example of the energy usage of a home computer. Each internal component transfers a variable amount of energy. If you left this PC on for an hour it would transfer;Energy = Power x TimeEnergy = 154W x 1 hourEnergy = 154W x 3600sE = 154J/s x 3600sE = JE = 554.4kJCurrent flow to the PC would then be…P = VI or P/V = 0.67A
11Resistance Heating...Charge carriers transfer kinetic energy to positive ions through repeated collisions.The pd across the material then provides an accelerating force to the charge carrierwhich then collides with another positive ion.V = I RP = I VP = V2REnergy per second transferredto the component P:P = I2 RIn the steady state this equalsthe heat transfer to the surroundings
12b) select and use power equations P = VI, P = I2R, P = V2/R TASK: Try and rearrange the equations and substitute so you can get from one to another!P = VIP = EtV = IRP = I2RP = V2/RQ = ItE = VitE = VQ
13c) explain how a fuse works as a safety device (HSW 6a);
15Facts Circuit Breakers Facts FusesFuses are thin pieces of wire which resist a large currentThey melt when too much current flowsCircuit is brokenHave to be replacedFacts Circuit BreakersCircuit breakers are designed to flip over when too large a current flowsThey break the circuitRely on magnetic forces
16d) determine the correct fuse for an electrical device; Domestic appliances are often fitted with a 3A, or a 5A or a 13 A fuse.If you don’t know which one to use for an appliance, you can work it out from thepower rating of the appliance and its potential difference (voltage).
17e) select and use the equation W = IVt; (SHC Practical) The specific heat capacity of a substance is the amount of energy required to change the temperature of one kilogram of the substance by one degree Celsius.E = mcE is energy transferred in joules, Jm is mass in kilograms, kg is temperature change in degrees Celsius, °Cc is specific heat capacity in J / kg °C
181 kg Metal blocks (Cu, Fe, Al) Equipmenttwo digital multimeters (V & A)power supply, 0 – 12 V DCthermometer, 0 – 100C12 V heaterDiagramstopwatch12VAVThermometerHeaterelectrical leadsMetal block1 kg Metal blocks (Cu, Fe, Al)
19Using a Multi-meter DC Volts Various Ranges DC mA various ranges Resistance ranges in Normal Currents up to 10AVolts or mA or
20Homework - Design a new experiment for water? power supply, 0 – 12 V DC?thermometer, 0 – 100CstopwatchDiagram?two digital multimeters (V & A)electrical leads
21Results.... A B C D E F G Metal (1kg) Start Temp / C End Temp / C Temp Rise / CCurrent /APD / VTime for rise / sSHC/ J /kg°C?21.53412.52.629.1420801AlFeCuQuoted....Aluminium (Al) is 902 J /kg°CIron (Fe) 450 J /kg°CCopper (Cu) is 385 J /kg°CAnalysis Thoughts?Are there any flaws with your experiment where electrical or thermal energy are lost and don’t raise the temperature of the block?What could you do to make the experiment more reliable?Can you think of a similar experiment to find the SHC of water?
22f) define the kilowatt-hour (kW h) as a unit of energy; The kilowatt hour, or kilowatt-hour, (symbol kW·h, kW h or kWh) is a unit of energy equal to 1000 watt hours or 3.6MJ.For constant power, energy in watt hours is the product of power in watts and time in hours.The kilowatt hour is most commonly known as a billing unit for energy delivered to consumers by electric utilities.The kilowatt-hour (symbolized kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power expended for one hour (1 h) of time.Inversely, one watt is equal to 1 J/s.One kilowatt hour is 3.6 megajoules, which is the amount of energy converted if work is done at an average rate of one thousand watts for one hour.
24g) calculate energy in kW h and the cost of this energy when solving problems (HSW 6a). Poor Dr Frankenstein did not look at his electricity bill and check the cost of each unit of electricity.1 Unit = 1KW hour of electricity.They are shown on the bill here..
25Plenary Question…. Answer…. Two resistors, A and B, have different resistances but otherwise have identical physical properties. E is a cell of negligible internal resistance.When the resistors are connected in the circuit shown in figure 1, A reaches a higher temperature than B. When connected in the circuit shown in figure 2, B reaches a higher temperature than A. Explain these observations fully, stating which resistance is greater. (6 marks)Answer….power determines heat produced (1)in series, current is same (1)I2 RA must be > I2 RB (1)in parallel, p.d. is same (1)iSlice<
26Can you connect the formulae... P = V2/RQ = ItE = VitE = VQP = VIP = EtV = IRP = I2R
27ConnectionConnect your learning to the content of the lessonShare the process by which the learning will actually take placeExplore the outcomes of the learning, emphasising why this will be beneficial for the learnerDemonstrationUse formative feedback – Assessment for LearningVary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sexOffer different ways for the students to demonstrate their understandingAllow the students to “show off” their learningConsolidationStructure active reflection on the lesson content and the process of learningSeek transfer between “subjects”Review the learning from this lesson and preview the learning for the nextPromote ways in which the students will rememberA “news broadcast” approach to learningActivationConstruct problem-solving challenges for the studentsUse a multi-sensory approach – VAKPromote a language of learning to enable the students to talk about their progress or obstacles to itLearning as an active process, so the students aren’t passive receptors