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2.2 Resistance G482 Electricity, Waves & Photons 2.2 Resistance G482 Electricity, Waves & Photons 2.2.5 Power Mr Powell 2012 Index.

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Presentation on theme: "2.2 Resistance G482 Electricity, Waves & Photons 2.2 Resistance G482 Electricity, Waves & Photons 2.2.5 Power Mr Powell 2012 Index."— Presentation transcript:

1 2.2 Resistance G482 Electricity, Waves & Photons 2.2 Resistance G482 Electricity, Waves & Photons Power Mr Powell 2012 Index

2 Mr Powell 2012 Index Power (Prior Learning) When we talk about Power what we mean is “the amount of energy delivered per second” 1 Joule / 1 Second = 1 Watt It then makes sense that the Power used by a component can be found from the product of current through and voltage across the component; Power = Voltage x Current P = V x I

3 Mr Powell 2012 Index Analogy Another way of thinking about it is saying that the current carries the energy; 4V 1A CCCC C C CC C 1J C As the Coulombs of Charge move they release their energy as heat and light (through the bulb) C 1J = 1 Coulomb of charge = 1 Joule of energy = 1 Second of time

4 Mr Powell 2012 Index Analogy 2 If the voltage increases, more energy is delivered so the power increases; 5V 1A CCCC C C CC C 1J C Power = 5V x 1A = 5J/s = 5W 1J C = 1 Coulomb of charge = 1 Joule of energy = 1 Second of time

5 Mr Powell 2012 Index Analogy 3 If the current increases, more energy is delivered so the power increases; 4V 2A CCCC C C CC C C 1J Power = 4V x 2A = 8J/s = 8W CCCC C C CC C C 1J C = 1 Coulomb of charge = 1 Joule of energy = 1 Second of time

6 Mr Powell 2012 Index Flow of Charge When an electrical appliance is on, electrons are pushed through the appliance by the potential difference The 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/s The 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

7 Mr Powell 2012 Index Energy and Potential Difference When 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. E = VQ Q=It E = VIt

8 Mr Powell 2012 Index Power Assessable learning outcomes a)describe power as the rate of energy transfer; b)select and use power equations P = VI, P = I 2 R, P = V 2 /R c)explain how a fuse works as a safety device (HSW 6a); d)determine the correct fuse for an electrical device; e)select and use the equation W = IVt; f)define the kilowatt-hour (kW h) as a unit of energy; g)calculate energy in kW h and the cost of this energy when solving problems (HSW 6a). Assessable learning outcomes a)describe power as the rate of energy transfer; b)select and use power equations P = VI, P = I 2 R, P = V 2 /R c)explain how a fuse works as a safety device (HSW 6a); d)determine the correct fuse for an electrical device; e)select and use the equation W = IVt; f)define the kilowatt-hour (kW h) as a unit of energy; g)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)

9 Mr Powell 2012 Index a) describe power as the rate of energy transfer;

10 Mr Powell 2012 Index a) HSW - PC Energy Use Question Energy = Power x Time Energy = 154W x 1 hour Energy = 154W x 3600s E = 154J/s x 3600s E = J E = 554.4kJ 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; Current flow to the PC would then be… P = VI or P/V = 0.67A

11 Mr Powell 2012 Index Charge carriers transfer kinetic energy to positive ions through repeated collisions. The pd across the material then provides an accelerating force to the charge carrier which then collides with another positive ion. V = I R P = I V P = I 2 R P = V 2 R P = V 2 R Energy per second transferred to the component P: Energy per second transferred to the component P: In the steady state this equals the heat transfer to the surroundings In the steady state this equals the heat transfer to the surroundings Resistance Heating...

12 Mr Powell 2012 Index b) select and use power equations P = VI, P = I 2 R, P = V 2 /R P = VI P = Et V = IR P = I 2 R P = V 2 /R Q = It E = Vit E = VQ TASK: Try and rearrange the equations and substitute so you can get from one to another!

13 Mr Powell 2012 Index c) explain how a fuse works as a safety device (HSW 6a);

14 Mr Powell 2012 Index Heater Circuit 3 Pin Plug / Mains Inside 3 Pin Plug Appliance

15 Mr Powell 2012 Index Facts Fuses  Fuses are thin pieces of wire which resist a large current  They melt when too much current flows  Circuit is broken  Have to be replaced Facts Circuit Breakers  Circuit breakers are designed to flip over when too large a current flows  They break the circuit  Rely on magnetic forces

16 Mr Powell 2012 Index d) 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 the power rating of the appliance and its potential difference (voltage).

17 Mr Powell 2012 Index e) 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 = mc  E is energy transferred in joules, J m is mass in kilograms, kg  is temperature change in degrees Celsius, °C c is specific heat capacity in J / kg °C 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 = mc  E is energy transferred in joules, J m is mass in kilograms, kg  is temperature change in degrees Celsius, °C c is specific heat capacity in J / kg °C

18 Mr Powell 2012 Index Diagram Equipment two digital multimeters (V & A) 12V Metal block Heater Thermometer A A V V 1 kg Metal blocks (Cu, Fe, Al) thermometer, 0 – 100  C 12 V heater power supply, 0 – 12 V DC electrical leads stopwatch

19 Mr Powell 2012 Index Using a Multi-meter Volts or mA or  DC Volts Various Ranges Resistance ranges in  DC mA various ranges Normal Currents up to 10A

20 Mr Powell 2012 Index Diagram Homework - Design a new experiment for water? two digital multimeters (V & A) thermometer, 0 – 100  C power supply, 0 – 12 V DC electrical leads stopwatch ? ?

21 Mr Powell 2012 Index Results.... ABCDEFG Metal (1kg) Start Temp /  C End Temp /  C Temp Rise /  C Current /A PD / V Time for rise / s SHC / J /kg°C ? Al Fe Cu Quoted.... Aluminium (Al) is 902 J /kg°C Iron (Fe) 450 J /kg°C Copper (Cu) is 385 J /kg°C Quoted.... Aluminium (Al) is 902 J /kg°C Iron (Fe) 450 J /kg°C Copper (Cu) is 385 J /kg°C Analysis Thoughts? 1.Are there any flaws with your experiment where electrical or thermal energy are lost and don’t raise the temperature of the block? 2.What could you do to make the experiment more reliable? 3.Can you think of a similar experiment to find the SHC of water? Analysis Thoughts? 1.Are there any flaws with your experiment where electrical or thermal energy are lost and don’t raise the temperature of the block? 2.What could you do to make the experiment more reliable? 3.Can you think of a similar experiment to find the SHC of water?

22 Mr Powell 2012 Index f) 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.

23 Mr Powell 2012 Index When Maths starts to hurt!

24 Mr Powell 2012 Index g) 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..

25 Mr Powell 2012 Index Answer…. Plenary Question…. 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) iSlice power determines heat produced (1) in series, current is same (1)  I 2 R A must be > I 2 R B (1)  in parallel, p.d. is same (1) <

26 Mr Powell 2012 Index P = VI P = Et V = IR P = I 2 R P = V 2 /R Q = It E = Vit E = VQ Can you connect the formulae...

27 Mr Powell 2012 Index Connection Connect your learning to the content of the lesson Share the process by which the learning will actually take place Explore the outcomes of the learning, emphasising why this will be beneficial for the learner Connection Connect your learning to the content of the lesson Share the process by which the learning will actually take place Explore the outcomes of the learning, emphasising why this will be beneficial for the learner Demonstration Use formative feedback – Assessment for Learning Vary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sex Offer different ways for the students to demonstrate their understanding Allow the students to “show off” their learning Demonstration Use formative feedback – Assessment for Learning Vary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sex Offer different ways for the students to demonstrate their understanding Allow the students to “show off” their learning Activation Construct problem-solving challenges for the students Use a multi-sensory approach – VAK Promote a language of learning to enable the students to talk about their progress or obstacles to it Learning as an active process, so the students aren’t passive receptors Activation Construct problem-solving challenges for the students Use a multi-sensory approach – VAK Promote a language of learning to enable the students to talk about their progress or obstacles to it Learning as an active process, so the students aren’t passive receptors Consolidation Structure active reflection on the lesson content and the process of learning Seek transfer between “subjects” Review the learning from this lesson and preview the learning for the next Promote ways in which the students will remember A “news broadcast” approach to learning Consolidation Structure active reflection on the lesson content and the process of learning Seek transfer between “subjects” Review the learning from this lesson and preview the learning for the next Promote ways in which the students will remember A “news broadcast” approach to learning

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