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2.2 Resistance G482 Electricity, Waves & Photons 2.2 Resistance G482 Electricity, Waves & Photons 2.2.2 EMF 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 2.2.2 EMF Mr Powell 2012 Index."— Presentation transcript:

1 2.2 Resistance G482 Electricity, Waves & Photons 2.2 Resistance G482 Electricity, Waves & Photons 2.2.2 EMF Mr Powell 2012 Index

2 Mr Powell 2012 Index 2.2.4 Resistivity Assessable learning outcomes There are opportunities for discussion of the factors that determine resistance including temperature, leading to superconductivity in some materials. Candidates should be able to: (a) define resistivity of a material; (b) select and use the equation..... (c) describe how the resistivity's of metals and semiconductors are affected by temperature; (d) describe how the resistance of a pure metal wire and of a negative temperature coefficient (NTC) thermistor is affected by temperature. Assessable learning outcomes There are opportunities for discussion of the factors that determine resistance including temperature, leading to superconductivity in some materials. Candidates should be able to: (a) define resistivity of a material; (b) select and use the equation..... (c) describe how the resistivity's of metals and semiconductors are affected by temperature; (d) describe how the resistance of a pure metal wire and of a negative temperature coefficient (NTC) thermistor is affected by temperature.

3 Mr Powell 2012 Index GCSE Background From GCSE physics we already know that; V = IR So simply in this circuit the resistance of the bulb is 12 . This is a measure of how much the bulb resists the flow of electrons.  But where does R and rho  come from and how are they linked?  To understand this topic clearly we need to delve into the structure of materials and their electronic configurations.  We must think about the chemistry behind things and the physical dimensions of materials…

4 Mr Powell 2012 Index http://phet.colorado.edu/simulations/index.php?cat=Physics http://www.batesville.k12.in.us/physics/PhyNet/e&m/current/ECurrent_Notes.htm Ohm’s law: The pd across a metallic conductor is proportional to the current through it, provided the physical conditions do not change Q. For the resistor opposite calculate: a) the resistance at this current b) the new pd when the current is 50  A 12V 2mA R ohm a) R = V = 12 = 6000  I 2.0 x 10 -3 b) V = I R = 50x10 -6 x 6000 = 0.3 V Resistance 1

5 Mr Powell 2012 Index Measurement of resistance: A V Record the pd across R for increasing values of current. ( change r to change the circuit current () r The ammeter has a very low resistance ( 0.2 ohm) The voltmeter has a very high resistance ( 20,000 ohm) Why? Pd / V I / A Gradient = V I Gradient = resistance R Resistance 2

6 Mr Powell 2012 Index Spark Version – m = Resistance! Current /A Voltage /V

7 Mr Powell 2012 Index Conductors Metals are generally known as conductors. Copper electricity cables is an example and what we mean is that they conduct electrons very well. Metals are conductors rather than insulators as they have a unique "property". This "property " is that the outermost electron on the atom is relatively loosely held and can hop from atom to atom when pushed!

8 Mr Powell 2012 Index Ionisation Energies Energy in J/mol Metaln=1n=2n=3Free Electrons Li5207298118151 K419305144121 Cu746195835541 or 2 Zn906173338332  The table below shows how much energy it takes to remove a whole mole of electrons from their respective atoms.  The blue shaded regions show which are the easy ones.  It is a similar type of thing that is happening in a conductor…. NB: Don’t need to learn these energies just be aware of the idea of the stripping away!

9 Mr Powell 2012 Index Modelling  So what we are talking about is an electron being ripped from its atom (home) and then moving through the structure of the other adjacent atoms in a form of drift or current.  This is an interesting motion where each electron gains some KE from the e.m.f. Jumps to another adjacent atom losing the KE and then repeats the process over and over….  Here is a simple example of how the process might work with boron NB: Boron was picked due to simple structure – it is not a good conductor! e-e- e-e- e-e- e-e- e-e-

10 Mr Powell 2012 Index Modelling II  Lets examine this rod of an elemental material  You can manipulate the rod and see how many atoms might look.  Then imagine how the electrons would move as on the previous slide  The harder it is to strip away that “free electron” the higher the resistance!

11 Mr Powell 2012 Index Other physical factors… e-e- e-e- e-e- e-e- e-e-  At any temperature above 0K atoms will jiggle around and impede any flow of electrons  Only the electrons move as they are about 2000 times lighter than the atoms they are attached to and pick up the e.m.f  In a Copper wire with 1 x 10 28 electron carriers per m 3 you would have to accelerate 1 x10 -26 kg's of electrons.  This is quite a mass of electrons and the more you have the more push you have to use overall to get them moving!  Area and length must also effect such a problem

12 Mr Powell 2012 Index Resistance or Resistivity  The formulae used to take into account how the physical factors of a wire effect resistance is chicken or egg as you can either consider using it from the R or the  perspective.  The way I remember is resistivity = RAL and of course remember to put the L underneath to make the units correct! NB: when A = 1m 2 and l = 1m  = R

13 Mr Powell 2012 Index Resistivity Formula There are two main principles at play here;  The resistance is proportional to length i.e. the longer the wire the more resistance there is  The resistance is inversely proportional to the area of the wire i.e. the bigger the area the smaller the resistance. where R = Resistance in ohms   = Resistivity in ohm metres  m A = Cross sectional area in metres squared m 2 l = length in metres m NB:  is the taken as the value of the Resistivity at room temperature 20  C

14 Mr Powell 2012 Index Facts and Figures TypeMaterial Resistivity in  m Metal Copper1.7 x 10 -8 Gold2.4 x 10 -8 Aluminium2.6 x 10 -8 Semiconductors Germanium (pure)0.6 Silicon (pure)1.7 x 10 3 Insulators Glass1.7 x 10 12 Perspex1.7 x 10 13 Polyethylene1.7 x 10 14 Sulphur1.7 x 10 15  This table shows some examples you should be familiar with.  Quite simply they mean that for each substance the resistivity will be ……  m (at 20  C) as its constant

15 Mr Powell 2012 Index Simple Example… Question Calculate the resistance of a one metre length of 24 SWG Nichrome wire. Answer The key to getting the correct value is to make sure you use consistent units for the values of resistivity, area and length. In this example SI units are used throughout. There is one other formula that is needed. To calculate the cross-sectional area of the wire we will assume that the wire is circular. So the cross-sectional area, A, is the area of a circle with a radius r. A = π × r² Resistivity of Nichrome = 110 × 10-8 Ω m Length = 1.00 m Diameter of 24 SWG wire = 0.559 mm Radius = 0.559/2 =.2795 mm = 2.795 × 10-4 m Cross-sectional area = π × (2.795 × 10-4)2 = 2.45 × 10-7 m² R = [(110 × 10-8) × 1.00] / [π × (2.795 × 10-4)2] R = 4.48 Ω

16 Mr Powell 2012 Index Simple Example… Question Calculate the resistance of a one metre length of 24 SWG Nichrome wire. Answer The key to getting the correct value is to make sure you use consistent units for the values of resistivity, area and length. In this example SI units are used throughout. There is one other formula that is needed. To calculate the cross-sectional area of the wire we will assume that the wire is circular. So the cross-sectional area, A, is the area of a circle with a radius r. A = π × r² Resistivity of Nichrome = 110 × 10-8 Ω m Length = 1.00 m Diameter of 24 SWG wire = 0.559 mm Radius = 0.559/2 =.2795 mm = 2.795 × 10-4 m Cross-sectional area = π × (2.795 × 10-4)2 = 2.45 × 10-7 m² R = [(110 × 10-8) × 1.00] / [π × (2.795 × 10-4)2] R = 4.48 Ω

17 Mr Powell 2012 Index Simple Application…. Material  Resistivity in  m Thickness cm Length cm Copper1.7 x 10 -8 0.120 Gold2.4 x 10 -8 0.210 Aluminium2.6 x 10 -8 0.412  Robert Oppenheimer was making a complex atomic bomb called “Fat Man”. He needed to use a certain type of wire in his detonator circuits to connect to the primary so that the overall resistance in that part of the microcircuit was exactly 0.0008 . Then his timing would be just right to produce the maximum number of neutrons possible and thus kill as many Japanese civilians in Hiroshima as possible. He had a choice of copper, gold or aluminium wire to use in the circuit. However, each wire was of a different thickness and length.  Without cutting the wires work out which one he could use…you need to do three separate calculations using the following data; Hint: tabulate your data or write it out for each question with conversions in full

18 Mr Powell 2012 Index Electrical Resistance Data SWG - Standard Wire Gauge

19 Mr Powell 2012 Index Material Resistivity at 20°C Ω·mµΩ·cm silver1.6 × 10 -8 1.6 copper1.7 × 10 -8 1.7 gold2.2 × 10 -8 2.2 aluminium2.7 × 10 -8 2.7 magnesium4.2 × 10 -8 4.2 nickel6.9 × 10 -8 6.9 iron10.1 × 10 -8 10.1 chromium13.2 × 10 -8 13.2 manganese160 × 10 -8 160 carbon (graphite)3 000 × 10 -8 3 000 Manganin44 × 10 -8 44 Constantan (Eureka)49 × 10 -8 49 Nichrome110 × 10 -8 110

20 Mr Powell 2012 Index This is the property of a material which is at or below a critical temperature T c where it has zero resistivity. Implications: Zero resistance no pd exists across a superconductor with a current flowing the current has no heating effects Properties of a superconductor:  material loses the effect above the critical temperature Tc.  If Tc is above 77K ( -196 C) it’s a high temperature superconductor  The highest Tc max = 150 K - 123C tempTcTc Superconductivity

21 Mr Powell 2012 Index Finding the Resistivity of a Wire  This new formulae has a simple characteristic which again would fit the; y = mx + c principle  One way of working out a resistivity of any material is to set up a circuit with a sample of the substance. Measure the area of the wire several times and take the average. Then measure the current flow through and potential difference across the wire. This enables you to work out R. Then repeat the experiment for several different lengths. NB:  is the taken as the value of the Resistivity at room temperature 20  C l R

22 Mr Powell 2012 Index Resitivity? Length in m Resistance in m  0.100.18 0.203.70 0.305.50 0.407.10 0.509.20 0.6011.00 0.7013.00 0.8014.50 0.9014.60 1.0018.00  Use the graphing technique to find out the resistivity of a sample wire which is 0.26cm thick. What metal is the wire made from? NB:  calculated at 20  C l R

23 Mr Powell 2012 Index Looks like Gold to me! 1.32739x10 -6 m 2 x 18.253x10 -3  m -1 =2.42 x 10 -8  m

24 Mr Powell 2012 Index d) Thermistor The resistance of a thermistor decreases as the temperature increases so if we look at it from the VI perspective it is the opposite of a bulb!

25 Mr Powell 2012 Index d) How do they work? The exact conduction mechanisms are not fully understood but metal oxide NTC thermistors behave like semiconductors, as shown in the decrease in resistance as temperature increases. The physical models of electrical conduction in the major NTC thermistor materials are generally based on this theory; A model of conduction called "hopping" is relevant for some materials. It is a form of ionic conductivity where ions (oxygen ions) "hop" between point defect sites in the crystal structure. The probability of point defects in the crystal lattice increases as temperature increases, hence the "hopping" is more likely to occur and so material resistivity decreases as temperature increases. Only need the outcome in red for AS Physics

26 Mr Powell 2012 Index d) Temperature Sensors? They are inexpensive, rugged and reliable. They respond quickly to changes and are easy to manufacture in different shapes. An example could be made from a combination of Fe 3 O 4 + MgCr 2 O 4 (metallic oxides) A NTC thermistor is one in which the resistance decreases with an increase in temperature. The circuit shows how you can use the thermistor as a potential divider. As the temperature changes the division of voltage or energy will change. You need the 5k resistor or the voltage would be that of the cell a constant 3V. A common use is the glass heat sensor in a car or the temperature sensor in a conventional oven.

27 Mr Powell 2012 Index LDR The resistance of a light- dependent resistor (LDR) decreases as light intensity increases. This is a similar process to a thermistor lux (symbol: lx) is the SI unit of illuminance and luminous emittance measuring luminous power per areaSIilluminanceluminous emittance

28 Mr Powell 2012 Index Mini Resistivity Question…. The wire is now stretched to twice its original length by a process that keeps its volume constant. If the resistivity of the metal of the wire remains constant, show that the resistance increases to 4R. A metal wire of length 1.4 m has a uniform cross-sectional area = 7.8 × 10 –7 m 2. Calculate the resistance, R, of the wire. Resistivity of the metal is = 1.7 × 10 –8 Wm iSlice Basic

29 Mr Powell 2012 Index Mini Resistivity Question…. The wire is now stretched to twice its original length by a process that keeps its volume constant. If the resistivity of the metal of the wire remains constant, show that the resistance increases to 4R. A metal wire of length 1.4 m has a uniform cross-sectional area = 7.8 × 10 –7 m 2. Calculate the resistance, R, of the wire. Resistivity of the metal is = 1.7 × 10 –8 Wm iSlices Basic

30 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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