G482 Electricity, Waves & Photons 2.2 Resistance G482 Electricity, Waves & Photons 2.2.2 EMF Ks5 OCR Physics H158/H558 Index Mr Powell 2012

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.

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…

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

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

Spark Version – m = Resistance! Voltage /V Current /A

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!

Ionisation Energies Energy in J/mol Metal n=1 n=2 n=3 Free Electrons 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…. Energy in J/mol Metal n=1 n=2 n=3 Free Electrons Li 520 7298 11815 1 K 419 3051 4412 Cu 746 1958 3554 1 or 2 Zn 906 1733 3833 2 NB: Don’t need to learn these energies just be aware of the idea of the stripping away!

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 e- e- NB: Boron was picked due to simple structure – it is not a good conductor!

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!

Other physical factors… 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 1028 electron carriers per m3 you would have to accelerate 1 x10-26kg'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 e-

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 = 1m2 and l = 1m  = R

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 m2 l = length in metres m NB:  is the taken as the value of the Resistivity at room temperature 20C

Facts and Figures 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 Type Material Resistivity in m Metal Copper 1.7 x 10-8 Gold 2.4 x 10-8 Aluminium 2.6 x 10-8 Semiconductors Germanium (pure) 0.6 Silicon (pure) 1.7 x 103 Insulators Glass 1.7 x 1012 Perspex 1.7 x 1013 Polyethylene 1.7 x 1014 Sulphur 1.7 x 1015

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

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

Simple Application…. 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; Material  Resistivity in m Thickness cm Length Copper 1.7 x 10-8 0.1 20 Gold 2.4 x 10-8 0.2 10 Aluminium 2.6 x 10-8 0.4 12 Hint: tabulate your data or write it out for each question with conversions in full

Electrical Resistance Data SWG - Standard Wire Gauge

Material Resistivity at 20°C Ω·m µΩ·cm silver 1.6 × 10-8 1.6 copper 1.7 × 10-8 1.7 gold 2.2 × 10-8 2.2 aluminium 2.7 × 10-8 2.7 magnesium 4.2 × 10-8 4.2 nickel 6.9 × 10-8 6.9 iron 10.1 × 10-8 10.1 chromium 13.2 × 10-8 13.2 manganese 160 × 10-8 160 carbon (graphite) 3 000 × 10-8 3 000 Manganin 44 × 10-8 44 Constantan (Eureka) 49 × 10-8 49 Nichrome 110 × 10-8 110

Superconductivity This is the property of a material which is at or below a critical temperature Tc 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 Tc temp

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. R l NB:  is the taken as the value of the Resistivity at room temperature 20C

Resitivity? Length in m Resistance in m 0.10 0.18 0.20 3.70 0.30 5.50 0.40 7.10 0.50 9.20 0.60 11.00 0.70 13.00 0.80 14.50 0.90 14.60 1.00 18.00 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? R l NB:  calculated at 20C

Looks like Gold to me! 1.32739x10-6m2 x 18.253x10-3 m-1 =2.42 x 10-8m

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!

Only need the outcome in red for AS Physics 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

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 Fe3O4 + MgCr2O4 (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.

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 area

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 m2. Calculate the resistance, R, of the wire. Resistivity of the metal is = 1.7 × 10–8 Wm Basic iSlice

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 m2. Calculate the resistance, R, of the wire. Resistivity of the metal is = 1.7 × 10–8 Wm Basic iSlices

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 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 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