1. Resistivity Physics – 12.1.4 Resistivity   Understand where the idea of resistivity comes from in a circuit  See derivation of resistivity formulae  Be able to apply the resistivity equation appropriately at room temperature

2. 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…

3. 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!

4. Ionisation Energies Energy in J/mol Metaln=1n=2n=3Free Electrons Li5207298118151 K419305144121 Cu746195835541 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!

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

6. 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!

7. 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

8. 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

9. 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

10. 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 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

11. 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

12. 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 cm Copper1.7 x 10 -8 0.120 Gold2.4 x 10 -8 0.210 Aluminium2.6 x 10 -8 0.412 Hint: tabulate your data or write it out for each question with conversions in full

13. 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

14. Resitivity?  Use the graphing technique to find out the resistivity of a sample wire which is 0.13cm thick. What metal is the wire made from? NB:  calculated at 20  C l R 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

15. Looks like Gold to me! 1.32739x10 -6 m 2 x 18.253x10 -3  m -1 =2.42 x 10 -8  m

16. Electrical Resistance Data SWG - Standard Wire Gauge