Background Knowledge: Metals conduct electricity because they have free electrons.

This means that they are able to move around the positive ions like a sea. The positive ions are unable to move. This “sea” of electrons is moved by the force of the voltage pushing it around the circuit. The electrons move from negative to positive, the opposing current moves from positive to negative.The formula for resistance is described as -Resistance = Voltage / CurrentOrR = V / I (R=resistanceV=voltageI=current)I will use this formula to work out the resistance of my wire each time. I will now need to select a variable.

The resistance of a metal is low, but not zero. A voltage difference is still required to generate current in the metal and the metal heats up when the current is flowing. This increase in temperature is one four possible variables that I have come up with…Possible Variables: WIDTHLENGTHTEMPERATURETYPE OF WIREIndependent: THE WIDTH OF THE NICHROME WIRE(input variable)Dependent (outcome variable): The resistance will be found by taking an Ammeter and then take one from a Voltmeter.

The resistance is found by dividing the voltage by the current ==>R = V / IControlled: I will keep everything the same except the width of the wire. This includes the length of the wire, the type of the wire and I will make sure that the wire is kept the same temperature – room temperature.Darren Cave 12B Page 1 5/2/2007Predictions: I predict that as I increase the cross-sectional area of the wire, the resistance of the wire will decrease. I have made this prediction because if the area is increased, then there will be more room for the current to flow through, therefore the current will be able to flow through quicker, in other words a lower resistance. If 500 pupils walked through a corridor, the quicker they got through would depend on the size of the corridor, the bigger the corridor the quicker they would move. This is the same as current in a wire…

the wider the wire, the quicker the current moves therefore the lower the resistance.Practical Procedure:Apparatus: Volt battery pack Nichrome wire (5 different widths)Voltmeter Connection leadsAmmeterSafety Precautions: If too much current is put through the nichrome wire at one time it could be dangerous because it could cause the wire to overheat, which creates the risk of the wire melting. The same result occurs when the current is allowed to flow through the wire for too long. These are the main safety points, also common sense.The apparatus mentioned in the list above should be set up as in the diagram, also shown above. We take a reading in Volts before the nichrome wire, using the Voltmeter.

After the nichrome wire we take a reading of current in Amps, using an Ammeter. The width of this nichrome wire should be the first of five – 0.32mm.

It is important that the length of each wire is 1Ocm.Then, once we have recorded these two readings we can use Ohm’s law to work out the resistance of the nichrome wire ==; R = V / I orResistance = Potential Difference / CurrentDarren Cave 12B Page 2 5/2/2007The experiment is then repeated but with the width of the nichrome wire slightly wider than the last – 0.37mm. Then the same experiment accept with the wire of width – 0.45mm.

Again, the experiment should be done with a slightly wider piece of nichrome wire – 0.56mm. Then lastly, the measuring should be taken with nichrome wire of width – 0.71mm in the circuit. I will repeat how it is essential that the wires are all 10cm in length to ensure a fair test. To make these results more accurate, this should all be repeated to give two Ammeter readings and two Voltmeter readings for each width. If the two readings are not similar, there is the option of a third attempt at the experiment.By this stage, at least ten results should have been recorded – two for each of the five different widths of wire and three where needed.

This is when we use Ohm’s Law to work out the average resistance.With my results, I will now form a series of tables which will lead to the drawing of numerous graphs of analysis. I will plot one of the tables in a graph of diameter of wire against average resistance. If the graph is not a straight line, I will try 1 / diameter of wire against average resistance. If this graph fails, my last attempt will be 1 / diameter of wire squared against average resistance.Darren Cave 12B Page 3 5/2/2007My first set of results are shown in the table below -WIRE WIDTH (mm)CURRENT(amps)VOLTAGE(volts)RESISTANCE(ohms)0.

320.7322.730.371.1421.

750.451.8721.070.562.6720.

750.715.0020.40The second set of results that I recorded are shown in the table below -WIRE WIDTH (mm)CURRENT(amps)VOLTAGE(volts)RESISTANCE(ohms)0.320.7322.730.

371.1221.790.451.6421.

220.562.7420.730.715.0020.40The not essential third set of results are shown in this table (dashes show where I felt a repetition was not required) -WIRE WIDTH (mm)CURRENT(amps)VOLTAGE(volts)RESISTANCE(ohms)0.

32—0.37—0.451.7421.150.56—0.71—This table is the one that will be used for the graph – diameter of wire in mm against average resistance in ohms -WIRE WIDTH (mm)AVERAGE RESISTANCE (ohms)0.322.

730.371.770.451.150.560.

740.710.40The graph is shown on the next page.Darren Cave 12B Page 4 5/2/2007This graph gave me a curve which will not help me in my attempt at finding a relationship between the two. Here is the table of the next graph that I will plot.

As planned, it is 1 / diameter of wire against average resistance -WIRE WIDTH (mm)1 / WIRE WIDTHAVERAGE RESISTANCE (ohms)0.323.1252.730.372.7031.770.

452.2221.150.561.

7860.740.711.

4080.40Page 6 shows the graph of 1 / diameter of wire against average resistance.Darren Cave 12B Page 5 5/2/2007Again, the graph gave a curve although this time with a positive gradient. This curve shows no consistent relationship between the resistance and the diameter of the wire so is therefore not the graph I was looking for. To stick with my previous proposal, I am now going to make the table and plot the graph of 1 / diameter of wire squared against average resistance. The graph follows on the next page.

WIRE WIDTH (mm)1 / WIRE WIDTH SQUAREDAVERAGE RESISTANCE (ohms)0.329.772.730.377.301.

770.454.941.

150.563.190.740.711.980.40Darren Cave 12B Page 6 5/2/2007It is clear from this graph that the resistance is directly proportional to 1 / diameter of wire squared. This means that I have found a relationship between the two, if you double one, the other one will be doubled also and so on.

Darren Cave 12B Page 7 5/2/2007Conclusion: Through my experiment, I discovered that as I increased the width of the wire, the resistance was less. This is exactly as I had predicted previously.If the width of the wire is huge, there will be a large amount of room for the current to flow through; therefore the resistance would be less. If the width of the wire was small, there would be a small amount of room for the current to flow through; therefore the current would be less. This can be backed up by the fact that.

..as I increased the width of the wire, the resistance decreased.

My three graphs were ==> diameter of wire against average resistance1 / diameter of wire against average resistance1 / diameter of wire squared against average resistanceIn the first two graphs I was unable to find a straight line or a connection between the two factors, they made curves. The third graph gave a straight line through the origin and found the connection for me. The resistance is directly proportional to 1 / the diameter squared.As I found a connection between the variable and the resistance, I consider this experiment to be a success so there is little I would change if I had another chance. The one problem I had was when I was trying to keep the wire 10cm in length. It was hard to get the wire completely straight so it could be measured accurately. Fortunately, this problem was not reflected in my results so I feel that the wires were measured accurately.

The aim of my coursework was to determine a formula linking the diameter of nichrome wire and the resistance of nichrome wire. So to conclude, as I successfully completed my objective, I consider the experiment to be a complete success.