I am going to investigate how the diameter, or gauge, of a wire affects its resistance. In order for me to do this, I will need to find as many different diameters of wire as I can. I will cut each one to a reasonable length (probably about 10cm), then connect it to a battery pack and measure the current and voltage (with an ammeter and voltmeter respectively).
From these readings I will calculate the resistance, using the formula: R = V / I (Resistance = Voltage / Current) The scale by which the diameter of wires is measured is called SWG (standard wire gauge) which confusingly goes up as the wire gets thinner. However I shall not use this in my graphs, as I do not know enough about SWG to make any worthwhile observations. The gauge of wires can also be measured by their cross-sectional area. Scientific Theory:
Some factors affecting the resistance of a wire are: ï¿½ its temperature ï¿½ the metal it is made from ï¿½ its diameter/cross-sectional area ï¿½ its length In metals, the structure of the atoms and their electrons means that there is a “sea” of free electrons (or delocalised electrons) on the surface of a wire that can move around easily, thus conducting the electricity.
This is because the outermost shell of a metal atom has a very weak force acting on its electrons.
The amount of electrons that can travel determine a material’s resistance. If there are two wires, one twice as thick as the other, but the voltage is the same on both, then (in the case of the thicker wire) there are twice as many electrons available and so the current goes up. This means that, in the case of the thicker wire, the same voltage is being divided by a larger current – therefore the resistance comes out as less. This is the same as if you were to connect several resistors in parallel in a circuit, as this reduces their resistance too. The greater the cross-sectional area (of a pipe or wire), the less the resistance. This is easily explained in a similar way to water flowing through a pipe. If a pipe has a large cross-sectional area, then more water can flow through at any given time, and it will flow through more easily. I will construct a graph of diameter against resistance and another of cross-sectional area against resistance. I shall expect them to be similar. Range of results: SWG Diameter (mm) 20 0.9 22 0.71 24 0.56 26 0.45 28 0.4 30 0.31 32 0.28 I will refer to the diameter of the wires with their diameter in millimetres as opposed to using SWG (Standard Wire Gauge). According to my scientific theory, this would give my graphs a negative correlation. Prediction: I believe that as the diameter increases, so the wire’s resistance will decrease, thereby showing a negative correlation. This is based on my scientific theory. Method / Apparatus: various different gauges of wire an ammeter a voltmeter a battery pack crocodile clips Firstly I found the different gauges of wire and cut a 15cm length of each. This was long enough to be useful, but not so long as to be unmanageable. Then I connected the battery pack to the ammeter – this was connected in series – and then the voltmeter – this was connected in parallel. I then used the crocodile clips to connect the wires from the ammeter to the ends of the length of wire. I observed the readings from the meters and recorded them, then used them to calculate the resistance of that particular piece of wire. I repeated this for each gauge of wire. Diagram: Fair Test: In order that my results were as accurate as possible, I made sure that all my wires were made of the same material, Constantan. This alloy (of nickel and copper) maintains the same resistance no matter what temperature it is – I used it so that the results would not be compromised by the wire heating up and changing its resistance. My wires were all cut to?2mm of the same length – this was because the length of the wire could affect the resistance, and the factor I was investigating wasn’t length. Table of Results: Run 1 Run 2 SWG Diameter Voltage (V) Amps (A) Resistance R=V/I Voltage (V) Amps (A) Resistance R=V/A 20 0.9 0.25 1.6 0.15625 0.25 1.6 0.15625 22 0.71 0.16 1 0.16 0.175 1.5 0.11666667 24 0.56 0.6 1.45 0.4137931 0.6 1.4 0.42857143 26 0.45 0.8 1.25 0.64 0.8 1.25 0.64 28 0.4 0.9 1.2 0.75 0.9 1.2 0.75 30 0.31 1.1 1 1.1 1.1 1 1.1 32 0.28 1.2 0.95 1.2631579 1.2 1 1.2 Run 3 Average SWG Diameter Voltage (V) Amps (A) Resistance R=V/A Voltage (V) Amps (A) Resistance R=V/A 20 0.9 0.25 1.6 0.15625 0.25 1.6 0.15625 22 0.71 0.17 1.2 0.14166667 0.168333 1.23333 0.13944444 24 0.56 0.6 1.4 0.42857143 0.6 1.41667 0.42364532 26 0.45 0.8 1.2 0.66666667 0.8 1.23333 0.64888889 28 0.4 1 1.2 0.83333333 0.933333 1.2 0.77777778 30 0.31 1 1.1 0.90909091 1.066667 1.03333 1.03636364 32 0.28 1.2 1 1.2 1.2 0.98333 1.22105263 Analysis: From my results I can gather that my prediction was correct – the diameter of a wire does affect its resistance, and in fact (generally) there is a strong negative correlation between the two. I observed this from my graph. I constructed my graph using diameter in mm on the x-axis and Resistance (?, or Ohms) on the y-axis. It showed a negative correlation, but had this weird thing going on at the end – an anomaly. I was able to draw quite a clear curve of best fit. The graph proved my theory correct, because as the diameter increased so the resistance of the wire decreased. Evaluation: I believe the experiment was a success, as it proved my theory correct and my graph showed a clear curve of best fit. However there was an anomalous reading, right at the end. This may have been a mistake in reading the meters, or a mistake in calculating the resistance. To improve the accuracy and fairness of my experiment, I could have done the following things: ï¿½ used a more reliable power source (the power flow from battery packs weakens with use), ï¿½ cleaned the connections within the battery pack (corroded or dirty connections can affect power flow), ï¿½ used more accurate am- and volt- meters (i.e. with a greater number of decimal places), ï¿½ widen my range of results by connecting the circuit to more than one cell of the battery pack, ï¿½ cut more lengths of wire from different rolls (one roll alone may not be representative of that particular gauge) as defects etc. may affect the results.