What affects the resistance of a piece of wire? - Sample Essay

Analysis From my results, I can see that at the length of wire doubles, so does the voltage (corresponding to my prediction) and therefore the resistance doubles. Another result that I obtained, was that resistance was double the voltage, the resistance doubles. I also predicted this. If we look at the graph, we have a visual representation of the results. We can see how resistance increases and so does voltage. The graph is a straight line graph, showing that length of wire, resistance and voltage are directly proportional to one another (the voltage doubles, so does the resistance).

Using this graph, we can predict the resistance of any length up to 100cm. To test if these results are accurate, we can compare the experiment we have performed, with those of previous resistance experiments. The book value, in ohms, for the resistance of a 100cm length of 28swg wire is 4. 4?. This convinces me that these results are accurate as our result of 4? is very similar. It may seem a way off the mark but considering the difference in resistance a couple of swg makes (e. g. the resistance of a 30swg wire is around 7. 5? ) the similarity ensures success.

The slight deviation may have been caused by a number of factors, like the variations of the power packs used and variations in the actual wire. I have recorded these results in a graph. From this graph, we can immediately see a few things. The line of best fit is a straight line through the origin with a positive correlation. This means that the resistance and voltage are directly proportional to one another, and that length of wire and resistance are directly proportional. This ties into ohms law. We can also measure the gradient of the graph.

This is done by taking a y value under the line of best fit, and dividing it by the corresponding x value. On my graph, the y value I have chosen is 0. 9? , and the x value 23cm. The formula for this is:  So the gradient is 0. 04. We can examine the results further by referring to the resistivity formula: Resistance, R = Resistivity, p (? m) x length, l (m) or R = p l (? ) Cross-sectional area, A (m2) A The diameter of a 28swg wire is 0. 367mm, so, using the formula: A = ? r2 We can figure out the cross sectional area.

Therefore, the cross sectional area is 0.105784493mm2. Converting this to metres squared, this is 0. 000105784493m2 And, since we know that the resistance is 4. 0? , we can rearrange the formula to find the resistivity of the wire: We rearrange from this: R = p l to this: p = R A A l Now we can find out the resistivity (p) of the wire, in ohm meters (? m). So… p = R A l = 4. 0 x 0. 000105784493 100 = 0. 00000423137972 ? m Resistivity is a useful figure to have (much like resistance) but can also be used to check our work. To check the gradient of the graph, we can divide the resistivity of the wire, by the cross-sectional area. So…

Gradient = p = 0. 00000423137972 = 0. 04 A 0. 000105784493 As 0. 04 is the same value I got for the gradient directly off the graph, I can rest assured that it is the correct value. As this is the correct value, I can work backwards to see that the previous calculations were correct. In order to obtain this gradient value, I used resistivity, cross-sectional area, and resistance. So we can be sure that the calculations for these is correct as the outcome is the same as the separate calculation for gradient, using the graph. So, from these results we can see that:  Resistivity = Gradient = 0. 04.

Cross-sectional area = 0. 000105784493m2.  Voltage is directly proportional to the length of the wire. * Resistance is directly proportional to the length of the wire.  As the length of the wire doubles, so does the voltage and so does the resistance. Conclusion From the consistency of the results I have gained, and the fact that proving the results using the gradient was successful, I can conclude that these results are reliable. The values that I noted were also very similar to the catalogued book values, which further supports the reliability of the experiment.

The drawn line of best fit is accurate as the gradient drawn from it is identical to the one I figured out using the formula. If I were to repeat this experiment, I would change it only slightly. I would allow more class time for the analysis and possibly allow a shorter time for the practical as it is very quickly finished (in fact, to do all the results in one lesson minimises the chances of getting irregular results, because you know that the conditions are the same). The whole thing worked very well. The experiment went smoothly, and the inclusion of a preliminary experiment helped a great deal.

The actual physics of this assignment were challenging, but understandable. It was a very well rounded task. Extension work that could be included may be to test the resistance of other wires and compare them, or to go up to 200cm and see if there’s any difference. Also, by using another wire, other than constantan, we could see how temperature affects resistance, although, one would need to have done a constantan experiment in order to understand the dynamics of the other factors, otherwise the experiment would be to complicated.