Length And Resistance In Experiment Two example essay topic

Aim: The resistance of a wire depends on certain factors. Investigate the effect of two of these factors - Planning Some variables that will be relevant to this investigation are: Length Thickness Temperature Voltage Resistance Material Of these the variables will be input and output voltages in experiment one, and length and resistance in experiment two. The other variables (temperature, material and voltage) will have to be kept constant in both experiments to make sure that only length, thickness and resistance are investigated. In experiment 1 the same bit of wire and the same thickness need to be kept constant. In experiment 2, the length will need to be kept constant to make sure only the variables indicated are investigated to ensure a fair test. Metals conduct electricity because the atoms in them do not hold on to their electrons very well, and so creating free electrons, carrying a negative charge to jump along the line of atoms in a wire.

Resistance is caused when these electrons flowing towards the positive terminal have to 'jumps' atoms. So if we double the length of a wire, the number of atoms in the wire doubles, so the number of jumps double, so twice the amount of energy is required: There are twice as many jumps if the wire is twice as long. The thinner the wire is the less channels of electrons in the wire for current to flow, so the energy is not spread out as much, so the resistance will be higher: We see that if the area of the wire doubles, so does the number of possible routes for the current to flow down, therefore the energy is twice as spread out, so resistance might halve, i.e. Resistance = 1/Area. This can be explained using the formula R = V / I where there is 2 X the current, and the voltage is the same, therefore R will halve. I did some research and in a book called 'Ordinary Level Physics' By A.F. Abbott, it says 'that doubling the area will therefore halve the resistance'- in other words the resistance of a wire is inversely proportional to its area, or R?

1/A, but we are measuring diameter, so if the area is: ? r 2 = ? (d? 2) 2 A = ? d 2? 4 Where A is area and d is diameter. Method Experiment One - First a length of wire over a metre long is sellotaped to a metre rule. The positive crocodile clip is attached at 0 cm.

And the negative is moved up and down the wire, stopping at 20, 40, 60, 80 and 100 cm. Each time reading the ammeter and voltmeter to work out resistance R = V / I. This is using 30 SWG wire. Other variables, voltage, thickness, and temperature will be kept constant, although the temperature will rise once current is passing through it, which will cause the atoms in the wire to vibrate, and so obstruct the flow of electrons, so the resistance will increase, creating an error. In both experiments constantan wire is used because it does not heat up as much as copper, so the resistance is not effected as much. Experiment Two - The circuit is set up is the same, as is the method apart from the length is constant at 50 cm, and the thickness is changed between 28, 30, 32, 34, 36, 38 and 40 sw g.

For both experiments the voltage will be kept the same at 2 V dc from a power pack. Both experiments will be done twice with different ammeters in case of any damaged or old equipment to gain more accurate results. Results Experiment 1 Length (cm) V 1 (volts) V 2 (volts) A 1 (amps) A 2 (amps) Average resistance (Ohms) 1001.001. 000.200.

205.00801. 001.000. 300.284. 00600.900. 900.400. 302.80400.

900.850. 500.401. 94200.700. 800.800. 750.94 Experiment 2 Thickness (mm) Area (mm 2) V 1 (volts) V 2 (volts) A 1 (amps) A 2 (amps) Resistance 280.360. 1070.80.

80.610. 591.30. 290.0660. 90.90. 490.511. 80.320.

250.0490. 90.90. 350.40. 240.340.

180.0251. 01.00. 250.251. 01.00. 250.40.

360.160. 0201.01. 00.160. 170.610. 380.120. 0111.11.

10.091. 000.091. 0011.6400. 050.00201. 2 Evaluation Evaluation Experiment one: This experiment was quite accurate, as when it is compared to the manufactures line which is on the same graph, we can see that this line is at most only 0.4? different form the manufactures line.

This is a percentage difference of approximately 8%, using the formula: Difference? original X 100 This shows that the results were good, as 8% is a very small margin of error. The error bars on the graph show that the most inaccurate result was the 60 cm result. This could be down to an error in the measurement of the wire or a temperature rise. The two results for 100 cm are exactly the same, and it is near to the manufacture's line, so this is the most accurate point. The other three readings have almost the same inaccuracy, an average of 10%, which again, is fairly accurate. The inaccuracy could have been because of the wire coming from a different manufacturer to the predicted results, as there is some discrepancy between the amount of copper and nickel in different brand's wire.

The ammeters and voltmeters could have been damaged and reading falsely on both the meters used. Measuring the lengths of the wire is also a inaccuracy as the rulers used are not exact, and it is difficult to get an accurate reading of length by eye, as the wire might not be completely straight, it may be of different thicknesses throughout the length. These would have contributed as well to the error. These results would be difficult to improve on as they are reasonably accurate, and there were no anomalous results. But if I were to do this experiment again, I would use newer, more accurate ammeters and voltmeters, a more accurate method of measurement, and take a much wider range of readings, and more readings so that a more accurate average can be taken. I would also investigate other factors, such as temperature, voltage and current, and see how these effect the resistance.

I would also do the experiments under different conditions such as temperature and pressure to see if it makes any difference to resistance. As these results had a range of only 5 readings, from 0-100 cm, and were only repeated twice, and that the results are not 100%, accurate due to the errors discussed earlier, then I would say that these results are not strong enough to base a firm conclusion on because there are so many sources of error, which are explained earlier. Experiment two - These results were not as accurate as experiment one. I had predicted that the resistance should halve as area doubles, which it does, however not to the predicted curve. When the resistance is 24 ohms, the % inaccuracy is 6%, and when the resistance is 6 ohms, the inaccuracy is 8%. These inaccuracies are fairly large.

The error bars, however, are too small to be drawn accurately on the graph. They are at most 3% inaccurate, using the same formula as before. This suggests that the inaccuracies were not experimental, but permanent errors due to problems with the measuring equipment. These results were this inaccurate as the tool used for measuring the diameter of the wire were very inaccurate due to a zero error on the screw reading, i.e. the mark given for zero mm was not the real mark, hence throwing all the results off by the same amount. There was one slightly anomalous result, at 0.25 mm 2. This could have been due to a unique error in the measuring and or reading of the meters, or a temperature change.

These results could be done better. If I were to do this experiment again, I would use newer, more accurate ammeters and voltmeters, a more accurate method of measurement, and take a much wider range of readings, and more readings so that a more accurate average can be taken. As these results had a range of only 7 readings, from 0.1 mm 2, and were only repeated twice, and that the results are not 100% accurate, due to the errors discussed earlier, then I would say that these results are not strong enough to base a firm conclusion on because there are so many sources of error, which have been explained earlier. Analysis The graph of experiment 1 is a straight line through the origin, which means R is directly proportional to L. This means that if the length is 40 cm, and resistance is 2? , then if length is doubled to 80 cm, resistance also doubles to 4.

This is because of the scientific idea, stated in the planning that if you double length, you double the number of atoms in it, so doubling the number of electron 'jumps', which causes resistance: The results support my predictions well, the results turned out the way I had expected, they match the predicted line well. I had predicted a straight line through the origin, which means R is directly proportional to L. The graph of experiment 2 is an inversely proportional curve. This is because R is directly proportional 1/A, this means when A doubles, R halves. for example when the Area is 0.025 mm 2 the resistance is 4.8. When A doubles to 0.05, R halves to 2.4? When A doubles again, R halves again to 1.2. This is because, as stated earlier: We see that if the area of the wire doubles, so does the number of possible routes for the current to flow down, therefore the energy is twice as spread out, so resistance might halve, i.e. Resistance is directly proportional 1/Area.