The second two were done using two digital Multimeters so presumably the results on the second two tables are more accurate. Test 1 ammeter/voltmeter: Length of constantan (cm) Potential difference (volts) Current (amps) Resistance (ohms) Ammeter/voltmeter repeat: Length of constantan (cm) Potential difference (volts) Current (amps) Resistance (ohms) Multimeter Test1 Length of constantan (cm) Potential difference (volts) Current (amps) Resistance (ohms) Multimeter repeat: Length of constantan (cm) Potential difference (volts) Current (amps).
Resistance (ohms) 1Average resistance over all four tests: Length (cm) Potential Difference (volts) Current (amps) Resistance (ohms).
Constantan theory resistances: Using the formula R=E(L/A) I have gathered the exact resistance of a length of SWG24 constantan regardless of any connections or battery faults: Length (m) Resistance (ohms) And this table shows the results of the computer simulation (the computer simulation incorporates the resistance of the connecting wires into the resistance): Length of constantan (cm) Potential difference (volts) Current (amps) Resistance (ohms).
We measured the resistance at 0 cm to be 0. 35 ohms. However this cannot possibly be true because the average at 10cm was 0. 2015 and the resistance Analysis This is the average power (IV) plotted against the average resistance: This shows that as the length of the constantan increases more power is required to drive the electricity through the wires and at the same time there is also more opportunity for the power to dissipate as heat and light.
This possess is used to good effect in electric fire elements where the wire is coiled to add greater length. My average results of our tests: The results from the preliminary computer tests: Conclusion: From my results and the computer graphs, I can see that the ratio between the length of the wire and its resistance is virtually linear and the theoretical resistance is exactly linier. This makes sense because the constantan wire is of uniform thickness. The resistance on my tests isn’t perfect and this is probably because of the heating effect as explained earlier.
Two things that we can see from this graph are that neither the voltage nor the current are linear. This is because the capacity of the batteries are finite and therefore, as the resistance increases to infinity; the current and voltage will level to the capacity of the battery, curving the graph. Still, you wouldn’t expect the voltage or to be linier in a circuit like this because of the internal properties of the batteries and the heating effect of the circuit. Another thing we can see is that as the resistance decreases to zero the current and voltage will decrease to zero.
However because the circuit will have it’s own fixed internal resistance including: The battery, the meters, the circuit wires, the contacts etc a theoretical zero resistance is not possible in this circuit. Also as the length of wire decreases to zero the results will become less accurate as the external resistances (mentioned above) become more significant and the temperature rises in the wire are more significant. For example if the internal resistance of the circuit we used totaled 3 ohms and we were testing 0. 001 cm of constantan equaling 0.001958 (if R=E*(L/A) is true) then the total resistance is 3. 001958 ohms, which is way off the actual total.
And as you make the circuit smaller there is less resistance so the current is higher meaning that more heat is generated. My resistance results on my hand drawn graph are very nearly linear and that supports my prediction that if you increase the length of wire then the resistance increases. As shown, the line of R=E*(L/A) is linear and if you add the internal resistance of the rest of the circuit to that, it is still linear, but appears lower on the graph.
If you take a point on the linear theory line and double its values then those values are true, meaning that in the lines formula X is directly proportionate to Y (this is safe to assume as there are no outside interferences i. e. battery, wires etc only the length and diameter of the constantan). The same is almost true (although shouldn’t be) for my total resistance average line. However, you cannot take the values of a point on the computer line, double them and expect them to be true.
This shows that for this line X isn’t directly proportionate to Y but is R=E*(L/A)+the internal resistance of the rest of the circuit. Evaluation Comparing my results to the computer generated ones; I’d say that the experiment went quite well. Although some off connections in our circuit were slightly rusty and damaged the results show that it wasn’t too much of a problem. The range of results obtained was sufficient to form good averages which were comparable to the results obtained from the computer and the results generated by R=E*(L/A), and the results supported my prediction.
I believe the results to be accurate because they are fairly consistent with each other. For example on the multimeter test at 50cm the resistance was 0. 9375 ohms and on the repeat I got 0. 88 ohms. This means that there was only a 0. 0575 ohm difference between them. However the results obtained from the voltmeter/ammeter tests were less accurate as it is very hard to read off them to more that 2dp. One strange thing with the results I obtained was that the resistance was much lower than I expected. This is shown by the graph.
I would have expected the line to be closer to the computer simulation than the R=E*L/A line because R=E*L/A doesn’t account for resistance where as the computer simulation does. My experiment certainly had resistance in the circuit and yet my experiments line falls almost on top of the one that doesn’t. There are a few ways to explain this. The current in the simulation was lower than the current in real life, the internal resistance of the circuit was very low (which I find hard to believe) or the measurements weren’t accurate enough, either way there is no definite answer.
Also on my hand drawn graph there are 3 obvious anomalous results, which didn’t fall on the line of best fit. They are likely to have been caused by poorer measurement, kinks in the wire or a bad connection (more sources of error have already been stated in the plan). However, I think that the evidence is still sufficient to support the conclusion. The way we set up and conducted the experiment was the standard however if I was to do repeat he investigation I would use much shorter circuit wires and use better connections to minimise circuit resistance.
It might also be a good idea to use a rectifier instead of batteries so that draining power isn’t an issue. You could also extend the investigation by measuring other variables such as temperature and wire diameter. Show preview only The above preview is unformatted text This student written piece of work is one of many that can be found in our GCSE Electricity and Magnetism section.