We calculate resistance using the formula R=V/I, where R=resistance, V=voltage and I=current. Voltage divided by current gives us the resistance, as voltage is the number of volts wanting to pass through the wire, whilst the current tells us how many are passing through at that point.
To change the wire diameter, we will use different wires. There are 8 wires we are going to use and they have diameters of:
0.315 0.559 0.376 0.234
0.712 0.193 0.457 0.274
During our experiments, we will note down specific details. We will take readings of the Wire number, diameter, Cross-sectional area, voltage and current. After we finish the experiments, we will then work out the resistance and average resistance, using R=V/I. We will make sure our results are accurate by carrying out 3 tests for each of the 8 wires, and taking the average resistance, which is why we have the final field-Average Resistance. This is because it is a useful indicator if we’ve got a result wrong, for whatever reason. If we find that two of the 3 testings have very different measurements of one of the fields, the test will be repeated, because it shows one of the two is incorrect.
Prediction: I predict that the larger the cross-sectional area, the smaller the resistance across the wire. This is because I know the larger the diameter, the more charge can pass through at the same time, so there is less resistance between electrons trying to pass through.
I.e.
I would expect a graph to look like this:
Area O- Obtaining results
Area A-Analysing Results.
My graph of resistance against cross-sectional area was a curve, meaning as the resistance increases, the area decreases and vice versa. Therefore, they are inversely proportional. I checked this by drawing a graph of resistance against by 1/Area. This is because (1/variable) makes an inversely proportional line proportional. If the line of best fit for resistance against 1/Area was a straight line, this proved the resistance to be inversely proportional to area, and it was.
In my prediction, I predicted that as the resistance went up, the area went down, and this was true so my prediction was correct.
This pattern is really quite logical. The larger the wire, the more charge can pass through. I know already that the larger the item (in this case the wire) the smaller the resistance.
So, therefore:
Resistance is inversely proportional to Area
Or:
Resistance is proportional to 1/Area
Area E- Evaluation
It was fairly easy to make a conclusion, since every part of our experiment was made accurate and precise. Therefore, the graph showed a clear line of best fit, into which most of the data fitted.
