Physics Coursework Analysing Data - page 4
Keywords: Physics Report Analysising Data Extracting Information Calculation Potato Batteries
By Jenny on 02/07/2009
Level: A Level (Year 13)
Page Number: 4 of 5 pages: 1 2 3 4 5Sit Back, Relax, and Get Paid for What You Think!
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for external resistance at the highest point for power is the internal resistance. This would allow me to check my findings from the other graph.
When the power is at its maximum the external resistance is 1148Ω. This means that the internal resistance of the potato battery is 1148Ω.
The outcomes of my analysis are that I have derived several more quantities from my initial data on current, potential difference and external resistance. I have worked out the errors on my initial data and the power, internal resistance and emf of my potato battery and their errors. I have discovered:
Internal resistance for my potato is (1335 +/- 109)Ω or 1148Ω.
The emf for my potato is (0.8864 +/- 0.0214)V.
My first value for the potatoes internal resistance was (1335 +/- 109) Ω, my second is only 78Ω (just 5.84% of that value) away from that values lower bound. This indicates that they are fairly accurate. I can also be quite confident my answers for emf are accurate as I originally found out that its theoretical value was 1.1V which is close (the difference from my upper bound is 0.1922V or 21.68% of my answer).
There are several important aspects that must be correct in order for my data to be correct. First any errors in the data itself from when it was collected would affect my results. I minimised this by taking 3 readings and using the average and by working out the error values to draw the error bars on the graph, allowing me to work out an error for my data.
Also the fact that all the points on my graphs fall in smooth lines without any obvious anomalous results indicates that the data is fairly accurate.
Also any errors in the construction of my graphs would affect my results and analysis. There was one significant error of this type made, when I was plotting my first graph of potential difference against current. Instead of putting down (mA) as my units on the x axis I just put (A). This meant that my result for internal resistance was out by a factor of 1000 with 1.335Ω rather than 1335Ω. This is a large error and it wasn’t until I noticed the large discrepancy between it and the internal resistance value I got from my graph of external resistance against power (1148Ω)that I realised something was wrong. I checked my workings for each method and noticed
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