To ensure a fair test I will try to keep the temperature constant in the circuit and in the room, also the equipment to be used and the degree of accuracy to which I will measure will be kept constant. However I do realise that I cannot control the temperature but ideally that is what I would like to do.
I should be able to measure the voltage to (+/- 0.05) +2% volts plus five per cent of the answer which is one hundredth of a volt plus five per cent due to the error in the machine from rounding etc. the current can be measured to one hundredth of an amp so once again the general rule of plus or minus a half scale division is applicable thus (+/- 0.05) amps. Other elements like parallax error when reading the analogue meter may affect the sensitivity of the results.
I will wait for the digital voltmeter to settle before I make a note of the reading. Before starting the experiment I will make note sure all the connections are connected, and such the like. Also to prevent parallax errors I will look directly on the top of the analogue ammeter.
To ensure the utmost safety while conducting my experiments I will not run excessive energy through the circuit, I will try not to let other students or teachers near my circuit to make sure it is entirely my responsibility. Also I won’t be passing enough volts or amps to short out the classroom or electrocute myself.
Proposed Analysis:
My research has told me that the graph I will plot using my results will show indirect proportionality between the voltage and the current.
Observations:
Here are the results obtained from my experiment together with the mean average.
All the above figures are rounded to two decimal places to show accuracy and consistency. The current was measured in milli-amps because this was the most appropriate measure. The current increases in whole milli amps also not in half milli amps, when I tested this out the differences were to small to take estimations from so some guesswork was involved as to the location of 0.11A on the analogue ammeter.
These results have been graphed (see G1 and G2) where I have calculated the emf and tir. I will now use these figures to interpret maximum power that could be delivered from the circuit.
Analysis:
The graphs I have drawn are straight line graphs thus the relationship between my points is a linear one. However, some of the points are slightly above or below the line of best fit. The best way for me to demonstrate this graphically is to draw error boxes for each point. The actual value could be anywhere in this box, the values which I have are mainly rounded to two decimal places.
The electro motive force of my circuit is 2.65 V and the total internal resistance is 15.63Ω, using these results I can predict the maximum power that could be delivered from the power supply.
To get the maximum power out of a circuit the total internal resistance must be equal to that of the external resistance, so supposing the internal and external resistance is the same and all equal to 15.625Ω then the total circuit resistance would then be 31.25Ω. The current is found when you divide the emf (2.65V) by the total circuit resistance (31.25Ω) which in this case gives and answer of 0.0848A. We can then use this to find the maximum power by multiplying the emf (2.65) by the current (0.0848) giving us an answer of 0.22472. So the maximum power that could be delivered from the said circuit is 0.22472 divided by two which equals 0.11236. Maximum power is 0.11W
If I were to try and obtain maximum power out of my circuit I would need first of all to have all the energy flowing through the circuit and not being lost through means of heat, sound, or light.
I am convinced that my graph is supposed to be a straight line, because all my points fit almost perfectly onto it, also because my measurements are accurate to ± half a scale division (and five per cent with the digital voltmeter) so I have taken the most accurate measurements that I could take. (Not taking into account the possible parallax error in the ammeter)
Evaluation:
My readings were taken three times so a reliable average could be taken. I calculated a mean value (because the mean is would take into account large anomalies) and found that there was little deviation from the mean for each value. The issues surrounding my accuracy are those like the parallax error in the analoghue ammeter. I may have looked at the ammeter at an angle which could have affected my result. Also the digital voltmeter which I used was only accurate to 100th of a volt. Because of this the actual result may have been rounded up or down which again may have affected the set of results I obtained.
I have mentioned that I took the mean of three results, for example for 0.9A I got the following results 1.38V, 1.37V, and 1.36V. These average out at 1.37V. I took the mean of a set of results because if I only took one reading I may gather some very un-reliable information and thus be misled as to the actual answer.
I drew two extra lines on my graph they indicate the line of most squares and that of the least squares; these show the uncertainty in my results. It shows almost 0.5V worth of error.
Conclusion:
In conclusion to this I can say that the Total Internal Resistance is 15.625Ω. The Electro Motive Force is 2.65V. The maximum power that could be derived is 0.11W.