I put the milliammeter on a suitable “scale” (eg so so it only went up to 10 ohms), so I could read the reading more accurately. I also put a voltmeter in the circuit, because the voltage would not always remain constant, even though the variable power supply read 2V constantly. If the voltage did change a bit, it would not matter because that change would be accounted for in the equation at the end, to work out the resistance.
I recorded the volts and milli-amps every 10 degrees, to give a good range of results.
Once the experiment had finished, the resistance was calculated by
dividing the volume by the current, because:
Resistance (Ω) = Voltage (V)
That is why I have changed the milliamps to amps on each of the tables above.
I have calculated the average for all of the measurements on each table, and the combined table of results (titled “average”) is after the other two tables.
Analysis and Conclusion
This graph is taken from the average of the two sets of data (“average” table is shown in observations).
In the planning, the prediction was that as the temperature of the thermistor inreased, its resistance will decrease as more electrons are released, increaing the current.
I have found that as the temperature of the thermistor has risen, the resistance of the circuit has fallen. However, the resistance has not fallen proportionately, but has reduced much less as the temperature became very high. That was not specified in the prediction. For example, between 0 and 10oC, the resistance fell alsmost 300Ω, whereas between 80 and 90oC the resistance fell just over 4Ω. Therefore the rate at which the resistance has fallen has decreased enormously as the temperature has become high.
Despite this, the resistance has done as predicted, and has fallen as the temperature has risen.
In a semi-conductor, the electrons are not free in the substance at low temperatures, because the electrons have no energy to move around, and so are usually bonded strongly with the substance. However, if these electrons are supplied with energy, then they can vibrate more, and move around, and so can break free from their bonds. This enables them to carry the electrical charge throughout the substance, which they would have been unable to do if they were in fixed positions.
This is where the heat comes in. Providing heat gives the electrons energy, so they can break free from the bonds. This is why a substance changes from a solid to a liquid, or a liquid to a gas, and becomes hotter. The particles in general have gained more energy, and can break free from their bonds, and so therefore spread out more. In this case, it is the electrons in the semi-conductor that are supplied with heat, and can therefore convert it into kinetic energy, and become more able to conduct the electricity. That is why the resistance decreases with temperature.
As noted before, however, the resistance does not decrease at a proportionate rate to the increase in temperature, but decreases much less as the temperature becomes higher. I think that this is because at a high temperature the particles are moving around so fast that they are near to boiling point. Therefore, by supplying more heat, it does not make a lot of difference to the “freedom” that the electrons have within the substance. So the higher the temperature already is, the less effect raising the temperature further will have on the movement of the electrons. That is why the graph starts off at a steep negative gradient, but “levels off” as the temperature rises.
Evaluation
The results which I have obtained are good in that they all show clearly the line of best fit as shown on the graph. Also, I do not have any obvious anomalies. However, there are many points which are not quite on the line. They could be like that for the following reasons.
Firstly, I could have not followed the exact procedure in ensuring a fair test, for example by not ensuring the temperature of the thermistor and the temperature of the thermometer were constant.
Also, there could have been faults in the wires, like faulty connections. The apparatus itself was accurate, and so should not have presented a problem.
Think that I could have obtained more accurate results if I had had a digital thermometer. Also, it may have been more accurate if I had started at the high temperature, and gone down. I would have started with boiling water, and taken the necessary measurements. Then, I would add ice gradually, to lower the temperature accordingly. I think that this would have been better because it would be easier to control the temperature, because when we heated it with a Bunsen burner, the temperature would continue to rise even after we had taken it away. However, by reducing the temperature using ice, we could regulate the temperature much more easily, and take bits of ice out if we added too much, whereas we can’t take away heat once we have heated it!
I could investigate the use of different thermistors (ie different semi-conductors) when heated, or possibly by varying the light intensity for something like an LDR.
