How does the power dissipated by a light bulb vary with voltage?

Variables

Controlled – I would like to control the temperature of the light bulb but I cannot because there is no easy way of cooling it down and keeping the same temperature. This will mean that the resistance of the light bulb may increase.

Independent – I will set the voltage passing through the circuit by changing the resistance of the rheostat. I will measure the voltage using a voltmeter, connected in parallel through the light bulb.

Dependant – My dependant variable will be the current passing through the circuit. I will find out the current by using an ammeter.

List of equipment

Voltmeter – The voltmeter will need to measure from 0V – 6V (because my light bulb will be a 6V one) so I will use the one which is 0 – 20V d.c. It will be precise to 2 decimal places.

Power pack – It will need to give out voltages of 0V – 6V (because my light bulb will be a 6V one).

Variable resistor or rheostat – The size of the rheostat I need will be 10 Ω and 5A. I found this out because I used this type of rheostat for my preliminary experiment and it gave me good results.

Wires

Light bulb – I will need a 6V light bulb.

Ammeter – The ammeter will need to measure from 0A to 0.29A (because from my preliminary measurements, at 6V, the current was 0.29A). So I will use the one which is precise to the nearest 1/10th of an amp and can measure from 0 – 10amps.It will be precise to 2 decimal places.

Circuit diagram

Procedure

1. I will set the circuit as shown in the circuit diagram.
2. I will set the voltage at the power pack on 1V.
3. I will change the position of the rheostat contact so it gives me a reading of about 0.5V.
4. I will record the voltage and current readings.
5. I will continue changing the voltage of the power pack and the position of the rheostat contact until I get 12 readings of voltage and current each, with the voltage readings at intervals of about 0.5V.

Conclusion

I found that as I increased the voltage through the light bulb, the power it dissipated increased as well at an increasing rate and the first graph clearly shows this. Since increasing the voltage means that the electrons are flowing faster around the circuit, the current also increases (because current is the amount of electric charge flowing through a point in a circuit in a given amount of time). So if both voltage and current increase, then according to the equation power = voltage x current, power should increase at an increasing rate.

I also found out from the gradient of the second graph that the relationship between voltage² and power is that voltage² increases with power but at a decreasing rate. The reason for this is that with the higher voltages, the current is less than it should be. This is because the electrons in the circuit collide more frequently with the ion cores in the tungsten of the light bulb. There is an increase in the resistance of the light bulb due to the fact that each of the electrons has more energy because of the increase in voltage. This means that the electrons collide more frequently with the ion cores which make the light bulb more resistive. Therefore this increase in resistance affects the shape of the curve in the graph because of the equation P = V²/R. As the voltage increases, the value of R also increases, making the value of P increase at a decreasing rate. This gives the graph a curved line. In an ohmic component, the resistance remains constant, whether or not the voltage increases or decreases. The reason that resistance increases in a light bulb is because the light bulb heats up. It heats up because as the electrons have more energy (as a result of the higher voltage), they flow faster through the tungsten of the light bulb, which means that they collide more frequently with the ion cores in the tungsten. With each collision, kinetic energy is converted into heat energy and therefore the ion cores vibrate faster, making it harder for the electrons to get past them and this increases the resistance.

In any ohmic component, the resistance of the component does not change even if the voltage through it is quite high. This means that if the value of R in the equation P = V²/R remains constant, then P and V² increase proportionally. This would mean that if a graph of P and V² is plotted for an ohmic component, the line would be a straight one. So if a graph of P and V² is plotted which includes lines for both types of component, the line for the light bulb would level off from the ohmic component line. Both lines would start off straight because the resistance of a light bulb does not change dramatically with relatively small voltages.

The results partly support my prediction, one of which was that as the voltage through a light bulb increases, the power dissipated by it also increases but at an increasing rate. However, my prediction of power and voltage² increasing proportionally was not supported by the evidence; instead my results showed that power increases with voltage² at a decreasing rate.

        In my prediction, I stated that both current and voltage increase at the same rate and that is why power increases at an increasing rate. This is absolutely true because if, for example, the current remained constant throughout, then power and voltage would become proportional because if the value I in the equation P = I.V remains constant, the values of P and V would increase together and therefore be proportional.

        I also stated that voltage2 and power would be proportional because P = V2/R and if the value of R remained constant, then P and V2 would increase proportionally. However, my graph of power over voltage2 clearly shows that resistance was increasing. The line seems to curve away; power is not increasing at the same rate. This can be explained from the equation P = V2/R. If the value of R increases each time V increases, then V is being divided by a larger amount each time and therefore power increases but with a decreasing rate.

Evaluation

Overall, I was very happy with how my experiment went. However, I could have improved my results and conclusion in other ways.

        The reason that my results were not perfect was that I could not control the temperature of the filament in the light bulb. As the voltage through it increased, it heated up which meant that the current through it was not as much as it could have been and this may have affected my results. However, it is almost impossible to keep the temperature of the filament constant, so I did not.

        However, I thought that by doing the experiment once and then repeating it twice more was a good idea. If I’d have done the experiment once, then I would not have known if the results were correct. If I’d have done it twice, and got two differing results, I would not have known which results were the correct results. However, by doing the experiment three times, and averaging the results, I ensured that if I had an anomalous result, I could check the two other results and dismiss the anomalous one and still get an accurate measurement.

        For the experiment, I used a 6V bulb and the power pack I used could give out 6V which was perfect for me, as the light bulb would not have worked with a voltage beyond 6V. I also found that I could reach up to exactly 6V so this ensured me that I could measure the maximum power dissipated by the light bulb. I also found that with 12 results, I could draw a line of best fit very easily. The line of best fit also went through the points smoothly – there were no points which were too far away from the best fit line.

        While doing the experiment, I decided to measure the current in the circuit at intervals of 0.5V. I thought this was the perfect as if I’d have measured more current measurement, I would not have found it any easier to draw my line of best fit because all of my 12 measurements lie near or on the line of best fit. If I’d have had fewer measurements, I would have found it much harder to draw a line of best fit because there would have been huge gaps between the points.

        However, there was one thing that I found hard to do during the experiment. I was using a rheostat to change the voltage in the circuit, but it was hard to change the voltage using the slider in the rheostat, because if I moved the slider back too much, the voltage would decrease by too much, and if I moved it forward too much, the voltage would increase too much. However, the rheostat was essential and it was a perfect way of varying the voltage and was extremely suitable for the experiment.

        I also found no anomalies in my results. In my results table, all of my current and power measurements (except for one) were exactly the same. The one different power measurement was only 0.01W less than the average so I decided that it was not anomalous. The maximum difference between my voltages was 0.03V but each voltage gave me exactly the same current so it did not greatly affect my results.

        When writing the conclusion, I found it easy because my results showed very clearly how power varies with voltage. For example, it was easy to mention what the curves of the two graphs looked like because all the points in the graphs were either on or very near the line of best fit. However, I found one power measurement that did not follow the trend of the other power measurements. All the other ones were exactly the same for their voltages except for the second power measurement of 4.49V which was 0.01W less than the average. This meant that this measurement was just 0.9% less than the average (0.01/1.04 x 100) so I did not count this as anomalous because it was too much of a small difference. Therefore, I decided that none of my results were anomalous and that anyone else would come out with the same conclusion because of how similar they are.

        The evidence that I wrote in the conclusion for how resistance increases with temperature came from an experiment that I did earlier in the year. In the experiment, I set up a circuit with a power pack, ammeter and a coil of wire placed in a non-conducting tin with two terminals. I connected a voltmeter parallel to the coil to work out the resistance. I poured hot water into the tin and measured both the current and voltage at intervals of 10 degrees Celsius. I found that as I decreased the temperature through the coil if wire, the resistance also decreased. So therefore, I concluded that in the light bulb, the exact same thing happens for a light bulb.

        However, in the light bulb, the filament is not made of copper but tungsten. This means that tungsten may not follow the same rules as the copper. This would mean that my whole conclusion could be different because it is based on how resistance increases in copper. To make a better conclusion, I would like to do the same experiment as with the copper, but with a coil if tungsten wire instead. This would ensure that the conclusion would be based on evidence for a light bulb, and I would therefore have a stronger conclusion.

This is what my results table would look like for the experiment: