# Efficency of a light bulb( Analysis of data)

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Introduction

Advancing Physics AS Coursework 3

ADVANCING PHYSICS – COURSEWORK 3

MAKING SENSE OF DATA

Introduction

Through this task I am going to investigate the relationship between the voltage and efficiency of the light bulb’s light emitting capacity. Efficiency is defined as useful energy output divided by the total electrical power consumed (a fractionalexpression).It is a dimensionless number with a value between 1 and 0, or when multiplied by 100 given a percentage.

Efficiency = Power out put / Power input

When lighting a light bulb, the electrical energy input is converted in to both light and heat energies. Here we are measuring the efficiency of the light bulb, i.e. the % of the electrical input energy that’s converted in to the light energy.

Experiment

The light energy emitted can be done by running various voltages through a circuit containing a light bulb and ammeter. The light bulb will be immersed upside down in a beaker containing water .Care was given, so that the socket did not get wet. The temperature of the water was also noted at each voltage.

Middle

7020

1.4 x 7.2 x 900

9072

1.6 x 8.5 x 900

12240

1.8 x 10.9 x 900

17658

2.1 x 12.0 x 900

22680

2.3 x 13.2 x 900

27324

Temperature difference, ΔT = Final temperature – Initial temperature

Since the light energy is difficult to calculate, heat energy is calculated first and is deduced from the total energy in (electrical energy).

The heat energy = Mass (200g) X Specific heat capacity of water (4.18 J) X Temperature difference

E= mCΔT

Here mass and the specific heat capacity is constant the only variable is change in temperature.200 X 4.18 = 836. So this can be written as 836 x ΔT

Now we have the total energy in (electrical energy), and the heat energy. So we can find the light energy.

Light energy = Total energy in (Electrical energy) - Heat energy

Conclusion

The experiment was not repeated, which makes it less reliable. Also no measurements were taken after 900 seconds.

For 6.0V the reading is far away from the line of best fit. This may be because of a number of reasons: The water may not have cooled completely between readings; there may have been a power spike during the reading meaning that the voltage was not actually 6.0V. The current could have been measured incorrectly influencing the initial equation for total electrical energy.

Conclusion

From my graph I can conclude that for these values the higher the voltage, the higher the efficiency of the light bulb, however this does not necessarily mean it will be true for all voltages. There has to reach a point when increasing the voltage will no longer increase the efficiency as the highest efficiency the bulb could theoretically attain is 100% efficiency and it is unlikely that it would even reach 100% efficiency. Also the bulb has a maximum voltage which once crossed will burn the bulb by burning out the filament, meaning the efficiency will be effectively 0%.

The average efficiency of the light bulb is 35.54

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This student written piece of work is one of many that can be found in our AS and A Level Electrical & Thermal Physics section.

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