# Efficency of a Light Bulb Essay Sample

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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 fractional expression).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. The temperature of the water may change as the heat energy emitted by the bulb is absorbed by the water. The light energy is very small and hence is difficult to measure it directly. So it shall be measured it by subtracting the heat energy given out from the total input energy.

The light bulb is submerged in 0.2kg water, and turned on for 900 seconds. The current and voltage of the electrical supply is noted and a table is made with initial findings.

Circuit diagram of the experiment

The law of conservation of energy states that energy may neither be created nor destroyed. Therefore the sum of all the energies in the system is a constant. So the energy input is equal to the energy output.

I.e.

Electrical energy = Heat energy + Light energy

The results obtained are shown below

Current(A)

Voltage(V)

Initial water temperature( (c))

Temperature after 900 sec(c)

1

3.0

17.0

20.0

1.1

4.4

17.0

21.0

1.3

6.0

17.0

22.5

1.4

7.2

17.0

24.0

1.6

8.5

17.0

26.0

1.8

10.9

17.0

29.0

2.1

12.0

17.0

32.0

2.3

13.2

17.0

33.0

These results were collected, so that I can measure any quantities from this data to find my ultimate aim – efficiency. I can manipulate this result to produce desired set of results.

The first step is to manipulate the table of results that we have, so as to find other quantities which are useful to find efficiency.

The electrical energy is calculated first,

Electrical energy = Voltage x Current x Time

Current (I) x Voltage (V) x Time (s)

Electrical Energy (J)

1.0 x 3.0 x 900

2700

1.1 x 4.4 x 900

4356

1.3 x 6.0 x 900

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, ?T = 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= mC?T

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 ?T

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

The table of results showing the manipulated data;

Current(A)

Voltage(V)

Initial water temperature (ï¿½C)

Temperature after 900 sec

Temperature difference (ï¿½C)

Total energy in(J)

Heat Energy(J)

Light energy(J)

1

3.0

17.0

20.0

3.0

2700

2508.00

192.0

1.1

4.4

17.0

21.0

4.0

4356

3344.00

1012.0

1.3

6.0

17.0

22.5

5.5

7020

4598.00

2422.0

1.4

7.2

17.0

24.0

7.0

9072

5852.00

3220.0

1.6

8.5

17.0

26.0

9.0

12240

7524.00

4716.0

1.8

10.9

17.0

29.0

12.0

17658

10032.00

7626.0

2.1

12.0

17.0

32.0

15.0

22680

12540.00

10140.0

2.3

13.2

17.0

33.0

16.5

27324

13794.00

13530.0

These values are calculated using the program Microsoft excel.

Now efficiency of the light bulb at different voltages can be calculated by the formula,

Efficiency = Useful energy (Light energy) / Total energy in (Electrical energy)

Since efficiency is a ratio it does not have any unit. The percentage can be calculated by multiplying it by 100.

Efficiency % = Useful energy (Light energy) / Total energy in (Electrical energy) x 100

Voltage (V)

Efficiency %

3.0

192 / 2700 * 100 = 7.11

4.4

1012 / 4356 * 100 = 23.23

6.0

2422 / 7020 * 100 = 34.50

7.2

3220 / 9027 *100 = 35.50

8.5

4716 / 12240 * 100 = 38.58

10.9

7626 / 17658 * 100 = 43.19

12.0

10140 / 22680 * 100 = 44.71

13.2

13530 / 27324 * 100 = 49.52

Graphs

This is my initial graph.

The graph showing Energy in (electrical energy) against useful energy

Above is my actual graph that I am taking results at.

Analysis of the graph

The graph clearly shows a trend that the higher the voltage the higher the Efficiency of the light bulb however as the line of best fit does not entirely cross all the points I believe that the readings are not exactly correct.

I theorize that for the majority of the readings the energy could also have heated the beaker and the table beneath losing part of the total energy, which was not measured therefore making all the readings incorrect (A systematic error).if the beaker was insulated ,we could have got more accurate results.

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%.