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# How does the power dissipated in a light bulbfilament vary with the applied voltage?

Extracts from this document...

Introduction

How does the power dissipated in a light bulb

filament vary with the applied voltage?

Apparatus (see later for details)

Rheostat                                12V Power Pack

Voltmeter                                Filament Bulb (12V)

Ammeter                                Several Wires

12V D.C

Theory and influence of preliminary exp.

To work out the power dissipated from the filament I will use the formula:

P=VI

(Where P = power

V = voltage

I = current)

To find the voltage I will put a voltmeter in parallel with the light bulb, as this will measure the voltage used by the bulb and not the whole circuit (resistance of the other wires, ammeter or rheostat may cause the circuit to draw slightly more voltage than the light bulb uses). This will help to give a more accurate and reliable answer. The ammeter should be in series, but does not necessarily need to be after the voltmeter. The rheostat allows me to alter the voltage to the level I want.

I carried out a preliminary experiment using the above circuit, to test that my method actually worked and to find out the range of voltages that the power pack could produce.

Middle

2.00

0.89

3.00

1.06

3.00

1.06

3.00

1.06

4.00

1.21

4.00

1.21

4.00

1.21

5.00

1.35

5.00

1.35

5.00

1.35

6.00

1.47

6.00

1.47

6.00

1.47

7.00

1.59

7.00

1.59

7.00

1.59

8.00

1.71

8.00

1.71

8.00

1.71

9.00

1.81

9.00

1.81

9.00

1.81

10.00

1.91

10.00

1.91

10.00

1.91

11.00

2.01

11.00

2.01

11.00

2.01

12.00

2.10

12.00

2.10

12.00

2.10

I then took the mean average of all the results to lessen the effect that anomalies might have and calculated power using the formula P=VI (the values are rounded to 2 d.p).

 Average Results Voltage (V) Current (Amps) Power (Watts) 0.00 0.00 0.00 1.00 0.69 0.69 2.00 0.89 1.78 3.00 1.06 3.18 4.00 1.21 4.84 5.00 1.35 6.75 6.00 1.47 8.82 7.00 1.59 11.13 8.00 1.71 13.68 9.00 1.81 16.29 10.00 1.91 19.10 11.00 2.01 22.11 12.00 2.10 25.20

I then plotted these results on a graph, which is shown on the next page. Analysis and Conclusion

To check whether my hypothesis was right I processed these results further, by making graphs of P against Vn where 1<‘n’<2 at 0.1 intervals. When I found a graph that looked the straightest, I stopped.

Conclusion

1.5 graph was straighter. But there is also another of checking my hypothesis. I could do this mathematically by using logarithms.

If :                    P  Vn

=>                P = KVn                                (K=constant)

Than :                ln P = ln K + ln (Vn)

Knowing that ‘’ln (Vn) = n ln V’’ allows me to substitute in to get:

ln P = ln K + n ln V

=>                n = ln P – (ln K)

ln V

This equation than allows me to calculate ’K’ when I have ‘n’. To find ‘n’ I will plot ‘ln P’ against ‘ln P’, and ‘n’ will be the gradient. This is shown below.

 LN P LN V * * -0.37591 0 0.576613 0.693147 1.156881 1.098612 1.576915 1.386294 1.909543 1.609438 2.177022 1.791759 2.409644 1.94591 2.615935 2.079442 2.790551 2.197225 2.949688 2.302585 3.09603 2.397895 3.226844 2.484907 I found the gradient of this line to be roughly 1.45 to 2 d.p. and therefore;

n = 1.45

This then allows me to calculate ‘K’, by substituting values into the formula above (in this case the values for 5V).

n = ln P – (ln K)

ln V

=>                1.45 = 1.909543 – (ln K)

1.609438

and                 ln K = 1.909543 – (1.45 x 1.609438)

=>                ln K = - 0.4241421

and                 K = 0.65 to 2 d.p

To conclude, after using natural logarithms to check my original hypothesis I found that the following formula was true:

P = 0.65V1.45

And I can say that these results (to 1 d.p) support my original prediction that the value was of n was 1.5.

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