Essay, Research Paper

Aim

I am going to be studying the resistance of wire. The purpose of this investigation is to see how length and thickness of wire affect the dependent variable, resistance. Prediction

I predict that, as the length of the wire doubles, the resistance will also double, but as the cross-sectional area of the wire doubles, the resistance halves. This means that the length will affect the resistance more than the thickness will. Hypothesis

Resistance is caused by electrons bumping into ions. If the length of the wire is doubled, the electrons bump into twice as many ions so there will be twice as much resistance. So

If the cross-sectional area of the wire doubles there will be twice as many ions and twice as many electrons bumping into them, but also twice as many electrons getting through twice as many gaps. If there are twice as many electrons getting through, as there is twice the current, the resistance must have halved. This means that

I am assuming that the temperature is kept constant and that the material is kept constant. We can include this in our equations by adding a constant

Method

Equipment needed:

1 x Power Pack (to give varied voltage)

1 x Voltmeter

1 x Ammeter

5 x wires (with crocodile clips)

wire of varied length and thickness

Controlled variables:

Temperature (room temperature)

Wire material

Dependent variable:

Resistance

Independent variables:

Thickness of wire

Length of wire

Circuit diagram First, set up the experiment as shown above. Turn on the power and set the power pack so that the voltmeter reads 0.1 volts. Take the reading from the ammeter recording both the current and the voltage. Then do exactly the same again but use voltages of 0.2 volts, 0.3 volts, 0.4 volts and 0.5 volts. This is so that when we work out the resistance (V/I) we will have five readings and can then take an average resistance. Then carry out the whole thing again, varying the length of the wire in intervals of 10cm from 10cm to 100cm. To do the thickness experiment, set up the equipment again as shown. Turn on the power and set the power pack to read 0.2 volts. Take the current reading then turn off the power and start again. Take four readings like this so that an average resistance can be found. Next, change the thickness of the wire and do the experiment again. Use the diameters 0.71mm, 0.56mm, 0.28 mm and 0.20mm. Although the diameters haven’t the same interval between them, once we have worked out the resistance, we can draw a graph to discover any relationship between the thickness and the resistance of wire. The equation for resistance = V/I Results

SWG

(thickness/mm) Voltage/volts Current/amps V/I=R/ohms Average R/ohms

22

(0.71) 0.2 0.29

0.28

0.29

0.28 0.69

0.71

0.69

0.71 0.70

24

(0.56) 0.2 0.23

0.24

0.24

0.24 0.87

0.83

0.83

0.83 0.84

32

(0.28) 0.2 0.09

0.08

0.08

0.08 2.22

2.5

2.5

2.5 2.43

36

(0.20) 0.2 0.04

0.04

0.04

0.05 5

5

5

4 4.75

Thickness investigation (Length kept constant at 15cms)

Graph 1 – relationship between the wire’s thickness and its resistance

Length/cm Voltage/volts Current/amps V/I=R/ohms Average R/ohms

100 0.1

0.2

0.3

0.4

0.5 0.04

0.08

0.13

0.17

0.22 2.50

2.50

2.31

2.35

2.27 2.386

90 0.1

0.2

0.3

0.4

0.5 0.04

0.09

0.14

0.19

0.24 2.50

2.22

2.14

2.11

2.08 2.21

80 0.1

0.2

0.3

0.4

0.5 0.05

0.10

0.15

0.21

0.26 2.00

2.00

2.00

1.90

1.92 1.964

70 0.1

0.2

0.3

0.4

0.5 0.05

0.11

0.17

0.23

0.29 2.00

1.82

1.76

1.74

1.72 1.808

60 0.1

0.2

0.3

0.4

0.5 0.06

0.13

0.20

0.26

0.33 1.67

1.54

1.50

1.54

1.52 1.554

50 0.1

0.2

0.3

0.4

0.5 0.07

0.15

0.22

0.30

0.38 1.43

1.33

1.36

1.33

1.32 1.354

40 0.1

0.2

0.3

0.4

0.5 0.09

0.18

0.26

0.36

0.44 1.11

1.11

1.15

1.11

1.14 1.124

30 0.1

0.2

0.3

0.4

0.5 0.1

0.21

0.32

0.43

0.54 1.00

0.95

0.94

0.93

0.93 0.95

20 0.1

0.2

0.3

0.4

0.5 0.15

0.26

0.40

0.54

0.67 0.67

0.77

0.75

0.74

0.75 0.736

10 0.1

0.2

0.3

0.4

0.5 0.17

0.34

0.50

0.68

0.85 0.59

0.59

0.60

0.59

0.59 0.592

Length investigation (Thickness kept constant at SWG24)

Graph 2 – relationship between the wire’s length and its resistance After doing the two graphs I have decided to do a graph of 1/thickness2, to see if thickness is inversely proportional to resistance.

Thickness/mm 1/Thickness2 Average resistance/ohms

0.71 1.98 0.7

0.56 3.19 0.84

0.28 12.76 2.43

0.20 25 4.75

Investigating thickness2 and resistance

Graph 3 – relationship between the wire’s thickness2 and its resistance Conclusion

I conclude that, as the length of a wire doubles, the resistance also doubles (provided the thickness of the wire is kept constant). I also conclude that, as the cross-sectional area of the wire doubles, the resistance halves (provided the length of the wire stays constant). I can conclude this because my graph shows that resistance is inversely proportional to 1/(thickness2). So The theory behind these conclusions are:

As the length doubles the resistance doubles. Resistance is caused by electrons bumping into ions. If the length of the wire doubles, the electrons bump into the ions twice as much so the resistance will double.

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