The Resistance Of A Wire

Aim: The resistance of a wire depends on certain factors. Investigate the effect of two of these factors -

Planning

Some variables that will be relevant to this investigation are:

Length

Thickness

Temperature

Voltage

Resistance

Material

Of these the variables will be input and output voltages in experiment one, and length and resistance in experiment two. The other variables (temperature, material and voltage) will have to be kept constant in both experiments to make sure that only length, thickness and resistance are investigated. In experiment 1 the same bit of wire and the same thickness need to be kept constant. In experiment 2, the length will need to be kept constant to make sure only the variables indicated are investigated to ensure a fair test.

Metals conduct electricity because the atoms in them do not hold on to their electrons very well, and so creating free electrons, carrying a negative charge to jump along the line of atoms in a wire. Resistance is caused when these electrons flowing towards the positive terminal have to 'jumps' atoms. So if we double the length of a wire, the number of atoms in the wire doubles, so the number of jumps double, so twice the amount of energy is required: There are twice as many jumps if the wire is twice as long.

The thinner the wire is the less channels of electrons in the wire for current to flow, so the energy is not spread out as much, so the resistance will be higher: We see that if the area of the wire doubles, so does the number of possible routes for the current to flow down, therefore the energy is twice as spread out, so resistance might halve,

i.e. Resistance= 1/Area.

This can be explained using the formula

R = V/I

where there is 2X the current, and the voltage is the same, therefore R will halve. I did some research and in a book called 'Ordinary Level Physics' By A. F. Abbott, it says 'that doubling the area will therefore halve the resistance'- in other words the resistance of a wire is inversely proportional to its area, or R ? 1/A , but we are measuring diameter, so if the area is: ?r2 = ?(d?2) 2 A= ?d2 ? 4 Where A is area and d is diameter.

Method

Experiment One - First a length of wire over a metre long is sellotaped to a metre rule. The positive crocodile clip is attached at 0cm. And the negative is moved up and down the wire, stopping at 20, 40, 60, 80 and 100cm. Each time reading the ammeter and voltmeter to work out resistance R = V/I. This is using 30 SWG wire. Other variables, voltage, thickness, and temperature will be kept constant, although the temperature will rise once current is passing through it, which will cause the atoms in the wire to vibrate, and so obstruct the flow of electrons, so the resistance will increase, creating an error. In both experiments constantan wire is used because it does not heat up as much as copper, so the resistance is not effected as much.

Experiment Two - The circuit isset up is the same, as is the method apart from the length is constant at 50cm, and the thickness is changed between 28, 30, 32, 34, 36, 38 and 40 swg. For both experiments the voltage will be kept the same at 2V dc from a power pack. Both experiments will be done twice with different ammeters in case of any damaged or old equipment to gain more accurate results.

Results

Experiment 1

Length (cm)

V1 (volts)

V2 (volts)

A1 (amps)

A2 (amps)

Average resistance (Ohms)

100

1.00

1.00

0.20

0.20

5.00

80

1.00

1.00

0.30

0.28

4.00

60

0.90

0.90

0.40

0.30

2.80

40

0.90

0.85

0.50

0.40

1.94

20

0.70

0.80

0.80

0.75

0.94

Experiment 2

Thickness (mm)

Area (mm2)

V1 (volts)

V2 (volts)

A1 (amps)

A2 (amps)

Resistance

28

0.36

0.107

0.8

0.8

0.61

0.59

1.3

0.29

0.066

0.9

0.9