The Aim of this experiment is to investigate the ways in which thickness (or rather the cross sectional area) affects the resistance of a wire, how it affects it, and why this happens.

Prediction

I predict that the thickness area of the wire will affect the resistance of the wire, as the thickness gets greater, the resistance decreases.

I also predict that the cross sectional area and resistance are inversely proportional to each other. Where the thickness is concerned, the thickness and resistance are directly proportional to each other, this I mean that as the thickness increases, the resistance decreases at the same rate. Therefore:

Thickness

Resistance

x

y

x + 1

y – 1

Theory

I am basing my prediction on what I know about how the cross sectional area affects the resistance of a wire. I

know that 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 ‘jump’ atoms.

The smaller the cross sectional area of the wire, the less channels of electrons in the wire for current to flow, as the energy is not spread out as much in a wire that has a larger cross sectional area, the resistance will be higher. This can be illustrated using a diagram:

Direction of current

SMALL CROSS SECTIONAL AREA- MORE RESISTANCE

Direction of current

LARGER CROSS SECTIONAL AREA- LESS RESISTANCE

The first diagram illustrates what happens when the cross sectional area is small, and therefore the resistance high. It shows that because it is smaller, and

therefore thinner, there is not as much space for all the current to get through, and therefore there is a lot of resistance.

The second diagram illustrates what happens when the wire has a large cross sectional area, it shows that twice the amount of current is able to go through, and therefore the resistance has decreased by half.

i.e. Resistance= 1/ Area

This can be explained using the formula:

R= I/V

The way in which the cross sectional area of a wire is found is the following:

Using a micrometer find the diameter of the wire

Half that by two in order to find the radius

Apply formula: A= ? rï¿½

The unit in which resistance is measured is called an ohm (?). Ohm’s Law gives the relationship between current strength and potential difference in a circuit. From Ohm’s Law it can be deduced that:

Potential Difference (volts V)

Resistance (ohm ?)=

Current (ampere A)

OR R= V/I

The definition of an ohm can be described as being:

A resistor offers a resistance of one ohm if a potential difference of one volt drives a current of one ampere through it.

There are also other factors that could possibly have an effect on the resistance of a wire. These are:

ï¿½length

ï¿½temperature

ï¿½material

The length affects the resistance due to the following things, metals conduct electricity because the atoms in them do not hold onto 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 ‘jump’ atoms. So if the length of the wire is doubled, the

amount of atoms in the wire double, so twice the amount of energy is required: there are twice as many ‘jumps’ if the wire is twice as long.

The material of wire could also affect the resistance of a wire, as some materials may vary from one another. The material must be chosen sensibly, as some materials may be more available, less expensive and easier in use than others.

As an obvious statement, when choosing the material you must make sure that the material actually conducts electricity!

Variables

Independent variable for 1st experiment- My independent variable for this first experiment (investigating the

appropriate length) is the length of the wire.

Dependent variables for 1st experiment-My dependent variables are the following:

ï¿½The material of the wire (constantan)

ï¿½The voltage

ï¿½The Ohmmeter, to prevent differences between two or more machines.

ï¿½The micrometer, to avoid differences between two or more different micrometers.

ï¿½The temperature

ï¿½The thickness (or rather cross sectional area)

Independent variable for the 2nd experiment- My independent variable for this experiment (investigating how the cross sectional area affects the resistance of a wire) will be the cross sectional area of the wire,

Dependent variables for the 2nd experiment-My dependent variables shall be the following:

ï¿½The material of the wire (constantan)

ï¿½The temperature

ï¿½The length

ï¿½The voltage

ï¿½The micrometer, in order to prevent differences occurring

ï¿½The Ohmmeter, in order to prevent differences between machines

Apparatus

For Experiment One:

?One multi-meter set on the setting for measuring in Ohms (?).

?One reel of constantan wire

?Two crocodile clips

?One metre ruler

?Two wires

For Experiment Two:

?One multi-meter, set on the Ohm (?) setting

?One micrometer

?One reel of constantan wire

?One metre ruler

?Two crocodile clips

?Two wires

Diagrams

For Experiment One:

– +

Ohmmeter

Metre ruler

Constantan

For Experiment Two:

– +

Ohmmeter

Constantan

Method

For Experiment One:

1. Sellotape a wire of 1metre long to a metre long ruler.

2. Set the multi-meter to the Ohmmeter setting.

3. Attach all the wires to the appropriate places.

4. Clip the positive crocodile clip to the 0 cm mark on the ruler.

5. Move the negative clip from 10cm, to 25 cm, to 40 cm,

to 50 cm, to 60 cm, and finally to 75 cm.

6. Note down all results carefully in results table each time a result is collected.

For Experiment Two:

1. Get a reel of constantan and mark it at the appropriate length.

2. Set the Multi-meter onto the Ohmmeter setting.

3. Attach all the wires to the appropriate places.

4. Attach the negative crocodile clip to one end of the wire and the other to the marked spot.

5. Note the results carefully into results table.

6. Fold the wire double the previous thickness and start again.

Using these two methods I will first investigate the appropriate length, and then investigate the effect of the cross sectional area on the resistance of a wire.

I believe that the results that I am collecting will provide me with a sufficient range for me to be able to make an educated statement about how the cross sectional area of a wire affects the resistance.

Data Presentation

I will present the results that I have collected first in two results tables (one for the appropriate length and one for the cross sectional area), then in a series of graphs that relate to these results.

Fairness

In order to keep my experiment fair, I will ensure that all the boundaries defining the variables are kept to and that the results are noted down carefully.

I will also make sure that all the equipment is in correct working order to ensure that no technical mistakes are made.

Safety

In order to ensure safety, all instructions must be followed carefully at all times. If this is done, then the experiment should be carried out without any problems to the persons safety what so ever.

Results

Length (cm)

Resistance (ohms)

10

20

0.94

40

1.94

50

60

2.80

75

Thickness

Resistance

0.32 mm

2.8

0.64 mm

1.5

1.28 mm

0.8

2.56 mm

0.5

5.12 mm

0.2

In order to investigate how the cross sectional area affects the resistance in a wire, I will now have to convert the thickness, into the cross sectional area of the wire.

Diameter= 0.32 mm

Radius= 0.16 mm

A= ? rï¿½

A= ? x 0.16ï¿½

A= 0.08 mmï¿½

Cross sectional area

Resistance

0.08 mmï¿½

2.8 ?

0.32 mm ï¿½

1.5 ?

1.29 mmï¿½

0.8 ?

5.2 mmï¿½

0.5 ?

20.6 mmï¿½

0.2 ?

Analysis

From the results that I collected I can conclude that what I stated in my prediction was correct. Using these results, I will now prove that the cross sectional area is inversely proportional to the resistance.

In order to do this I must now compose another Graph, the graph of Resistance over 1/Area:

From this Graph I am able to verify that my statement is correct.

Evaluation

Overall the results that I collected in both the experiments were as I had expected. There were no problems when collecting my results. There were however slight anomalies such as the resistance being o.1 of an Ohm

off the expected results. But these differences are negligible.

It was also hard at times to measure the resistance at the right length as the wire had to be folded double. This caused me to have to measure to see if it was still the correct length each time.