# What Factors affect the resistance of a wire?

Extracts from this document...

Introduction

GCSE Physics Courseworkby Anna Lambregts

What Factors affect the resistance of a wire?

## Aim

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.

Middle

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= π r²

The unit in which resistance is measured is called an ohm (Ω). 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 Ω)=

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.

Conclusion

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= π r²

## A= π x 0.16²

A= 0.08 mm²

Cross sectional area | Resistance |

0.08 mm² | 2.8 Ω |

0.32 mm ² | 1.5 Ω |

1.29 mm² | 0.8 Ω |

5.2 mm² | 0.5 Ω |

20.6 mm² | 0.2 Ω |

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.

This student written piece of work is one of many that can be found in our GCSE Electricity and Magnetism section.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month