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# The Electrician's dilemma.

Extracts from this document...

Introduction

Name:   Ahmed Yassin

Physics Gcse Coursework

An electrician has been asked to repair an old radio. He believes that he needs a resistance of 1.9 ohms which is not easily obtained.

You are provided with some reels of resistance wire and are asked to plan and carry out an investigation to help the electrician solve this problem.

When your investigation has been concluded you discover that a resistance of 28.5 ohms is also required to complete the repair. Show how you use the data from your investigation to also obtain this resistance.

As mentioned in the previous page, an electrician wants two resistances: 1.9 ohms and 28.5 ohms. I am to find out how to do this.

I will be provided with different reels of wire of constantan, each reel having a different diameter. I will, for three different diameters, find the resistance of the wire at different lengths. For example, I will take the 24 gauge wire (0.56mm diameter) and find its resistance when it is at 100cm, 80cm, 60cm, 40cm. Exactly how I will do this will be in my method.

I will then plot a graph of resistance (Ω) against length (cm). From the graph I should be able to read off what length is needed to create 1.9Ω. Then finding and using the equation of the graph I will be able to find what length is need to create 28.5Ω.

I will then be able at the end to say:

“ you need ….. cm of ……gauge constantan wire to create a resistance of 1.9Ω/28.5Ω

I will through this investigation also identify patterns and etc. of how length affects resistance. In addition I will also hopefully find how diameter affects resistance.

To solve the electrician’s problem, I will investigate relationships between length of wire and resistance.

Middle

If temperature increase means resistance increase, temperature decrease must mean resistance decrease. Metals such as lead and aluminium become super conductors when cooled to near absolute zero. Current will continue to flow through it for ever as no energy is turned to heat ( i.e. no resistance) so long as the metals stays near absolute zero.

How the type of material affects resistance.

The type of material will affect the amount of free electrons which are able to flow through the wire. The number of free electrons depends on the outer shell of the atoms, so if there are more, there must be more electrons available. If the material has a high number of atoms  there will be a high number of electrons causing a lower resistance due to the increase in electrons. Also if the material’s atoms are closely packed then the electrons will collide more frequently therefore resistance will increase.

How wire length affects resistance.

As wire length is increased the electrical resistance is also increased.

Why?

Electricity is the flow of electrons. As electrons pass through a wire , they collide with the atoms of the wire. This slows the flow of electricity, creating resistance. Also if the length of the wire doubles the resistance should double. This is because there will be roughly twice as many atoms to collide with. Increasing length is like increasing resistance in series. (                                                                            )

If we double the length……

How Cross Sectional Area affects Resistance

According to one of Ohm’s laws , resistance is inversely proportional to the cross sectional area. This means that as the cross sectional ( or you could say diameter or radius) increases the resistance decreases.

Conclusion

Graphs follow:

I am now going to plot a graph of resistance against length. I should then be able to read off, from the graph, the length needed to create 1.9Ω.

I will then find and use the equation of the graph to find what length is needed to create 28.5Ω (using 32 swg constantan wire).

 Length (cm) Resistance (cm) 100 7.6 80 6.4 40 3.25

Graph follows:

This time a problem arises because we cannot read 1.9Ω off the graph . So instead we will use the equation of the graph to find what length of 32swg constantan wire is needed to create 28.5Ω and 1.9Ω.

Summary of results

These are in essence the answers to the Electricians Dilemma:

24swg ( 0.56mm dia.)

• of 24 swg Constantan 95 cm is needed to create a resistance of 1.9Ω
• of 24 swg Constantan 14.25m is needed to create a resistance of 28.5Ω

28swg (0.37mm dia.)

• of 28 swg Constantan, 41 cm is needed to create a resistance of 1.9Ω.
• of 28 swg Constantan,6.59m is needed to create a resistance of 28.5Ω

32swg (0.28mm dia.)

• of 32 swg Constantan, 23.85cm is needed to create a resistance of 1.9Ω.
• of 32 swg Constantan, 3.56m is needed to create a resistance of 28.5Ω

1)

I noticed clear patterns from my results ( in the form of graphs) for 24swg, 28swg and 32swg.

I noticed that all my Voltage against resistance graphs were directly proportional i.e. the line was a straight one going through the origin and that as current doubled voltage doubled. Their general shape was like this:

2)

I noticed that my resistance against length graphs showed that as length increased resistance increased. They also showed that as length was doubled resistance was roughly doubled. I also found that the intercept of that graph was 0 or very close to 0. This all shows that my graphs show that resistance is directly proportional to length.

Explanation of results

All the voltage against resistance

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

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