- Level: AS and A Level
- Subject: Science
- Word count: 1047
Investigating how the length of a wire affects its resistance
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Introduction
Physics Coursework
Investigating how the length of a wire affects its resistance
Sean Connolly 12M
I have decided for my coursework that I will investigate how length affects the resistance of a wire.
Background information on resistance
In 1826, George Ohm discovered that:
The current flowing in a metal is proportional to the
Potential difference across it providing the temperature
remains constant.
He then developed a formula: -
Resistance (R) = Voltage (V) ÷ Current (I)
Resistance is a measure of how easily the electrons can move through a metal. Therefore a low resistance means that the electrons can move more easily.
Aim
In my investigation I shall try to see if the length of a piece of wire affects its resistance.
Prediction
I predict that the length of wire will affect its resistance. I am making this prediction because all metal s contain electrons in their outer shell. The larger the surface
Middle
For each length of wire I will use a table like this: -
Setting | Voltage (V) | Current (A) | Resistance (Ω) |
1 | |||
2 | |||
3 | |||
4 | |||
5 |
I will then get the average resistance by adding all five resistances up and diving by 5.
I will then draw 5 separate graphs for each length showing voltage against current.
Then I will incorporate all 5 lengths into one graph to show the changing trends.
Finally my last graph will include length against average resistance.
Table of Results
10 CM
Setting | Current (A) | Voltage (V) | Resistance (Ω) |
1 | 0.6000 | 0.1500 | 0.2500 |
2 | 0.5000 | 0.1255 | 0.2510 |
3 | 0.4000 | 0.1015 | 0.2538 |
4 | 0.3000 | 0.0760 | 0.2533 |
5 | 0.2000 | 0.0510 | 0.2550 |
Average Resistance = 0.2526Ω
20 CM
Setting | Current (A) | Voltage (V) | Resistance (Ω) |
1 | 0.4500 | 0.1500 | 0.3333 |
2 | 0.3500 | 0.1180 | 0.3371 |
3 | 0.2800 | 0.0920 | 0.3259 |
4 | 0.2000 | 0.0653 | 0.3265 |
5 | 0.1200 | 0.0400 | 0.3333 |
Conclusion
X1= 0.3000, Y1= 0.0760, X2= 0.6000, Y2= 0.1500
Y2-Y1÷X2-X1
= 0.1500-0.0760÷0.6000-0.0760
=0.0740÷0.3000
= 0.2467Ω
My average resistance for 10Cm was 0.2526Ω. My answer for the resistance of the graph was 0.2467Ω. This shows that my graphs and tables are very accurate and that the straight line on the voltage-current graph represents resistance.
I feel that my results gained were reliable as all my results tallied. But at times the voltage and current readings on the voltmeter and ammeters flickered giving me at times, inaccurate results. If I were to repeat the experiment I would use much more up to date meters. I feel that some of my graphs did not turn out the way I would have liked them because I didn’t take enough readings in the practical. If I had have taken more results them my graphs would have been much more accurate. I do though however feel that the evidence is sufficient to support my predictions as my results turned out to be accurate and they supported my predictions well.
This student written piece of work is one of many that can be found in our AS and A Level Electrical & Thermal Physics section.
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