Apparatus
The wire will be made of: Nichrome
The Voltmeter will be set at: 20V
The Ammeter will be set at: 10 A
Method
I shall take five different lengths of wire: 10CM, 20CM, 30CM, 40CM and 50CM.For each piece of wire I will take five different readings of voltage against current get the resistance of each reading and then get the average resistance of each of the five lengths.
Results
For each length of wire I will use a table like this: -
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
Average Resistance = 0.2526Ω
20 CM
Average Resistance: 0.3312Ω
30 CM
Average Resistance: 0.4188Ω
40CM
Average Resistance: 0.5344Ω
50 CM
Average Resistance: 0.7545Ω
Graphs
Evaluation and Conclusion
I predicted that length does affect resistance from my results, I can conclude that this is correct.
In each of my lengths of wire, the average resistance increased each time the length was increased, starting with 0.2526Ω at 10CM and rising to 0.7545Ω at 50CM. This proves that atoms are made up of a positively charged nucleus surrounded by negatively charged electrons, that collide with each other which slows down the electron and current is therefore dropped. The larger the surface area (length) the more electrons and ions there are present so therefore the longer the piece of wire, the more resistance.
In each of the graphs I drew, I got straight lines. This proves Ohm's law; the current flowing in a metal is proportional to the potential difference across it providing the temperature remains constant.
I plotted the graph of 10CM Voltage vs. Current. The straight line I got from this represents resistance. I will now use the formula y2-y1÷x2-x1 to get the resistance of the voltage-current graph. If I have gained accurate results the answer I get from this formula should match closely the result of the average resistance of a 10CM piece of nichrome wire.
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.
