Objectives
Abstract
The total length of the wire affects the amount of resistance. The longer the wire is the more resistance there will be. This is because resistance occurs as result of collisions between charge carries. More collision along the wire means more resistance. Secondly the diameter will affect the amount of resistance. A bigger diameter means a wider cross-sectional area, thus less resistance there will be to the flow of electric current. Last but not least temperature affects resistance. An increase in temperature of the wire will lead to an increase in resistance.
Equipment
Procedure 1
Data
Result table
Material | Brass | Copper | Aluminum | steel |
Current (A) | Voltage (v) | voltage | voltage | voltage |
0.2 | 0.004 | 0.002 | 0.004 | 0.065 |
0.4 | 0.009 | 0.003 | 0.007 | 0.131 |
0.6 | 0.013 | 0.004 | 0.011 | 0.196 |
0.8 | 0.017 | 0.005 | 0.015 | 0.262 |
1 | 0.021 | 0.006 | 0.018 | 0.328 |
Diameter(mm) | 1.019 | 1.017 | 1.012 | 1.014 |
Analysis and Calculations.
Resistance calculations:
V=IR V2= 0.009, I2= 0.4
V= 0.004, I= 0.2 R2= 0.009/0.4
R= 0.004/0.2 R2= 0.02Ω.
R= 0.02 Ω
Resistivity (p) = RA, Where R is the resistance and A is the cross-section area
A= πr^{2}, r= 1.019/2= 0.51 mm or 0.00051m
A= 3.14*0.00051^2= 8.15*10^{-7}m^{2}
P= 8.15*0.02*10^-7= 1.63*10^{-8}Ωm^{2}
V=0.002,I=0.2
R= 0.002/0.2
R=0.01Ω
Resistivity
A=πr^{2},r= 1.017/2=0.5085 mm 0r 0.0005085
A= 8.12*10^-7m^{2}
P= AR
P= 8.12*10^{-9}Ωm^{2}
V=0.004, I=0.2
R= 0.004/0.2
R= 0.02Ω
Resistivity
A=πr^2, r= 1.012/2=0.506 mm 0r 0.000506m
A= 8.04*10^{-7}m^{2}
P= AR
P= 1.61*10^{-8} Ωm^{2}
V=0.065, I= 0.2
R=0.065/0.2
R=0.325
Resistivity
A=πr^2, r= 1.014/2= 0.507 mm or 0.000507
A= 8.07*10^{-7}m^{2}
P= AR
P= 2.62*10^{-7}Ωm^{2}
From the above calculations and analysis of the data obtained from the experiment, it is clear that steel has the highest resistance and resistivity, thus cannot be used in making electrical wires. On the other hand copper has the lowest resistance and resistivity thus can be used in making electrical wires. This shows that the material of a substance will affect charge flow, Not all materials are created equal in terms of conductive ability. As seen from the experiment some materials are better conductors than others.
Procedure 2, The effects of conductor length on resistance
I= 1A | Dia= 1.019 | Resistance= V/I |
Length cm | Voltage v | |
2 | 0.002 | 0.002 |
4 | 0.003 | 0.003 |
6 | 0.005 | 0.005 |
8 | 0.007 | 0.007 |
10 | 0.009 | 0.009 |
12 | 001 | 0.01 |
14 | 0.012 | 0.012 |
16 | 0.014 | 0.014 |
18 | 0.016 | 0.016 |
20 | 0.017 | 0.017 |
22 | 0.019 | 0.019 |
24 | 0.021 | 0.021 |
As observed from the data above an increase in length leads to a correspondent increase in resistance.
Graph of resistance vs Length.
Procedure 3, The Effects of cross sectional area on resistance
Data
Diameter mm | Area | Voltage v | Resistance Ω |
0.508 | 2.03*10^{-7}m^{2} | 0.087 | 0.087 |
0.804 | 5.07*10^{-7}m^{2} | 0.036 | 0.036 |
1.019 | 8.15*10^{_7}m^{2} | 0.021 | 0.021 |
1.263 | 12.5*10^{-7}m^{2} | 0.015 | 0.015 |
As observed from the above data analysis an increase in cross-sectional area leads to a decrease in resistance of charge flow through the material.
The y intercept is 0.0843 as seen from the graph, R= 0.878.
Conclusion
The experiment was a success, although the data was not 100% accurate it was near perfect to the theoretical value known. All the experiments followed Ohm’s law.