The factors affecting the resistance of a metalic conductor.
INVESTIGATION: THE FACTORS AFFECTING THE RESISTANCE OF A METALLIC CONDUCTOR
Metals conduct electricity because the electrons in the metal can move about inside the structure. These electrons are called free electrons.
Electricity is conducted through a conductor by means of free electrons. Atoms consist of protons, electrons and neutrons. The protons and neutrons make the nucleus of an atom while the electrons circle the outer area of the atom. Electrons in metal are able to move freely and are used as current in an electric circuit. This is because they carry a charge and can move all around the circuit with this charge. While these electrons are travelling around the circuit, atoms are sometimes in the way, causing the two to collide. This takes out some of the energy from the electron and transfers it to the atom. This is how resistance occurs. The number of free electrons depends on the material and the more the free electrons in a substance the better the material as a conductor.
All conductors offer resistance to the flow of current. The conductor's atoms determine this resistance. For example copper atoms offer negligible resistance to an electric current because a significant proportion of its electrons are free to move from electron to electron. Thus copper is commonly used as a conductor.
Current, is the flow of electrons around a circuit. Those materials, which have a lot of "free" electrons, will make it a lot easier for current to flow through, and so there is low resistance. That's why not all metals are equally as good at conducting electricity. Therefore materials with a low resistance are called good conductors and those with a high resistance are called good insulators.
In many materials there is a simple relationship between the voltage across two points and the resulting current. Such materials are called ohmic materials and obey what is known as ohms law, which sates that:
"The current through a metallic conductor at a constant temperature is proportional to the potential difference (voltage)" i.e.
V I
V = IR
R= V/ I V= Potential difference in volts (V) I = Current in amps (A) R= Resistance in a unit is called an ohm ()
The graph below shows the relation between the current and the voltage of a conductor, showing that as the current increases the voltage increases. This is what is termed as Ohm's Law.
In order to find the resistance of a conductor, the gradient of the graph is found out using the formula:
Gradient of the graph = I/V
Whereby R = 1/gradient
FACTORS AFFECTING THE RESISTANCE OF A WIRE:
There are four external factors that influence the resistance in a conductor. Thickness (cross sectional area of the wire), length, the material of the wire being used and temperature all have some effect on the amount of resistance created in a conductor.
LENGTH:
The length of a conductor is similar to the length of a hallway. A shorter hallway would allow people to move through at a higher rate than a longer one. When an atom has more energy it begins to vibrate and as it receives more energy, it vibrates more. As it vibrates it is hitting other atoms around it and passes on its energy. This has a knock on effect and energises all atoms around until this energy is lost through heat. As the atoms vibrate they are more likely to hit electrons because the electrons now have less space to move through. If this is over a shorter distance there will be less atoms for the electrons to hit. If this is carried out over a larger distance there will be more atoms for the electrons to hit. Therefore resistance is proportional to the length of the conductor, providing the temperature, cross sectional area and material of the conductor are kept constant. The graph below shows this:
CROSS SECTIONAL AREA:
The cross-sectional area of a conductor (thickness) is similar to the cross section of a hallway. If the hall is very wide, it will allow more people through it, while a narrow hall would be difficult to get through due to it's restriction to a high rate of flow. If you consider this situation in terms of electrons, a wire with a small cross sectional area would have a smaller number of electrons flowing through it while a wire with a larger area would have a greater amount of electrons flowing through it thus allowing a higher current to flow through it. Thus showing that the resistance is less in a wire with a larger cross sectional area, which would mean that the resistance of the conductor will be inversely proportional to the cross sectional area of it, providing the temperature, length and material of the conductor are kept constant. The graph below shows this theory:
TEMPERATURE:
The temperature of a conductor has a less obvious effect on the resistance of the conductor. It would be as hard to apply the hallway analogy as it is hard to say whether a hot hallway would make us move faster or slower than a cold hallway. To truly understand the effect you must picture what happens in a conductor as it is heated. Higher temperature means more vibrations. Imagine a hallway full of people. Half of the people (the electrons) are trying to move in the same direction you are and the other half (the protons) are evenly spaced but stationary in the hallway. This would represent a cold wire. Since the wire is cold the protons are not vibrating much so the electrons can run between them fairly rapidly. As the conductor (hallway) heats up, the protons start vibrating and moving slightly out of position. As their motion becomes more erratic they are more likely to get in the way and disrupt the flow of the electrons. As a result, the higher the temperature, the higher the resistance and we can say that Ohm's law holds true unless temperature changes. Which leads us to conclude that the resistance of the wire is proportional to the temperature. The graph below shows this:
MATERIAL:
The fourth factor is the conductivity of the material we are using. Some metals are just more electrically conductive than others. This however, is considered an internal factor rather than an external one. The resistivity is a value that only depends on the material being used. For example, gold would have a lower value than lead or zinc, because it is a better conductor than they are as it has more free electrons flowing within it thus has a lower resistance.
Another factor that affects the resistance of a metallic conductor is the type of ...
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MATERIAL:
The fourth factor is the conductivity of the material we are using. Some metals are just more electrically conductive than others. This however, is considered an internal factor rather than an external one. The resistivity is a value that only depends on the material being used. For example, gold would have a lower value than lead or zinc, because it is a better conductor than they are as it has more free electrons flowing within it thus has a lower resistance.
Another factor that affects the resistance of a metallic conductor is the type of circuits being used to conduct the current through the conductor.
There are 2 types of circuits used:
* CIRCUITS IN SERIES
* CIRCUITS IN PARALLEL
SERIES CIRCUITS:
A series electrical circuit is a circuit where the components share the current flowing around the circuit.
These resistors are said to be in series, and as indicated, it is possible to consider them as one single equivalent resistor R. To find this equivalent resistor, we exploit the fact of length being proportional to the resistance of the wire, which by using ohm's law becomes:
V = V1 +V2
IR = IR1 + IR2 R = R1 + R2.
In relation to length this would become:
L = L1 + L2
Owing to the fact that the circuit is in series, there are actually two wires in use that have been connected (using a series connection) the length of the wire is actually increased and the electrons flowing through it will have a longer distance to travel and thus the resistance will be increased.
PARALLEL CIRCUITS:
In a parallel electrical circuit, the components share an energy source but do not share the same current.
These resistors are said to be in parallel, and as before, it is possible to consider them as one single equivalent resistor R. To find this equivalent resistor, we exploit the fact from Ohm's law, that the potential difference across R1 and R2 is the same, we find this becomes:
1/R = 1/R1 + 1/R2
In relation to the cross sectional area of the wire this would be calculated as:
Since area is inversely proportional to resistance,
1/A = 1/A1 + 1/A2
which would be equal to,
1/R = 1/R1 + 1/R2
Owing to the fact that the circuit is in parallel, there are actually two wires in use that have been connected (using a parallel connection) the cross sectional area of the wire is actually increased and the electrons flowing through it will have a shorter distance to travel and thus the resistance will be decreased.
Below are graphs that have the current (I) against the voltage (V) of the factors affecting the resistance of the wire.
Effect of length [cm] on resistance:
This graph shows 3 wires of different lengths, and shows us that as the length of a wire increases the resistances increases as seen in the steepness of each slope. The steeper the slope, the higher the resistance. You can see that the 15cm wire is the steepest line, the second is the 30cm wire and the least steep line is the 45cm wire. The steeper the line, the higher the gradient. In this graph the gradient is equal to I / V, and we know that Ohm's Law states that R= V/I therefore I can say that the resistance of the wire from the graph is 1 / gradient.
Effect of cross sectional area [mm] on resistance:
This graph shows 3 wires of varying thickness' and shows us that as the area increases the resistance decreases as seen in the steepness of each slope. The steeper the slope, the lower the resistance. You can see from the graph that the 0.8 mm line has the steepest slope, followed by the 0.6mm and the least sleep slope is the 0.4mm line.
Effect of material on resistance:
This graph shows 3 different wires of different materials and shows us that the copper wire has a lower resistance than the silver wire and the iron wire has a greater resistance than both these wires. You can see form the graph that the copper wire line has the steepest slope, followed by the silver wire line and the iron wire line has the least steep slope. The steeper the line, the higher the gradient. In this graph the gradient is equal to I / V, and we know that Ohm's Law states that R= V/I therefore I can say that the resistance of the wire from the graph is 1 / gradient.
FAIR TEST on the factors affecting the resistance of a wire:
APPARATUS:
* DC power supply of 5V
* Crocodile clips
* Rheostat
* Ammeter within a 0 to 2 amps range
* Voltmeter within 0--5V range
* Wires of lengths: 15cm, 30cm, 45cm
* Wires of areas: 0.4mm, 0.6mm, 0.8mm
* A copper wire [which I will be investigating]
I will set up the experiment as shown in the circuit diagram below, then a current will be passed through the wire. I will then record the readings of the ammeter and the voltmeter; these results will be used to record the resistance of the wire (R= I/V). I will then change the length of the wire and record the different results for each length. I will only change the length of the wire keeping the material and cross sectional area of the wire the same. The experiment will then be repeated twice more as above but instead of changing the length; I will change the cross sectional area and then the material of the wire.
I predict that my results will obeys Ohm's Law, which means that the current will increase in proportion to the voltage. Therefore if you double the voltage, the current will also double.
I also predict that as the length of the wire gets larger so will the resistance. So a plotted graph of the resistance compared to the length would also go directly through the origin and therefore be directly proportional.
Here is a table of how I plan to present my results and includes the voltages I plan on using to achieve this. I will have separate tables for length, cross sectional area and the materials being used.
Serial number
Voltage [v]
Current [amps]
Resistance
R=V/I
0.1
2
0.2
3
0.3
4
0.4
5
0.5
6
0.6
7
0.7
PRECAUTIONS:
To ensure that I carry out the experiment in the safest way possible as well as accurately the following will have to be done:
* In this experiment we will only change one factor, the length of the wire. This should affect the resistance of the wire in the ways stated in my hypothesis.
* I must keep the surrounding room temperature the same or the particles in the wire will move faster (if the temperature is increased) and this will therefore have an effect on the resistance.
* I must avoid parallax error by looking straight down at the reading on the ruler and not form below the ruler or form its side.
* I must avoid zero error by measuring the 1cm mark as the beginning of my ruler just in case the zero has been rubbed off due to daily usage.
* I will switch off the apparatus when not in use to prevent overheating.
PRIOR TEST:
* I got all the equipment together that would be required; a power supply, a rheostat, voltmeter, ammeter, nichrome wire, leads and crocodile clips and checked that all of these worked correctly before setting up circuit as I illustrated previously.
* I Checked that the circuit worked correctly and that all wires were connected correctly, positive to negative and in correct places
* Throughout the experiment, I used wire that is made of nichrome as I had planned earlier
* I used a battery of 12v, which was of a DC supply
* Starting with a length of wire that was 50 cm [as opposed to using a 30cm wire as I had planned] long, I connected this with crocodile clips at each end of the wire into the circuit in the place as shown in the circuit diagram.
* I then Repeated this process of 8 readings with lengths of wire that were 50,100,150 and 200cm long, making sure that I took readings as accurately as possible to make sure that final results were precise. I first took the readings in ascending order then in descending order to make sure there was no change in the current readings.
* I then Put all results into a table, calculating resistance and checking all calculations to make sure no mistakes had been made
* I used a digital ammeter and voltmeter therefore I did not have to take parallax and zero errors into consideration.
* I then did the same for the cross sectional area readings of the wire and used wires that were 0.4mm, 0.56mm, 0.71mm and 0.91mm thick as compared to using wires of 0.6mm and 0.8mm.
* I then set up the experiment to find out how the material affected the resistance of the wire.
Below are the results of my testing in tabular and graphical form:
RESULTS FOR TESTING FACTORS AFFECTING THE RESISTANCE OF A WIRE:
. Length:
Thickness of wire is 0.4mm
Material is nichrome
Length = 50cm
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
R = 5.06
Length = 100cm
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
R = 9.83
Length = 150 cm
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
R = 14.63
Length = 200cm:
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
R = 19.87
2. Cross-sectional area
Length of wire is 100cm
Material is nichrome
Thickness = 0.4mm:
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
Thickness = 0.56mm
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
Thickness = 0.71mm
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
Thickness = 0.91mm
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
4. Type of circuit:
Length = 20' cm 30'cm
Thickness = 0.4mm
Series circuit:
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
Length = 100cm
Thickness= 0.4mm
Parallel circuit:
Sr. no
Voltage [v]
Current
[Amps]
Current [amps]
R = V/I
I1
2
As I have found from my experiments, the results I obtained show that the factors I predicted of affecting the resistance of a wire have proved true.
Firstly, the factor of length increasing and resulting in an increase of resistance of the wire held true, because, as I varied the lengths of the wire form 50cm to 200cm the resistance increased from approximately 5 to 20.
Secondly, the factor of the cross sectional area increasing and resulting in a decrease of the resistance of the wire also held true, because as I varied the cross sectional areas of the wire from 0.4mm to 0.91 mm, the resistance went from approximately 5.4 to 0.9.
The above-mentioned results occur due to the fact of how many free electrons are being conducted through the wires of different lengths and areas.
* In the case of length; the resistance increases as the length increases because a wire with a shorter distance will have a smaller amount of atoms for the free electrons to hit. Whereas a wire with a longer distance will have more atoms for the electrons to hit and the resistance gradually increases as a result.
* In the case of the cross-sectional area; a wire with a greater area allows more electrons to flow through it as well as having more channels for the electrons to flow through it. As a result the resistance decreases, because current can flow without any restrictions. Whereas a wire with a smaller area all a smaller number of electrons to flow through it resulting in diminished flow of current through the wire. Thus the resistance increases because the current is flowing with restrictions.
From the graphs on the following pages, I have illustrated my results obtained from my experiments and done so with the relationship between the current [I] and the voltage [V].
As predicted earlier, the graphs explain how, as the length of the wire increases, the resistance also increases, as shown in the graph I. As well as explaining how, as the cross sectional area of the wire increases, the resistance decreases, as seen in graph II. I have also plotted a graph to the effect on the resistance of the wire when a specific type of circuit is being used. A series circuit effect is seen in graph III and a parallel circuit effect is seen in graph IV.
GRAPH I:
* Length as a factor affecting the resistance of a wire:
From the graph, we can see that the shorter the wire, the steeper the slope. Each line represents the wires I used that were of lengths ranging from 50cm to 200cm. The 50cm wire line has the steepest slope, thus we say it has the lowest resistance, while the 200cm wire has the least steep slope and we say it has the highest resistance. However to verify this fact I calculated R from the graph and found:
For the length of the wire:
> 50 cm
Gradient = 0.28 - 0.20/ 1.4 - 1.0 = 0.08/0.4 = 0.2
Therefore; R = 1/0.2 = 5
> 100cm
Gradient = 0.12 - 0.10 / 1.2 - 1.0 = 0.02/0.2 = 0.1
Therefore; R = 1/0.1 = 10
> 150cm
Gradient = 1.4 - 0.10 / 1.07 -0.08 = 0.02 / 0.26 = 0.077
Therefore; R = 1/0.77 = 13
> 200cm
Gradient = 0.04 - 0.02 / 0.8 - 0.4 = 0.02 / 0.4 = 0.05
Therefore; R =1/0.05 = 20
Below is my observation table that compares the value of R from my tabular calculations and the value of R from my graphical results:
Length [cm]
Calculated value of R []
Graphical value of R []
R/L []
50
5.06
5
00
9.83
0
50
4.63
3
200
9.87
20
This graph shows, that the length of the wire is directly proportional to the resistance.
GRAPH II:
* Cross - sectional area as a factor affecting resistance of a wire:
From the graph, we can see that the thicker the wire the steeper the slope. Each slope represents the various thickness' of wires I used ranging from 0.4mm to 0.91mm. The 0.4mm wire has the least steep slope, and we say that it has highest resistance, whereas the 0.91mm wire line has the steepest slope and thus we say it has the lowest resistance.
However I must verify these facts form my graph and calculate the value of R from the graph. Here are my results:
For the thickness of the wire:
> 0.4mm
Gradient = 0.25 - 0.15 / 1.25 - 0.82 = 0.1/ 0.43 = 0.21
Therefore; R = 1/0.21 = 4.8
> 0.56mm
Gradient = 0.57 - 0.45 / 1.61 - 1.25 = 0.12/0.36 = 0.33
Therefore; R = 1/0.33 = 3
> 0.71mm
Gradient = 0.99 - 0.80/ 1.6 - 1.25 = 0.19 /0.35 = 0.54
Therefore; R = 1/0.54 = 1.8
> 0.91mm
Gradient = 1.03 - 0.81/ 1.0 - 0.8 = 0.22/0.2 = 1.1
Therefore; R = 1/1.1 = 0.9
Below is my observation table that compares the value of R from my tabular calculations and the value of R from my graphical results:
Area [m ]
Calculated value of R []
Graphical value of R []
R x A []
0.12
5.49
4.8
0.25
2.86
3
0.39
.62
.8
0.65
0.97
0.9
This graph shows that the cross sectional area of the wire is inversely proportional to the resistance of the wire.
GRAPH III:
* A series circuit as a factor affecting the resistance of a wire:
From the graph, we can see that when we use a wire of length 50cm, and pass current through it, via a series circuit the resistance is slightly less, than when we pass current through two wires of lengths 20cm and 30 cm and connect them with a series connection then the resistance is slightly higher. This is because we're doubling the length of the resistor, thus we say that the resistance increases with the total length of the resistors.
To verify my findings I have calculated the value of R of the graph:
> 20'30 cm
Gradient = 0.19 - 0.11 / 1.2 - 0.65 = 0.08/0.55 = 0.145
Therefore; R = 1/0.145 = 6.8
> 50 cm
Gradient = 0.25 - 0.21/1.25 - 1.03 = 0.035/0.22 = 0.159
Therefore; R = 1/0.159 = 6.2
Below is my observation table that compares the value of R from my tabular calculations and the value of R from my graphical results:
Length [cm]
Calculated value of R []
Graphical value of R []
20'30'
6.04
6.8
50
5.06
6.2
GRAPH IV:
* A parallel circuit as a factor affecting the resistance of a wire:
From the graph, we can see that when we use a wire with a smaller cross sectional area and pass current through it via a parallel circuit, the resistance is slightly higher than when using a normal circuit. This is because placing resistors in parallel is equivalent to increasing the cross-sectional area A through which current can flow. In my graph I have used a wire of thickness 0.4mm and passed current through it using a parallel circuit, and the resistance is lower. The resistance for the 0.56mm wire when passing current through it using a normal circuit is higher.
However to verify my findings, I calculated the value of R from my graph and found:
> 0.4mm
Gradient = 0.52 - 0.49/1.4 - 1.3 = 0.3
Therefore; R = 1/0.3 = 3.3
> 0.56mm
Gradient = 0.49 - 0.42/1.4 - 1.2 = 0.07/0.2 = 0.35
Therefore; R =1/0.35 = 2.8
Below is my observation table that compares the value of R from my tabular calculations and the value of R from my graphical results:
Area [m ]
Calculated value of R []
Graphical value of R []
0.12
2.62
3.33
0.25
2.86
2.8
MATHEMATICAL DEDUCTIONS TO FURTHER PROVE MY RESULTS:
Consider a wire of length 'l'
Where;
A = cross -sectional area / number density of electrons
n = electron density =number of free electrons per unit volume
If voltage V, is applied to the wire, the electrons will drift to the positive terminal with a velocity, v.
Volume of the wire = Al
Number of free electrons in conductor = n x A x l = nAl
Total charge that is free to move = n x A x l x e =nAle
Current = charge/time = Q/t
Time required for all electrons to emerge out of the end of the conductor = l/v
Therefore; I = Q/t = nAle/l/v = nAve
Drift velocity (from battery cell) is the EMF,
Therefore; Force = mass x acceleration (force to move current)
Acceleration = velocity/time
L = m x v / t
Work is done by voltage in moving electrons i.e. acceleration:
Work done = Force x Distance
Distance = l
Therefore; work done/electronic charge = w/e
Work done per unit charge = V = w/e = l
Therefore; V = l x m x v/et
R = V/I therefore; V = l x m x v/et = l x m x et
I = nAve nAe
From my analysis I can conclude that as the length of a wire increases, so does the resistance. This is because there is a larger amount of wire to travel up and therefore there will be more factors to increase resistance
I can now sat that I believe my experiments were quite accurate as I performed them fairly and properly, this is demonstrated in the good results I have obtained. My measurements were accurate enough as I used digital ammeters and voltmeters, making them more reliable. In doing so I also avoided the possibility of parallax errors as well as zero errors.
Were my results accurate to draw a conclusion?
I was able to draw a valid conclusion for the measurements of current and voltage, as they were more or less what I expected to attain. I managed to prove that resistance is proportional to length - as length increases, the resistance of the wire increases - and that resistance is inversely proportional to the cross sectional area - as cross sectional area increases, the resistance of the wire decreases.
As well as that, for 2 wires connected via a series circuit combination, the resistance pattern will be the same as that when length of a wire is varied. Whereas for 2 wires connected via a parallel circuit combination, the resistance pattern will be the same as that for when area of the wire is varied.
My results did not agree fully with my heat theory as they showed slight variations, such as a 14.63 result instead of a result close to 13 for a wire of length 150cm. And a 19.87 result instead of a result close to 18 for a wire of length 200cm. These were my anomalous results. This was probably due to the temperature variation of the wire. However these anomalous results were not big enough to change my final reading.
I believe my results allowed me to cover a wide range of factors affecting the resistance of a wire, because I took a total of 4 lengths and eight readings for each length, giving me enough to analyse. I performed the experiment once, but I did take the reading of the current twice. Once in an ascending order, then in a descending order, thus I had two sets of results, which improved my accuracy. I also did a fair test because I followed the precaution of using the same equipment each time the experiment had to be carried out.
Thus I can regard my results as being reliable values, as when compared to actual values, such as getting a 5 resistance for the 50cm wire as compared to supposedly having to get a 4.5 resistance, or a 9.83 resistance for a 100cm wire as compared to having to obtain a 9 resistance.
I don't have any outstanding anomalous values that showed up on my graphs, only a few points did not quite fit on my line of best fit, which were quite close to it anyway. As I mentioned earlier these could have been caused by the heating effect of the equipment which resulted in slight variations of my readings.
OTHER EXPERIMENTS TO MEASURE A CURRENT - VOLTAGE RELATIONSHIP:
Testing a silicon diode:
I could connect a battery, a lamp, and a diode in series. Then connect the narrow end of the diode nearest to the negative terminal of the battery. Using an analog VOM type meter, I would set the meter to one of the lower ohms scales, say 0-2K, and measure the resistance of the diode both ways. If I get zero both ways, the diode is shorted. If I get INFINITY both ways, the diode is open. If I get INFINITY one way but some reading the other way (the value is not important) then the diode is good and I can measure the current and the voltage.
As the graph shows, almost no current flows if the voltage applied is in the reverse direction.
Testing the transistor:
Testing a unijunction transistor (UJT) is a relatively easy task if you view the UJT as being a diode connected to the junction of two resistors, as shown in figure 4-21. With an ohmmeter, measure the resistance between base 1 and base 2; then reverse the ohmmeter leads and take another reading. Both readings should show the same high resistance regardless of the meter lead polarity. Connect the ohmmeter's negative lead to the UJT's emitter. Using the positive lead, measure the resistance from the emitter to base 1, and then from the emitter to base 2. Both readings should indicate high resistances approximately equal to each other. Disconnect the negative lead from the emitter and connect the positive lead to it. Using the negative lead, measure the resistance from the emitter to base.
From my mathematical deductions (to further prove my investigation) in my analysis section I calculated the value for resistivity for a nichrome wire and compared the calculated value with the actual standard value of rho. That is;
LENGTH [cm]
R from my experimental results
R from my graphical results.
R/L [/cm]
50
5.06
5
0.10
00
9.83
0
0.099
50
4.63
3
0.092
200
9.87
20
0.099
AREA [m]
R from my experimental results []
R from my graphical results []
R X A [m ]
.26 x 10
5.49
4.8
6.47
2.5 x 10
2.86
3
7.32
3.96 x 10
.62
.8
6.77
6.5 x 10
0.97
0.9
6.04
= RA/L therefore;
Material
Length [m]
Area [m ]
Resistance []
Calculated [m]
Standard [m]
Nichrome
.50
.26 x 10
3.81
16.0 x 10
10 x 10
Nichrome
Nichrome
Nichrome
Thus we see that my results were not so different from the actual standard value of resistivity, and this is mainly because of the temperature variations that occurred during my experiment. This evidence does support a firm conclusion that if someone was to repeat the same investigation I would expect the to receive the same results.