Determining the resistivity of a piece of wire Free Essay Example

Resistivity is a property of a wire linking the cross sectional area and lengths of a piece of wire. Resistivity is the measure of resistance to flow on a molecular level.

“It is a measure of the resistance to the flow of current from a microscopic level, that is, as explained in terms of the atoms, the basic building blocks of all solid materials. The resistivity, ?? (Greek letter rho), depends on the behaviour and number of free, or conduction, electrons and not on the shape of the conductor, as does resistance.

”

(http://mst-online.nsu.edu/mstonline/electronic/electronic2.htm)

As explained above it differs from resistance as it takes into account the properties of the material. Resistance does not do this it is merely the ratio of Voltage to current through a material/component.

Resistivity (?) is defined through the equation below where resistance is R length is l and cross sectional area is A.

? = R A

l

The unit of resistivity can be show below

? = R A

l

? = ? x m2

m

? = ?m

So the Si unit of resistivity is ohm meters (?m)

The piece of wire to be used will be an alloy called Nichrome.

It is an alloy of Nickel and Chromium. The exact compositions of the alloy Nichrome are 80% Chromium and 20% Nickel. This alloy is designed to have a high resistivity which is generally around 100×10-8 ?m. Nichrome also has a very high melting point which means it is excellent for high resistance wire.

Resistance is very important in resistivity and is defined as the ratio of voltage to current as shown below in the equation.

R = V

I

Resistance is the property of a body to resistance the flow of electricity around a circuit.

As coulombs of charge pass through a body they must pass through channels of free electrons. These delocalised electrons can move relatively freely through the material but must pass around nuclei and must avoid other electrons. This produces a difficult pathway for a coulomb of charge to pass through and it is this that is defined as resistance.

The diagram below is a simplified version of the pathways coulombs must take to move through a material.

(http://micro.magnet.fsu.edu/electromag/java/filamentsistance/)

As can be seen the nuclei are represented as large orange/yellow ‘dots’. The coulombs of charge are represented by the small red dots and the small orange/yellow dots are the fixed electrons.

The nuclei resist the flow of coulombs and thus produce electrical resistance.

Resistivity is a very important property and involves cross sectional area of a material. Changing the area will not reduce/increase the resistivity but will change the resistance. Therefore the Area must be measured accurately so this can be accounted for in the formula which involves resistance and length as well.

Micrometer readings of the wires’ diameter are used to calculate the cross sectional area by using the formula

A = ? d2

2

Results

The results table shown overleaf contains the raw and manipulated data from the experiment. During the experiment the current was not altered and was always kept at 0.50A. This was a suitable value as in a previous experiment the current was kept at 0.50A and this did not heat the wire suitably to alter the results. The length was however changed for each trial. The length was measured at 10cm intervals. Voltage changed was measured. There were two trial runs for each length of wire. The voltages were measured and recorded as V1 and V2. As the error on the ruler was �0.001m the maximum and minimum values that the reading could’ve been are shown. This also applied to the voltmeter and ammeter which had errors of �0.01V and �0.01A respectively. Knowing these errors a maximum and minimum were recorded taking into account both V1 and V2.

E.g. in the fist reading at l=1.00m V1=7.74V and V2=7.23V this meant accounting the random error of �0.01V the minimum and maximum voltages are 7.72V and 7.75V respectively. All raw data must be recorded to 2 decimal places as this is the greatest accuracy of the equipment. Minimum and maximum also have 2 decimal places as they are not manipulated data.

Once the minimum and maximum voltages and currents have been found the minimum and maximum resistances can then be found using the formula

R = V

I

As resistances are calculated this is manipulated data and allows for an extra decimal place i.e. three decimal places.

So to for the first reading to find the maximum resistance would be the maximum voltage over the minimum current i.e. R=7.75/0.49=15.816?.

The minimum resistance would be the minimum voltage over the maximum current

i.e. R=7.22/0.51=14.157?

This was then done for all the trial data.

Lastly the systematic errors of the equipment must be added but, as there were no systematic errors in any of the equipment this does not need to be accounted for.

A graph of these errors was plotted with length against Resistance

The formula linking ?, A, L and R is shown below

R = (?l)/A

where ? is the resistivity, R is the resistance, A is the cross sectional area and l is the length of the wire.

Below is the equation rearranged in the graphical form y = mx + c

R= ? l

A

Therefore Resistance (R) is represented on the y-axis and length (l) is represented on the x-axis.

The gradient is denoted by ?/A there fore the gradient multiplied by A will give the resistivity of the wire.

As the error bars are plotted on the graph two gradients can be drawn a maximum and minimum shown below are the gradient calculations. As gradients are manipulated data they allow for an extra significant figure so they are to 3 significant figures similar to the resistance.

Gardientmax = ?y/?x = 14/880 = 0.0159

Gradientmin = ?y/?x = 12.6/910 = 0.0139

Now that minimum and maximum gradients have been found an average and an error can be calculated.

Gradientavg = (gradientmax+gradientmin)/2 = 0.0149

Error in gradient = (gradientmax-gradientmin)/2 = �0.001

As the line of best fit is linear and the gradient is positive it shows that R is directly proportional to l

R ? l

R = l x (a constant)

R = (a constant)

l

This constant is our gradient ?/A

From this we can calculate the resistivity using the formula

m = ?/A

where m is the gradient but first the cross sectional area must be found.

Shown below is the raw data where the Area was calculated using A = ?(d/2)2

As area is manipulated data it can be recorded to 3 significant figures

D1/m �0.01m

D2/mm �0.01m

Dmax/m

Dmin/m

Amax/m2

Amin/m2

0.32

0.32

0.31

0.33

0.0755

0.0855

Now that area is known the resistivity can be found

Maximum

m = ?/A

0.0159 x 0.0855 = ?

?max = 0.00136Nmm

Minimum

m = ?/A

0.0139 x 0.0755 = ?

?min = 0.00105Nmm

No that a minimum and maximum resistivity has been found an average and error can be found

?avg = (?max+?min)/2 =0.00121Nmm

Error in ? = (?max-?min)/2 = �0.000155Nmm

Converting Nmm into Nm is a simple case of multiplying the answer by 10-3

Therefore our resistivity is 1.21×10-6Nm �1.55×10-7Nm

However the raw data we collected was only accurate to two significant figures therefore our final result can only be that accurate. Therefore our final determined resistivity is 1.2×10-6Nm �1.6×10-7Nm

This is close to the accepted value of 1×10-6Nm

Evaluation

A few problems arose during the experiment. These were mainly to do with the difficulty of setting the current to 0.50A each time. Towards the larger lengths of wire the current was increasingly difficult to alter accurately. Once a suitable current value was set the voltage reading then was not static. It was constantly changing making it very difficult to measure. This problem was not solvable and thus the error bars for the higher resistance values are fairly large. One further difficulty was the necessity to measure the wire and alter the current and hold the jockey at the same time. This was near impossible at first where the wire was loose next to the ruler. To overcome this problem sellotape was used to affix the wire to the ruler this then made it relatively easy to measure.

If I did the experiment a further time I would change only one piece of equipment and that would be the power supply. If this was changed it would maybe keep the voltage at a much more constant level. The rest of the experimental procedure was very substantial however. Therefore if the experiment was repeated the method and the equipment (power supply excluded) need not be changed.

For an extension to the experiment the resistivity of other wires could be measured. This could include copper, steel and other metal/metal alloy wires. Another extension would be to change the diameter of the wire to see what affect that had on the resistivity.