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To investigate how the resistance, R, of a length of wire, l, changes with diameter, D and determine the resistivity of the material the wire used.

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A level Physics  

Resistance coursework by Priyesh Patel 12O

Resistance of a Wire


   To investigate how the resistance, R, of a length of wire, l,changes with diameter, D and determine the resistivity ρ of the material the wire used.


   In this experiment, I am investigating the resistance, length and diameter to find the resistivity of a wire. Hence, we use the resistance formula to calculate this:


R is the resistance of the conductor in Ohms (Ω)
A is the cross sectional area in m2
l is the length of the wire in metres (m)
ρ is the resistivity of the material in Ohm metres (Ωm)

   Three external factors influence the resistance in a conductor. Thickness (cross-sectional area of the wire), length, and temperature all have some effect on the amount of resistance created in a wire. The fourth factor is the resistivity of the material we are using. 

The wires which are available for use are:

Constantan (mm)

Nichrome (mm)

Cross sectional area (x 10-8m2) 2.d.p



























   I found that it would be better to use the constantan wire because of the range of diameters, hence a wide range of cross sectional areas of wire are available. The temperature coefficients of the resistance for a constantan and nichrome wire are shown in the table below:


Temperature coefficient of resistance (K-1)




Standard resistors



Heating elements

The resistance of most materials varies with temperature, in a metal, an increase in the temperature causes a greater vibration of the atoms in the fixed framework of atoms, and hence there is more interaction between these vibrating atoms and the flowing electrons. This results in the resistance of the metal rising as the temperature increases.

...read more.


If I double the cross sectional area of the wire, then I will cut the resistance by half. This is because a wide wire would allow a high current to pass through it, while it would be more difficult for the current to flow in a narrow wire due to it’s restriction to a high rate of flow.The resistance would be higher for a thin wire compared to that of a thick wire because of the increase in collisions of electrons and the metal ions.

   I will make my experiment as accurate as I can by using accurate measuring apparatus, such as the micrometer, a digital multi-meter instead of the traditional analogue voltmeter, as the scale divisions are much smaller on the digital meter than on the analogue meters.

   I will try to make sure there are no kinks in the wires as this increases the thickness of the wire. I shall measure the diameter from a number of places on the wire so that I can see that the wire is of one diameter on the length.

   I will have to take account of possible errors, such as the zero error in equipment, and other random and systematic errors, which can occur.

   I will try to avoid making the parallax errors, (the error which occurs when the eye is not placed directly opposite a scale when a reading is being taken). This can be made on reading off a ruler. The reading errors (the error due to the guess work involved in taking a reading from a scale when reading lies between the scale divisions, and the zero error (the error which occurs when a measuring instrument does not indicate zero when it should)

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   So if I were to find the uncertainty in the value for the resistivity, I would have to use the following formula: -

% Uncertainty


(½(max gradient-mingradient))



Gradient of line of best fit

% Uncertainty


(½(50.00 x 10-8Ωm -46.80 x 10-8Ωm))



48.60 x 10-8Ωm

% Uncertainty


1.60 x 10-8Ωm



48.60 x 10-8Ωm







   However, as the gradient on the graph was equal to the resistivity x length, I have to add the error in length to find the possible error in the overall experiment.

Total % Uncertainty


% Uncertainty in resistivity + % Uncertainty in length


3.29% + 0.50%



   Now, I have to calculate the error from this percentage uncertainty. You do this by calculating 3.29% of the gradient of the line of best fit.

3.79% of 48.60 x 10-8Ωm


0.0379 x 48.60 x 10-8


1.84 x 10-8Ωm



(48.60 x 10-8 ± 1.84 x 10-8) Ωm

   This % Uncertainty of 3.79%, tells me that the procedure or method I used is quite accurate, by looking at the size of the possible error of ± 1.84 x 10-8Ωm. However, it also tells me that there is some room for improvement, such as repeating readings for accuracy and taking the average. The largest % Uncertainty was in the current, this being because I had used an analogue ammeter, if I had used a digital ammeter, the % Uncertainty would have reduced to 1.56%.

   Overall, on looking at the actual resistivity, from “Physics For You”, I believe that my results were accurate, but taking the errors into account there is a possible range in my resistivity being between: 46.76 x 10-8 Ωm – 50.44 x 10-8 Ωm. This tells me that the resulting error is a major contribution to the resistivity, as this error can have a huge effect on the resistivity of the constantan wire I used. However, on doing this experiment again, I would know what to look out for, i.e. specific errors, change of equipment, i.e. the steel rule and digital meters.


“Physics For You”


Keith Johnson

“Dictionary of Science”


C. Stockley

Simmone Hewett

C. Oxlade

Sue Holt

J. Weitheim

John Miller

...read more.

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Here's what a teacher thought of this essay

4 star(s)

This is a well structured report.
1. There is a running commentary through this investigation that should be removed.
2. The report needs to have a 'Variables' section that contains information is that spread throughout the report.
3. The report is incomplete and needs to be finished.

Marked by teacher Luke Smithen 05/09/2013

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