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investigating the relationship between the diameter and the current in a wire at its melting point

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Introduction

Investigation Report

Aim –

Theory –

When current is passed through a wire resistance is created. This resistance is depends on the resistivity of the wire, current, voltage, wire length and cross-sectional area.

So…

R = ρ . l

        A

R = Resistance.

Ρ = Resistivity.

L = Length.

A = Cross sectional area.

(Pg. 200, Accessible Physics, 1995, MacMillan Press ,ISBN  0333-627-80-6)

(Pg. 47, Advanced Physics Revision Handbook, 1996, Oxford University Press, ISBN 0-19-914640-3)

Looking for a formula for resistance -  

R = ρ . l        ( Resistance is the product of resistivity and length

         A                       per unit cross sectional area.)

R = Resistance.

Ρ = Resistivity.

L = Length.

A = Cross sectional area.

(http://physics.bu.edu/~duffy/PY106/Resistance.html)

Looking for a formula for voltage –

V = I . R  (Voltage is the product of current and resistance)

V = Voltage

I = Current

R = Resistance

(http://www.allaboutcircuits.com/vol_1/chpt_2/1.html)

Looking for a formula for Cross sectional area –

A = Πr2   (Cross sectional area is the product of pie and radius                                      

                                                squared)

Or. A = Π image03.png2   ( Cross sectional area is the product of pie and

                                        diameter per unit 2 all squared)

A = Cross sectional area

Π = Pie

D = Diameter

(http://www.equationsheet.com/sheets/Equations-20.html)

So putting these formula together-

A = Π image03.png2     And        R = ρ . l

                                                 A        

R =  ρ . l

     Π (d/2)2

R = ρ . l

     Π . d2/4

R = 4 . ρ . l

        Π . d2

R = 4 . ρ . l   .     1

            Π            d2

R           =            4 . ρ . l           .            1                     + 0

                                Π                           d2image00.pngimage00.png

image01.png

          Y          =             m                       x                 +c

References to the specification –

Aim Of My Investigation –

...read more.

Middle

By using a digital ammeter and voltmeter the uncertainties in the measurement will be kept as low as possible. In these cases both uncertainties are about one percent. If the resistance alone was measured the uncertainty would be a lot higher and therefore the results would be of decreased effectiveness.

Determining the additional resistance:

An additional resistance of three ohms will be added to the circuit. This will lower the current to around one amp which will be enough to protect the ammeter and prevent any major heating in the wire that could, potentially, change the resistance. Since it was found that a thin nicrome wire with two amps passing through it was hot enough to glow red, which would affect the resistance significantly, the current needs to be kept to around one amp. This should still provide a good reading on both the ammeter and the voltmeter while removing the problems faced by temperature on all the test wires.

‘The resistance of a metal increases with an increase of temperature’

(Pg. 87 ‘Essential AS Physics for OCR’ By Jim Breithaupt, Nelson Thornes, ISBN 0 7487 8507 8)

image02.png

Ohm meter (Ω) – o-200Ω

Constantan Wire – 30 swg.

Resistance – 3.2 Ωhms.

Some after heat change – Temperature changes of less than 10 oc (i.e. changes in room temperature) have little or no effect on resistance.

Determining the wires material:

...read more.

Conclusion

This can be accounted for by reference to the % uncertainty in my voltage, current, resistance and diameter.

Sources of error –

In my experiment there where several sources of error. Firstly, the diameter of a thinner wire would be more accurate than that of a thicker wire. This is because when they are made they have to be made to a certain percentage accuracy of there thickness, thus a thicker wire will have a bigger margin if error in production. I could improve the accuracy of the diameter of the wire by checking it myself with a micrometer.

Secondly, when measuring out the metre length of wire, the thinner wire will be a more accurate length. This is because the thicker wire is full of kinks and thus is almost impossible to get perfectly straight. One way I could improve the accuracy of measuring the wire is by using sellotape to hold the wire to the meter rule while I measured the wire.

Thirdly, the voltage and current I recorded in my results is particularly accurate. This is because the ammeter and voltmeter is not very accurate. The only way to improve these readings is to use a more accurate ammeter and voltmeter.

Fourthly, I only did one set of results for my experiment. This means I could not take mean results. If I was doing this experiment again I would take multiply results and take a mean of these results.

...read more.

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