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# Describe how you would investigate the properties of a piece of wire so as to enable an electrician to make a resistor of the required value.

Extracts from this document...

Introduction

Physics Coursework

The Problem

In catalogues such as R.S components, Malpin, Cpc etc resistors are listed with certain fixed values, e.g.) 47Ώ.

While repairing an old piece of electrical/electronic equipment an electrician discovered that he needed to replace a resistor with a value not listed in any of the catalogues. Describe how you would investigate the properties of a piece of wire so as to enable him to make a resistor of the required value.

Background Knowledge.

To carry out an investigation dealing with resistance and electricity, we firstly need to know about what resistance is. The resistance is the opposition of a conductor to current. It occurs when electrons travelling along the wire collide with the ions in the wire. It could also be described as a measure of how hard it is to move the electrons/ions through a wire.

The current is the rate of the flow of charge and the voltage is the energy transferred per unit charge.

To measure the resistance you need to know the current flowing through a circuit and also the voltage. These can be measured by using an ammeter for current and a voltmeter for voltage.

The relationship between voltage, current and resistance is known as Ohms law. Resistance is measured in Ohms.

Ohms law states that the current through a metallic conductor is directly proportional to the voltage across it providing its temperature and physical conditions remain the same.

The equation connecting resistance to voltage and current is V=IR, where V is voltage, I is current and R is resistance. Conductors intended to have resistance is just the sum of all the resistances.

In a parallel circuit the total resistance is always less than the branch with the smallest resistance. The equation for this is R=R1+R2

Middle

Ohms law states that the current through a metallic conductor is directly proportional to the voltage across it providing its temperature and physical conditions remain the same.

The equation connecting resistance to voltage and current is V=IR, where V is voltage, I is current and R is resistance. Conductors intended to have resistance is just the sum of all the resistances.

In a parallel circuit the total resistance is always less than the branch with the smallest resistance. The equation for this is R=R1+R2+R3.

Another equation which is also useful is the equation for Resistivity. This is:

Resistance also depends on temperature, because if the wire is heated up the ions in the wire will start to vibrate because of their energy increase. This causes more collisions between the electrons and the atoms as the atoms are moving into the path of the electrons. This increase in collisions means that there will be an increase in resistance.

In planning for this experiment I came across yet another equation which may be useful. This is the equation for Drift Velocity:

I=nAeV

Where:

n= number of electrons per m3

A=cross sectional area

E=charge on the electron (-1.6×10-19)

V=drift velocity.

I also know that as the drift velocity decreases the current decreases and so the resistance decreases. This happens with a longer wire.

For this experiment the independent variable is the CSA (cross sectional area). The dependant variable is the resistance and the controlled variables are the temperature, material and the length.

My prediction for this is that as the area increases the resistance will decrease. Therefore I predict that the resistance will be inversely proportional to the CSA.

Conclusion

I can also see from my results that as the area doubles the resistance decreases by roughly ½.

As the Resistance is inversely proportional to the Area:

R    1/A

R=k×1/A

K=PL

As the length is always 1m:

K=P

So again the constant is the Resistivity.

We can go on to form an equation for Resistance:

Resistance=area × 1/K

I can also determine the resistance from the graph of resistance against 1/area.

Resistance =Resistivity x length

CSA

Resistance = Resistivity x length x 1/CSA

The length is constant; 1m

Therefore the gradient of the line = y2-y1             4.3-2.1        x 10-5

X2-x1             81-37.7

Also Ώm2       = Ώm

M2

The resistance is 5x 10-7m.

I think that this was the best way of doing this experiment because if we had of done it manually we could have made mistakes very easily for example reading the results wrongly. It would have also taken far longer. You could also have made a mistake drawing the graph or calculating the gradient wrongly and it’s also hard to know where to draw the line of best fit.

The way we did the investigation using the computer eliminated many of these problems. Although there is still a chance of making a mistake.

However the computer method is very accurate.

If I were to do the experiment again I would use more wires. In this experiment I would use more than 6 to get a clearer result. I would also use thicker or thinner wires and increase the range of SWG of the wires.

I could also take more values of voltage and current which may perhaps add more values to the graph making it even more accurate and hence the gradient i.e. the resistance.

From doing this experiment I am confident that the graph you would plot had you done the experiment correctly would prove that the resistance is inversely proportional to the CSA.

This student written piece of work is one of many that can be found in our GCSE Electricity and Magnetism section.

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