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The silicon chip has been the cornerstone of the IT revolution for the past several decades. Over the years we have seen the size of devices such as microprocessors shrink and their speed increase considerably.

Extracts from this document...

Introduction

Tarique Sabah        Physics Coursework        

Task

The silicon chip has been the cornerstone of the IT revolution for the past several decades. Over the years we have seen the size of devices such as microprocessors shrink and their speed increase considerably. However to enable miniaturisation to proceed as it has electronic device engineers have had to integrate discrete devices such as resistors, capacitors and inductors into the chips themselves. To achieve this they have had to make these components essentially out of the silicon itself.

To enable them to do so they have had to understand what factors affect, for example, the resistance of the material that they used. In a similar manner, see if you can discover what factors affect the resistance of a metal wire in the laboratory.

Introduction

To understand resistance and be able to calculate it many things must first be understood, such as: conduction in solids, current, charge, potential difference and how they connected to finding resistance.

For conduction to take place there must be a supply of charge carriers in the solid and be able to move freely inside it. It is believed, in solids that the carriers of the supply is loosely held electrons (This experiment will be based on this theory).

Current is the amount of movement of negatively charged electrons. An ammeter measures current. Current is measured in amps (A).

Potential difference is the amount of electrical energy transformed into other forms of energy when unit charge passes from one point to the other. A Voltmeter measures potential difference. The unit of potential difference is Volts (V).

Resistance is the opposition to the current. The freely moving electrons that carry the charge have to move through the metal in-between the gaps between the ions.

...read more.

Middle

The shorter the resistor the less times the electrons is knocked off course, which lowers the resistance because the electrons pass through the resistor quicker.

The longer the resistor the more times the electrons are likely to be knocked off course which increases the time the electrons flow through the resistor and thus increases the resistance.

We can also put into context that if you double the amount of ions, by doubling the length of the wire, then you double the resistance because the number of obstacles (ions) is doubled. If you go to the next extreme, if you triple the length of wire then the resistance is tripled. From this theory we can derive:

Resistance is directly proportional to Length.

R α L

Apparatus

  • 1.5V Batteries x 2
  • Electronic Ammeter x 2
  • Electronic Voltmeter x 2
  • 1m ruler x 1
  • 1m Constantan wire (0.02 diameter)
  • Sellotape
  • Electrical leads x 6
  • Crocodile clips x 2

Diagram

Circuit Diagram

Method

  1. Set-up apparatus as shown above in the diagrams. There must be a gap in the circuit so it can be turned on and off.
  1. The Constantan wire show be sellotaped taut again the ruler and that the wire spans the whole length of the ruler.
  1. Connect one crocodile clip on the wire at the 0cm mark on the ruler.
  1. Connect the second crocodile clip at wire on the 100cm mark.
  1. Connect the wires to the 2 x batteries so that a full circuit i.e. a current flow is flowing.
  1. Record the current using the ammeter.
  1. Record the potential difference across the wire using the voltmeter.
  1. Disconnect the wires from the batteries as quickly as possible so that the circuit is broken.
  1. Then repeat points 4-8 but this time change the length of the wire by moving the second crocodile clip (at point 4)
...read more.

Conclusion

Therefore, R = x L/A

Where ‘x’ is a constant called the resistance of the material (for a fixed temperature and other physical conditions).

“The resistance of a material is numerically the resistance of a sample of unit length and unit cross-section area, at a certain temperature.”

To find ‘x’ we can rearrange the equation “R = x L/A” to get “x = AR/L”. Thus, to find the resistance (x) of the Constantan wire used in this experiment we must substitute for A, R and L in the equation x = AR/L.

  • The wire being used in the investigation should have a uniform cross-sectional area, but, to confirm this, the diameter of the wire can be measured using a micrometer. In this investigation the diameter of the wire was 0.02mm and so the cross-sectional area of the wire can be estimated, by assuming the wire’s cross-section is circular, using the equation:

Cross-sectional area = πr2

 Where ‘r’ is the radius of the circular cross-sectional area, which is half of the diameter

  • Other ways to further the experiment would be to use wires made from different materials to find differences in resistance that each wire produced. It could then be decided which of the wires was the best conductor.

Cross-sectional area could also be investigated, if the experiment was furthered, and it could be investigated whether the resistance of a wire is inversely proportional to its cross-sectional area. To investigate the effect of cross-sectional area on resistance of similar wires (i.e. wires of the same length, material, etc.) with different cross-sectional areas will be used.

  • The effect of temperature on a wire could also be investigated.

I believe that the experiment was performed successfully and that the results obtained were accurate.

The predictions that were made were also confirmed by the results and the wire obeyed the rules that it was expected to.This experiment we can confirm that the resistance of a wire is directly proportional to the wire’s length.

...read more.

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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