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We are going to set up the diagram above and measure the resistance across the wire by measuring the current and voltage then working out the resistance using R=V/I.

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Introduction

Wires Coursework for GCSE.

Area P-Planning

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We are going to set up the diagram above and measure the resistance across the wire by measuring the current and voltage then working out the resistance using R=V/I.  The variables that will change are the resistance setting on the protective resistor and the cross sectional area of the wire-we will change the thickness of the wire and see how the resistance is affected.  When the protective resistance is changed the current and voltage will change but the resistance should remain the same.  The thicker the wire, the less resistance.

        I think the things that might affect the resistance of the wire are the area, the number of items (eg bulbs), the brightness or power needed to operate

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Middle

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 I would expect a graph to look like this:

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Area O- Obtaining results

Wire

Diameter (mm)

Cross sectional area (mm)

Voltage (V)

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Conclusion

        In my prediction, I predicted that as the resistance went up, the area went down, and this was true so my prediction was correct.

This pattern is really quite logical.  The larger the wire, the more charge can pass through.  I know already that the larger the item (in this case the wire) the smaller the resistance.image45.pngimage38.pngimage40.pngimage47.pngimage41.pngimage48.pngimage38.pngimage48.pngimage40.pngimage48.pngimage43.pngimage44.pngimage49.pngimage42.pngimage45.pngimage45.png

So, therefore:

Resistance   is inversely proportional to  Area

Or:

Resistance  is   proportional to   1/Area


Area E- Evaluation

It was fairly easy to make a conclusion, since every part of our experiment was made accurate and precise.  Therefore, the graph showed a clear line of best fit, into which most of the data fitted.

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