• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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.

Extracts from this document...

Introduction

Wires Coursework for GCSE.

Area P-Planning

image39.pngimage46.pngimage20.pngimage28.pngimage10.pngimage16.pngimage00.pngimage01.png

image50.pngimage03.pngimage51.pngimage02.png

image04.pngimage04.png

image05.pngimage06.pngimage05.pngimage07.png

image11.pngimage09.pngimage08.png

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

...read more.

Middle

image13.pngimage13.pngimage13.pngimage18.pngimage15.png

image21.pngimage13.pngimage13.png

 I would expect a graph to look like this:

image27.pngimage26.pngimage25.pngimage24.pngimage23.pngimage22.pngimage29.png


Area O- Obtaining results

Wire

Diameter (mm)

Cross sectional area (mm)

Voltage (V)

...read more.

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.

...read more.

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

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Electricity and Magnetism essays

  1. Investigate the resistance of different wires and how at different lengths the voltage increases ...

    fell outside the line of best fit as they would have suffered the affects of the heating affect but My tertiary set of data (Focus Educational Software) was 100% accurate. There were no anomalous results, I think this is because the software was done really accurately as they had the right equipment to prevent the heating affect.

  2. relationship between voltage and current

    two ends of a wire-the negatively charged electrons will flow toward the positively charged end of the wire, creating electric current. Direct current (DC) is the flow of electricity in one direction. Alternating current (AC) intermittently reverses direction because of the way it is generated.

  1. Length vs Resistance

    by the ammeter being read wrongly since the current value in the first table for 40cm with the power pack set to 2V, was much lower than the other values. So I had to test the two other thicknesses of wire, 0.25 mm and 0.31 mm with the results in the table below.

  2. Determining Voltage, Resistance and Current in a Parallel, Series and Series-Parallel Circuit.

    3.80 = 7.70 ? Power Supply Voltage = 6.12V Voltage of Bulb 1 = 3.00V Voltage of Bulb 2 = 2.95V ? Total Voltage of the Bulbs = 3.00 + 2.95 = 5.95V Calculating the Current of Bulb 1 and bulb 2 Using Ohm?s Law, given V = I x

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work