• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11

Investigate the effects of resistance on a nichrome wire

Extracts from this document...

Introduction

Physics Coursework : Resistance of a wire

Aim:

To investigate the effects of resistance on a nichrome wire

Planning

My planning will begin with providing a background understanding on the theory of resistance and how it participates in the overall process of the electric current. Reistance is simply a force which challenges the flow current. Resistance is like car racers racing round a track, the more friction there is due to the conditions, and the material of the tyres, the more slower the car will move, similarly an electric current with a more resistant material, causes more resistance thus makes the current move slower. The electrons are like the cars, as they move round the circuit they collide with positive ions and the current slows down creating resistance.

The unit of resistance in an electric current is the ohm, (or known as the greek letter omega). The formula for resistance is as follows:

Resistance ( R ) = Voltage(V) / Current (I)

The investigation is split into 3 experiments, the experiment will be on the length of a nichrome wire, the second experiment will be on the length of a copper wire, and finally the third experiment shall be on the cross sectional area of the nicrome wire.

Equipment

  • Nicrome wire, in 3 different cross sectional areas
  • Copper wire
  • Battery
  • Crocodile clips
  • Ammeter
  • Voltmeter
  • Pliers
  • 1 metre ruler
  • Micrometer

These are the constant variables that I will test in the first experiment

  • The wire will be nichrome and will stay as nichrome
  • The voltage
  • Cross sectional area
  • Room temperature

These values must be kept at a constant, to keep the experiement as a fair test.

...read more.

Middle

Obtaining Evidence

Length of the nicrome wire

This is the table for the results of resistance against the length of the nichrome wire, the cross sectional area was 1.11 x 10-06m2

Length of the wire(cm)

Current1 (amps)

Current2 (amps)

Voltage1 (volts)

Voltage2 (volts)

Resistance1 (ohms)

Resistance2 (ohms)

Average Resistance (ohms)

10

0.92

0.90

0.10

0.13

0.11

0.11

0.11

20

0.87

0.87

0.21

0.20

0.23

0.23

0.23

30

0.85

0.85

0.33

0.23

0.35

0.26

0.30

40

0.82

0.80

0.39

0.36

0.48

0.45

0.47

50

0.80

0.75

0.43

0.37

0.57

0.49

0.49

60

0.75

0.71

0.54

0.49

0.67

0.69

0.68

70

0.72

0.67

0.61

0.43

0.83

0.64

0.75

80

0.71

0.67

0.65

0.51

0.93

0.75

0.84

90

0.69

0.66

0.71

0.57

1.01

0.86

0.93

100

0.65

0.65

0.80

0.60

1.23

0.92

1.08

Results of resistance for the length of a copper wire, the cross sectional area was 3.07 x 10-06 m2

...read more.

Conclusion

2I will need to divide the radius by 1000, then use the area of a circle formula , to find out the cross sectional area.

Therefore:

0.595mm / 1000 = 0.000595 m

π(0.000595)2= 1.11 x 10-06 m2

Cross Sectional Area and wire no.

Current1 (amps)

Current2 (amps)

Voltage1 (volts)

Voltage2 (volts)

Resistance1 (ohms)

Resistance2 (ohms)

Average Resistance (ohms)

Wire 1 1.11 x 10-06 m2

0.80

0.75

0.40

0.37

0.50

0.49

0.49

Wire 2

3.7 x 10-07 m2

0.56

0.58

0.72

07

1.28

1.20

1.24

Wire 3

1.0 x 10-07 m2

0.37

0.37

1.60

1.50

4.32

4.05

2.23

Analysis

These are the Graphs for each table of resistance.

image00.png

image01.png

image02.png

Conclusion

In my conclusion, I conclude that the wire with the least length had the least resistance, and this is evident from the graph. You can see that at 10 cm the resistance was about 0.10 or 0.11 ohms,  also the wire with the longest length had the most resistance which in accordance to graph is at the length of 100 cm and at the resistance of 1.5 or 1.8 ohms . This also applies to the copper wire.

In regard to the cross sectional area, the more bigger the cross sectional area was , the lower the resistance was. From the graph and the tables we can see that at the lowest cross sectional area of 1.0 x 10-07 m2

The resistance was 2.23 ohms.

Evaluation

I think that the experiment was a success, the results were correct in accordance to the background knowledge and also the predictions made at the start

I think the few ways that the experiment could have been improved is a longer wire could have been used, about 2m, also I think that if I were to use different materials the theory would have been more accurate. Checking the wire with the micrometer in 10 different places wouldve also helped. However despite these facts I belive that generally the experiment was successful.

...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 how the length of wire effects its resistance.

    will be using a variable resistor to decrease or increase the amount of current flowing through the circuit. This maybe useful when the current is low to get an accurate reading of the voltmeter and Multimeter, and also if the current is too high.

  2. Resistance of a Wire Investigation

    decreasing the distance from the light source to the plant Output variables - volume of oxygen produced (rate of photosynthesis) is to be measured by finding the volume of oxygen produced in a minute, and thus finding the rate of photosynthesis Control variables -Light wavelength (colour)

  1. An experiment to find the resistivity of nichrome

    To measure the wire width I would use different widths of the same length and same material of wire e.g. thin , medium and thick copper wire with thin and thick constantin wire. To record the difference in widths I would use the same voltage and measure the resistance for each thickness.

  2. To investigate how the length (mm) and the cross-sectional (mm2) area of a wire ...

    Therefore, my predictions will be based on these ideas. 1.9.1. LENGTH: I predict that the resistance will be directly proportional to the length of the wire, i.e. the resistance will increase as the length increases. Since there is no limit to a length of a wire, there would consequently be no limit to the resistance of a wire.

  1. To investigate how the length of a wire effects the resistance of it.

    The electrons carrying the charge try to move through the wire however the because the wire is full of atoms which are constantly colliding and getting in the way, the electrons then have to use up more energy. To ensure that the experiment is as safe as possible I will

  2. "Are rechargeable batteries more economical than alkaline batteries?"

    Cutaway view of a rechargeable battery (Nickel-Cadmium/ Ni-Cad Battery). A rechargeable battery works the same way as normal battery, but in the rechargeable batteries the reactions that take place can be reversed. In chemistry this term is called electrolysis, it is where external electric current is given, to force the reaction to reverse.

  1. Resistance of a nichrome wire.

    This law may be expressed as; Potential difference = constant Current For a given potential difference, a high resistance will pass a small current and a low resistance will pass a large current. There for the value of the constant in the above equation which is high when the current

  2. The resistance of 'nichrome' wire.

    How is it measured? The resistance of a length of wire is calculated by measuring the current present in the circuit (in series) and the voltage across the wire (in parallel). These measurements are then applied to this formula: V = I � R where V = Voltage, I =

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