# An Investigation to Calculate the Resistivity of a Piece of Wire.

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Introduction

James Rollinson12 March 2003

## An Investigation to Calculate the Resistivity of a Piece of Wire.

## Aim

To calculate the resistivity of a piece of wire by passing an electrical current through it. I will be using the formula-

R = ρ L/A

Where-

R = Resistance, measured in Ohms (Ω)

L = Length of wire, measured in Meters (M)

A = Cross-sectional area of the wire measured in (M2)

ρ= Resistivity, which is a constant for the material.

## Background Knowledge

There are four main factors that affect the resistance of a piece of wire, temperature, material of the wire, cross-sectional area of the wire and length.

The flow of electricity needs a flow of electrons through a material.

+++++++++++++++++++++++++++++++

Factors-

Temperature – if a piece of metal wire is heated up then the atoms in the wire start to vibrate more because of the increase I energy they have, this increases resistance because it makes it harder for the electrons to flow through the material, this also increases collisions between the atoms in the metal and the electrons making it harder for them to flow.

Wire Material/Density – If the material is denser then the atoms are more closely packed together, this means that the resistance will be higher because it will be harder for the electrons to flow through the material.

Middle

Wire = 0.315

Measures = …………..

Power pack = 2x 1.5v cells

Safety Precautions

- Keep a low voltage on the power pack (about 3v) as this will stop the wire from heating up, also don’t touch the wire when the power is connected as it may be hot.
- Be careful when cutting and handling the wire.

Results

Table of results collected from experiment-

Length of wire (m) | Circuit Current at A (Amps) | Circuit Voltage at V (Volts) | Calculated Average Resistance of Wire (Ohms) |

20 | 1.51 | 1.97 | 1.305 |

30 | 1.15 | 2.19 | 1.904 |

40 | 0.91 | 2.33 | 2.560 |

50 | 0.76 | 2.42 | 3.184 |

55 | 0.71 | 2.47 | 3.479 |

60 | 0.66 | 2.51 | 3.803 |

65 | 0.61 | 2.51 | 4.114 |

70 | 0.57 | 2.54 | 4.456 |

75 | 0.54 | 2.55 | 4.722 |

80 | 0.51 | 2.58 | 5.059 |

85 | 0.48 | 2.59 | 5.396 |

90 | 0.47 | 2.62 | 5.574 |

95 | 0.43 | 2.62 | 6.093 |

Tables of results to show highest and lowest values for Resistance when taking into account percentage errors-

Lower Values – using highest possible values for current and lowest possible values of voltage, within the percentage error for the apparatus, with the formula-

R (Lowest possible value) = V (value + 0.01)

I (value – 0.01)

Length of wire (m) | Upper Value for Current (Amps) | Lower Value for Voltage (Volts) | Lowest Possible Values for Resistance (Ohms) |

20 | 1.52 | 1.96 | 1.289 |

30 | 1.16 | 2.18 | 1.879 |

40 | 0.92 | 2.32 | 2.522 |

50 | 0.77 | 2.41 | 3.130 |

55 | 0.72 | 2.46 | 3.417 |

60 | 0.67 | 2.50 | 3.731 |

65 | 0.62 | 2.50 | 4.032 |

70 | 0.58 | 2.53 | 4.362 |

75 | 0.55 | 2.54 | 4.619 |

80 | 0.52 | 2.57 | 4.942 |

85 | 0.49 | 2.58 | 5.265 |

90 | 0.48 | 2.61 | 5.438 |

95 | 0.44 | 2.61 | 5.932 |

Upper Values

Conclusion

(Ohms)

Highest Values for Resistance

(Ohms)

Difference in highest and lowest

(Ohms)

20

1.305

1.289

1.320

0.031

30

1.904

1.879

1.930

0.051

40

2.560

2.522

2.600

0.078

50

3.184

3.130

3.240

0.110

55

3.479

3.417

3.543

0.126

60

3.803

3.731

3.877

0.146

65

4.114

4.032

4.200

0.168

70

4.456

4.362

4.554

0.192

75

4.722

4.619

4.830

0.211

80

5.059

4.942

5.180

0.238

85

5.396

5.265

5.532

0.267

90

5.574

5.438

5.717

0.279

95

6.093

5.932

6.262

0.330

The table shows that the difference between the high and low calculations for resistance increases as the

Error Bars

I will be able to plot a graph showing the error in reading on the instruments, the difference between lowest and highest values will be the height of the error bars on the graph and the average will be the mid-point for the error-bar the width of the error bar will be much simpler, as this will just be the error reading the meter ruler

Graph Gradients

Upper Gradient =

Mid Gradient =

Lower Gradient =

Cross-sectional Area of Test-wire

I measure the wire in five different places with a micrometer along the 1m section I was using. I will then be able to calculate the cross-sectional area of the wire using the formula-

Area = π r2

As my measurement of the wires diameter is in mm I must divide the answer by 1000 to get metres.

Results-

Measurement of Diameter (mm) | Average Diameter (mm) |

0.318 | 0.3176 |

0.316 | |

0.318 | |

0.317 | |

0.319 |

Radius of test wire = 0.318 +/- 0.01 mm (to 4 s.f.)

Upper Cross-Sectional Area-

π × ((0.328/2)/1000)2 = 8.44×10-8m2 (to 3 s.f.)

Lower Cross-Sectional Area

π × ((0.308/2)/1000)2 = 7.44×10-8m2 (to 3 s.f.)

Actual Cross-Sectional Area = Upper Value + Lower Value / 2

Actual Cross-Sectional Area = 7.94×10-8m2 +/- 0.5×10-8m2 (to 3 s.f.)

Resistivity

Upper Value = ρ = m (upper value) × A (Upper value)

Upper Value = ? × 8.44×10-8 =

Lower Value = ρ = m (lower value) × A (lower value)

Lower Value = ? × 7.44×10-8 =

Actual Resistivity = Lower Value + Upper Value × 2

Actual Resistivity =

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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