# What can affect the resistance of nichrome wire?

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Introduction

Investigating Resistance – What Can Affect the Resistance of Nichrome wire?

Background Knowledge: Metals conduct electricity because they have free electrons. This means that they are able to move around the positive ions like a sea. The positive ions are unable to move. This “sea” of electrons is moved by the force of the voltage pushing it around the circuit. The electrons move from negative to positive, the opposing current moves from positive to negative.

The formula for resistance is described as –

Resistance = Voltage / Current

Or

R = V / I (R=resistance

V=voltage

I=current)

I will use this formula to work out the resistance of my wire each time. I will now need to select a variable.

The resistance of a metal is low, but not zero. A voltage difference is still required to generate current in the metal and the metal heats up when the current is flowing. This increase in temperature is one four possible variables that I have come up with…

Possible Variables: WIDTH

LENGTH

TEMPERATURE

TYPE OF WIRE

## Independent: THE WIDTH OF THE NICHROME WIRE

(input variable)

Dependent (outcome variable): The resistance will be found by taking an Ammeter and then take one from a Voltmeter. The resistance is found by dividing the voltage by the current ➔

R = V / I

Controlled: I will keep everything the same except the width of the wire.

Middle

With my results, I will now form a series of tables which will lead to the drawing of numerous graphs of analysis. I will plot one of the tables in a graph of diameter of wire against average resistance. If the graph is not a straight line, I will try 1 / diameter of wire against average resistance. If this graph fails, my last attempt will be 1 / diameter of wire squared against average resistance.

Darren Cave 12B Page 5/2/2007

My first set of results are shown in the table below –

WIRE WIDTH (mm) | CURRENT (amps) | VOLTAGE (volts) | ## RESISTANCE(ohms) |

0.32 | 0.73 | 2 | 2.73 |

0.37 | 1.14 | 2 | 1.75 |

0.45 | 1.87 | 2 | 1.07 |

0.56 | 2.67 | 2 | 0.75 |

0.71 | 5.00 | 2 | 0.40 |

The second set of results that I recorded are shown in the table below –

WIRE WIDTH (mm) | CURRENT (amps) | VOLTAGE (volts) | ## RESISTANCE(ohms) |

0.32 | 0.73 | 2 | 2.73 |

0.37 | 1.12 | 2 | 1.79 |

0.45 | 1.64 | 2 | 1.22 |

0.56 | 2.74 | 2 | 0.73 |

0.71 | 5.00 | 2 | 0.40 |

The not essential third set of results are shown in this table (dashes show where I felt a repetition was not required) –

WIRE WIDTH (mm) | CURRENT (amps) | VOLTAGE (volts) | ## RESISTANCE(ohms) |

0.32 | - | - | - |

0.37 | - | - | - |

0.45 | 1.74 | 2 | 1.15 |

0.56 | - | - | - |

0.71 | - | - | - |

Conclusion

My three graphs were ➔ diameter of wire against average resistance

1 / diameter of wire against average resistance

1 / diameter of wire squared against average resistance

In the first two graphs I was unable to find a straight line or a connection between the two factors, they made curves. The third graph gave a straight line through the origin and found the connection for me. The resistance is directly proportional to 1 / the diameter squared.

As I found a connection between the variable and the resistance, I consider this experiment to be a success so there is little I would change if I had another chance. The one problem I had was when I was trying to keep the wire 10cm in length. It was hard to get the wire completely straight so it could be measured accurately. Fortunately, this problem was not reflected in my results so I feel that the wires were measured accurately.

The aim of my coursework was to determine a formula linking the diameter of nichrome wire and the resistance of nichrome wire. So to conclude, as I successfully completed my objective, I consider the experiment to be a complete success.

Darren Cave 12B Page 5/2/2007

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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