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# RESISTIVITY OF A MATERIAL

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Introduction

IB Physics-The Investigation

Data Set: Thursday 15th Nov 2007

Data Due: Thursday 22nd Nov 2007

RESISTIVITY OF A MATERIAL

ALBERT.LIU

11B

Aim:

The aim of the investigation is to determine the resistivity of a material by changing the length/ diameter of the conductor and compare/ evaluate the results obtained.

Theory:

The resistance of a conductor depends on four main factors:

• Length
• Cross-sectional area
• Resistivity
• Temperature

It can be shown that when temperature is kept constant Where R is the resistance in , is the resistivity in m is the length of the conductor in m, and A is the cross-sectional area of the conductor in .

The resistivity is specific to the type of material being used as resistance is affected by the nature of delocalised electrons and the positive ions within the material. The resistivity of a material is the resistance across opposite faces of a cube with sides of one metre.

Circuit diagram: Data Collection:

The data is which we collected at the experiment.

Table 1:

 Length (m) Diameter(mm)±0.01mm Current (A)±0.01A Voltage (V)±0.01V 2.5 0.26 1.98 1.84 0.28 1.93 2.26 0.61 4.19 1.02 0.63 4.22 1.00 0.89 5.20 0.83

Table 2:

 Diameter (mm) Length (m) Current (A)±0.01A Voltage (V)±0.01V 0.61 2.0 4.45 0.9 2.5 4.40 1.0 3.5 4.10 1.22 4.0 4.00 1.38

Middle

1. Set the range of the voltmeter in “0-20V”, and set the range of the ammeter in “0-20A”
2. Connect wire number 1 in the circuit.
3. Record the current and potential difference across the wire.
4. Record the values for current and voltage in the table.
5. Repeat step number 4 and 6 until finish with all wires.

Data processing and conduction:

Form the formula I= , we can know R= , so when we calculate the resistance, just put the values of Voltage and Current into the formula. The way to calculate the area of wires’ cross section is π( )2.

Table 3:

 Length (m) Diameter(mm)±0.01mm Current (A)±0.01A Voltage (V)±0.01V Resistivity (Ω)±0.01Ω Area(,3.m.f) × 2.5 0.26 1.98 1.84 0.93 2.65 0.28 1.93 2.26 1.17 3.08 0.61 4.19 1.02 0.24 14.6 0.63 4.22 1.00 0.24 15.6 0.89 5.20 0.83 0.16 31.1

Because of , and the length is constant.

Conclusion

Ways to decrease percentage error:

• When using the copper wire as a conductor for this experiment to find out the resistivity, make sure that there is no rust on the wire, the rust can affect the current pass it. So, before use the copper wire. We must use sand paper to rub two sizes of the wires. This is to make sure a better and a more accurate reading.
• We must make sure all the wire connected completely, if not, the figure of the meter will always change.

Conclusion:

From the experiment we can see: the first experiment is when the length is constant, the experiment of that error is 70.6%. The figure is much bigger that normal. So I think is the errors I shown above happened, so the experiment is so much. The second experiment is when the diameter is constant, the experiment of that error is 23.5%, the figure is not so difference as normal, but still have experiment error, I think why this like this is: the room temperature is changed, so the resistance is change. And the other wires also have resistance, but we couldn’t calculate it.

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