# Effect of changing the temperature on the resistance of a thermistor

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Introduction

Name: Ray Li

Physics Coursework

Aim: Effect of changing the temperature on the resistance of a thermistor

I am going to find out has the temperature effects the thermistor’ resistance in the circuit.

Planning

Method:

- Clear the table and take off the needless thing
- Use the wires to link all the equipment to complete the circuits.
- Put about 25cm3 water in the beaker.
- Turn on the gas to boiling the water.
- Look at the thermometer and record the p.d and current each 5 C
- Repeat this experiment.
- Put all data into table.
- Calculation the resistance and draw the graphs.

The equation is: Resistance(R) = Potential different (p.d.)

Current (I)

Appliance list:

Wires- Use these wires to link all appliances to complete the circuit.

Power Supply- Use these supply the energy into circuit.

Thermistor- I use this to change the resistance in the circuit.

Voltmeter- Measuring the volt in the circuit across the thermistor.

Ammeter- Measuring the current in the circuit through the thermistor.

Beaker- Have thermistor and water made above at same temperature.

Bunsen burner- I am boil the water and increasing the temperature of thermistor.

Thermometer- Measuring the temperature of the thermistor.

Middle

45℃

0.0017

1.2

45℃

0.002

1.6

50℃

0.0018

50℃

0.0022

1.2

55℃

0.0021

1.2

55℃

0.0024

1.2

60℃

0.0022

1

60℃

0.00245

1.1

65℃

0.0023

1

65℃

0.0026

1

70℃

0.0025

0.95

70℃

0.0028

1

75℃

0.0025

0.9

75℃

0.003

0.9

80℃

0.0028

0.8

80℃

0.00325

0.9

85℃

0.003

0.8

85℃

0.0033

0.8

90℃

0.00315

0.75

90℃

0.0035

0.8

95℃

0.0033

0.75

95℃

0.0037

0.8

100℃

0.0034

0.75

100℃

0.0039

0.75

There have list of equation:

I will use this equation find out the resistance

Resistance(R) = Potential different (p.d.)

Current (I)

I use equation find out the average of resistance:

(Resistance1 + Resistance 2)÷2=Average of resistance.

Conclusion

Slope A:

(1000-570) / (55-35) = 21.5

The gradient is 21.5

Slope B:

(285.7-220.6) / (100 -80) = 4

The gradient is 4

In the second experiment graph:

Slope A:

(997 – 500) / (55 – 35) = 23.5

The gradient is 23.5

Slope B:

(242.4-192.3)/ (55-35) = 4.23

The gradient is 4.23

In the average of resistance

Slope A:

(970.55-535)/ (55-35) = 21.78

The gradient is 21.78

Slope B:

(280-206.45)/ (55-35) =3.68

The gradient is 3.68

The average gradient | |

Slope A gradient | 21.78 |

Slope B gradient | 3.68 |

In the graph if the gradient is large, there have large current flow in and the low volt, the resistance will high decrease. Also the gradient is small, the resistance will decrease is low.

All the graphs in 35℃ to 55℃ decrease very faster, but the end 80℃to 100℃ is decrease quite slow. This is same with my prediction graph.

Evaluation

In my experiment I should repeat few more times to have more data. This could make the result fair.

In my graphs have some points did not on the best-fit line. I could have some errors in my experiment. It could be read the voltmeter or ammeter not ready correct have some error. I should read quickly. And than the thermometer is not ready measuring the thermistor. Because the thermometer did not close the thermistor. So there have a little error to change the result. So I should use the stand to keep thermometer close the thermistor.

This student written piece of work is one of many that can be found in our AS and A Level Electrical & Thermal Physics section.

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