At the end of this lesson, the students should be able to:
• Define temperature coefficient of resistance
• Draw the graph of resistance against temperature
• Solve simple problems on temperature coefficient of resistance
2.12 TEMPERATURE COEFFICIENT OF RESISTANCE
If the resistance of a coil of insulated copper wire is measured at various temperatures up to say 200°C, it is found to vary as shown in the figure 10.1 below.
Fig 10.1 Variation of resistance of copper with temperature
The resistance at 0°C been taking as 1 . The resistance increases until it reaches 1 426 for an increase of 100°C in the temperature.
We may define temperature coefficient of resistance as “the increase in resistance per original resistance at 0°C per temperature change”
Mathematically α = R▼1 –R▼o
(10.1) R▼o (θ▼1- θ▼0)
The unit of temperature coefficient of resistance is °C^-1
Graph of resistance against temperature
The graph of resistance against temperature is shown in Fig 10.2. below.
 |
| Fig 10.2 |
θ▼2 – θ▼1 θ▼1 - θ▼o
Temperature coefficient of resistance = slope of resistance/ temperature Resistance
i.e. α1 = R▼2 – R▼1 (10.2)
R▼1 (θ▼2 .. θ▼1)
α▼0 = R▼1 – R▼2 (10.3)
R▼o (θ▼1 – θ▼o)
From (10.3), θ▼o =0
=> α▼0 = R▼1 –R▼2
R▼0 θ▼1
= >α▼0 = R▼o (1 + α θ▼1), where α▼0 =α
where R1=Resistance at temperature θ▼1
R2= Resistance at temperature θ▼2
α = Temperature coefficient of resistance
Example 10.1
A coil has a resistance of 50 at 0°C{0 degree Celsius}, if the temperature coefficient of resistance for the coil is 0.0043 / °C{0.0043/degree Celsius}, determine the resistance of the coil at 25°C{25 degree Celsius}
Solution
R▼θ = R▼o (1+αθ) = 50(1+0.0043X25) =55.38
Example 10.2
The resistance of a coil of a copper wire at the beginning of a heat test is 173 , the temperature being 16°C{16 degree Celsius}. At end of the test, the resistance is 212 . Calculate the temperature raise of the coil. Assuming the temperature coefficient of resistance of copper to be 0.00426/°C{0.00426/degree Celsius} at 0°C{0 degree Celsius}.
Solution
R▼1=173 ,R▼2=212 ,θ▼1=16°C, θ▼2=?,α=0.00426/°C
R▼1=1+αθ▼1
R▼2 1+αθ▼2
θ▼2 = 0.30896= 72.520C
0.00426
Raise in temperature = θ▼2 - θ▼1=72.5-16 = 56.5°C
Example 10.3
A copper wire has a resistance of 85Ω at 15°C. Calculate its resistance at 90°C. Assuming the temperature coefficient resistance of copper to be 0.0043/°C,
Solution
θ▼1 = 150C, θ▼2 = 900, R▼1 = 85Ω and α▼0 = 0.00-43/°C,
→R▼15 = 1 + α▼0 θ▼1 = 1 + 0.0043 X 15 R▼90 1 + α▼0 θ▼2 1 + 0.0043 X 90
i.e. R▼90 = R▼1
1 + 0.0043 X 90 1 + 0.0043 X 15
∴ R▼90 = 110.75Ω
NOTE: This sign "▼" indicate that the next character is a subscript of the previous.
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