Exercise EE1-04.04 – Temperature coefficient measurement (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-26)

Relates to EE1-04 Resistance

Task

During the quality control of a wire made of an unknown material, two resistance measurements are carried out at different temperatures. The first measurement at $T_1=25°C$ gives a resistance of $R_1=510\Omega$, the second measurement at $T_2=60°C$ a resistance of $R_2=580\Omega$.

a) Calculate the temperature coefficient ${\alpha }_{20}$ of the wire material.

b) Determine the resistance $R_{20}$ at the reference temperature of $20°C$.

c) Identify the material of the wire from the temperature coefficient you have determined.

Solution

a) $4.0\cdot 10^{-3}{\mathrm{K}}^{-1}$
b) $500\mathrm{\Omega }$
c) Aluminium

▶️ Full worked solution

◀️ Back to the task

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Solution

Given

Given explicitly:

• Temperature of the first measurement: $T_1=25°C$
• Resistance at the first measurement: $R_1=510\Omega$
• Temperature of the second measurement: $T_2=60°C$
• Resistance at the second measurement: $R_2=580\Omega$

Known:

• Temperature dependence of the resistance: $R(T)=R_{20}(1+{\alpha }_{20}\cdot \Delta T)$
• Reference temperature: $T_0=20°C$

Required

a) Temperature coefficient ${\alpha }_{20}$ in ${\mathrm{K}}^{-1}$
b) Resistance $R_{20}$ at $20°C$ in $\Omega$
c) Identification of the material

a) Temperature coefficient

Setting up the equations for both measuring points:

$$R_1=R_{20}(1+{\alpha }_{20}\cdot (T_1-20°C))=R_{20}(1+{\alpha }_{20}\cdot 5\mathrm{K})$$ $$R_2=R_{20}(1+{\alpha }_{20}\cdot (T_2-20°C))=R_{20}(1+{\alpha }_{20}\cdot 40\mathrm{K})$$

Dividing the second equation by the first:

$$\frac{R_2}{R_1}=\frac{1+{\alpha }_{20}\cdot 40\mathrm{K}}{1+{\alpha }_{20}\cdot 5\mathrm{K}}$$

Inserting the measured values:

$$\frac{580\Omega }{510\Omega }=\frac{1+{\alpha }_{20}\cdot 40\mathrm{K}}{1+{\alpha }_{20}\cdot 5\mathrm{K}}$$ $$1.1373=\frac{1+{\alpha }_{20}\cdot 40\mathrm{K}}{1+{\alpha }_{20}\cdot 5\mathrm{K}}$$

Rearranging by multiplying out:

$$1.1373\cdot (1+{\alpha }_{20}\cdot 5\mathrm{K})=1+{\alpha }_{20}\cdot 40\mathrm{K}$$ $$1.1373+5.686\cdot {\alpha }_{20}\mathrm{K}=1+40\cdot {\alpha }_{20}\mathrm{K}$$ $$0.1373=(40-5.686)\cdot {\alpha }_{20}\mathrm{K}$$ $$0.1373=34.314\cdot {\alpha }_{20}\mathrm{K}$$ $${\alpha }_{20}=\frac{0.1373}{34.314\mathrm{K}}=0.00400{\mathrm{K}}^{-1}=\underline{4.0\cdot 10^{-3}{\mathrm{K}}^{-1}}$$

b) Resistance at 20 °C

From the equation of the first measurement:

$$R_1=R_{20}(1+{\alpha }_{20}\cdot 5\mathrm{K})$$ $$510\Omega =R_{20}(1+0.004{\mathrm{K}}^{-1}\cdot 5\mathrm{K})$$ $$510\Omega =R_{20}\cdot 1.02$$ $$R_{20}=\frac{510,\Omega }{1.02}=\underline{500\Omega }$$

Check with the second measurement:

$$R_2=R_{20}(1+{\alpha }_{20}\cdot 40\mathrm{K})=500\Omega \cdot (1+0.004\cdot 40)=500\Omega \cdot 1.16=580\Omega$$

✓

c) Identification of the material

The temperature coefficient you have determined, ${\alpha }_{20}=4.0\cdot 10^{-3}{\mathrm{K}}^{-1}$, corresponds exactly to the literature value for aluminium.

The wire material is therefore pure aluminium, which is frequently used in electrical engineering as a low-cost conductor, particularly for overhead power lines and in power engineering.

Aluminium as a conductor material After copper, aluminium is the second most important conductor material in electrical engineering:

• Temperature coefficient: ${\alpha }_{20\mathrm{Al}}=4.0\cdot 10^{-3}{\mathrm{K}}^{-1}$
• Resistivity: ${\rho }_{20\mathrm{Al}}=0.0287\frac{\Omega \cdot \mathrm{m}{\mathrm{m}}^2}{\mathrm{m}}$
• Advantages: low weight, low cost
• Disadvantages: higher resistivity than copper, tendency to oxidize

Typical applications: overhead power lines, busbars, cables for the transmission of electricity

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1. Dimension the cross-section so that $\Delta U\le 3\%$ at $U_{\mathrm{rated}}=230\mathrm{V}$, $I=12\mathrm{A}$, $l=30\mathrm{m}$ (Cu).
2. A copper winding has $R_0=12.0\Omega$ at $20^{\circ }\mathrm{C}$. During operation you measure $R=13.2\Omega$. Estimate $T$ (take ${\alpha }_{20}=3.93\cdot 10^{-3}/\mathrm{K}$).
3. Determine the operating point of a linear source ($U_0=9\mathrm{V}$, $R_{\mathrm{i}}=1.0\Omega$) connected to $R_{\mathrm{L}}=10\Omega$ graphically and by calculation. (Hint: internal resistance model + straight lines in the $U$–$I$ diagram.)