Exercise EE1-04.03 – Temperature coefficient of an aluminium electric line (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-27)

Relates to EE1-04 Resistance

---

Task

The resistance of an aluminium electric line is $2.20\Omega$ at $12.0°C$.

a) Calculate the resistance of the electric line at the reference temperature of $20°C$.
b) At what temperature has the resistance increased by $6\%$ compared with the value at $12°C$?
c) What resistance would the electric line have at $-10°C$?

You can find the values of the temperature coefficient on the detail page on resistance.

Solution

a) $2.27\Omega$
b) $26.5°C$
c) $2.0\Omega$

▶️ Full worked solution

◀️ Back to the task

---

Solution

Given

Given explicitly:

• Resistance at $T_1=12°C$: $R_1=2.2\Omega$
• Material: aluminium
• Increase in resistance: $6\%$

Known:

• Temperature coefficient of aluminium: ${\alpha }_{20\mathrm{Al}}=4.0\cdot 10^{-3}{\mathrm{K}}^{-1}$
• Temperature dependence: $R(T)=R_{20}(1+{\alpha }_{20}\cdot \Delta T)$
• Reference temperature: $T_0=20°C$

Required

a) Resistance $R_{20}$ at $20°C$ in $\Omega$
b) Temperature $T_2$ for a $6\%$ increase in resistance, in $°C$
c) Resistance $R_{-10}$ at $-10°C$ in $\Omega$

a) Resistance at 20 °C

From the temperature dependence for $T_1=12°C$:

$$R_1=R_{20}(1+{\alpha }_{20,Al}\cdot (T_1-20°C))$$ $$2.2\Omega =R_{20}(1+0.004{\mathrm{K}}^{-1}\cdot (12-20)\mathrm{K})$$ $$2.2\Omega =R_{20}(1+0.004\cdot (-8))$$ $$2.2\Omega =R_{20}\cdot 0.968$$ $$R_{20}=\frac{2.2\Omega }{0.968}=2.273\Omega \approx \underline{2.27\Omega }$$

b) Temperature for a 6 % increase in resistance

The increased resistance:

$$R_2=R_1\cdot 1.06=2.2\Omega \cdot 1.06=2.332\Omega$$

Using the $R_{20}$ determined above:

$$R_2=R_{20}(1+{\alpha }_{20}\cdot (T_2-20°C))$$ $$2.332\Omega =2.273\Omega \cdot (1+0.004{\mathrm{K}}^{-1}\cdot (T_2-20°C))$$ $$\frac{2.332}{2.273}=1+0.004\cdot (T_2-20°C)$$ $$1.026=1+0.004\cdot (T_2-20°C)$$ $$0.026=0.004\cdot (T_2-20°C)$$ $$T_2-20°C=\frac{0.026}{0.004}=6.5\mathrm{K}$$ $$T_2=20°C+6.5\mathrm{K}=\underline{26.5°C}$$

Check: Temperature difference from $12°C$ to $26.5°C$ = $14.5\mathrm{K}$ Expected change in resistance: $\Delta R/R=\alpha \cdot \Delta T\approx 0.004\cdot 14.5=0.058\approx 6\%$ ✓

c) Resistance at -10 °C

$$R_{-10}=R_{20}(1+{\alpha }_{20}\cdot (T-20°C))$$ $$R_{-10}=2.273\Omega \cdot (1+0.004{\mathrm{K}}^{-1}\cdot (-10-20)\mathrm{K})$$ $$R_{-10}=2.273\Omega \cdot (1+0.004\cdot (-30))$$ $$R_{-10}=2.273\Omega \cdot 0.88$$ $$R_{-10}=2.00\Omega \approx \underline{2.0\Omega }$$

At $-10°C$, the resistance drops to $2.0\Omega$.

Temperature dependence in practice

The temperature dependence of resistance has practical consequences:

• Winter operation: At low temperatures, overhead power lines have a lower resistance → lower losses
• Summer operation: At high temperatures, the resistance increases → higher losses and additional heating

In the example, warming from $12°C$ to $26.5°C$ already means $6\%$ more power loss!