Exercise EE1-04.R4 (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-12-30)

Relates to EE1-04 Resistance

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Task

During earthworks in summer ($T=20°C$), a two-core copper telephone line laid in the ground has been damaged. It connects the house connection box to the main distributor at the roadside.

At one end of the electric line (at the main distributor), a resistance of $6.1\Omega$ is measured between the two core ends. At the other end (in the house connection box), a resistance of $11\Omega$ is measured. You measure the diameter of each copper core as $0.7\mathrm{mm}$.

You suspect a short circuit between the two cores and now want to locate the point of damage precisely.

a) Sketch the arrangement with the short circuit and explain which current paths the current flows through in the two measurements.

b) Calculate the distance $l_1$ between the main distributor and the point of damage.

c) Calculate the distance $l_{\mathrm{tot}}$ between the house connection box and the main distributor (single cable length).

For the calculation, assume a conductivity of $56\frac{\mathrm{S}\cdot \mathrm{m}}{{\mathrm{mm}}^2}$ for copper at $20°C$.

Solution

$l_1\approx 66\mathrm{m}$
$l_{\mathrm{tot}}\approx 184\mathrm{m}$

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Solution

Given

Given explicitly:

• Temperature: $T=20°C$
• Resistance measured at the main distributor: $R_1=6.1\Omega$
• Resistance measured at the house connection box: $R_2=11\Omega$
• Core diameter: $d=0.7\mathrm{mm}$
• Material: copper

Known:

• Conductivity of copper (20 °C): ${\sigma }_{\mathrm{Cu}}=56\mathrm{S}\cdot m/m{\mathrm{m}}^2$
• Resistivity: $\rho =\frac{1}{\sigma }$
• Conductor resistance: $R=\rho \frac{\ell }{A}$
• Cross-section of a round conductor: $A=\frac{\pi d^2}{4}$

Required

a) Sketch and explanation of the current paths
b) Distance from the main distributor to the point of damage: $l_1$ in $\mathrm{m}$
c) Distance between the house connection box and the main distributor (single cable length): $l_{\mathrm{tot}}$ in $\mathrm{m}$

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a) Sketch and current paths

Principle of fault localization:

A short circuit between the two cores forms a conductive connection at the fault location. When you measure the resistance at the ends of the electric line, you obtain different loop lengths:

• Measurement at the main distributor: The measuring current flows through core 1 to the point of damage, across the short circuit into core 2, and back through core 2. Loop length: $2\cdot l_1$

• Measurement at the house connection box: The measuring current flows through core 1 to the point of damage, across the short circuit into core 2, and back through core 2. Loop length: $2\cdot (l_{\mathrm{tot}}-l_1)$

b) Distance to the point of damage

Cross-section of the electric line:

$$A=\frac{\pi d^2}{4}=\frac{\pi \cdot (0.7\mathrm{mm}{)}^2}{4}=\frac{\pi \cdot 0.49\mathrm{m}{\mathrm{m}}^2}{4}=0.385\mathrm{m}{\mathrm{m}}^2$$

Resistivity of copper:

$${\rho }_{\mathrm{Cu}}=\frac{1}{{\sigma }_{\mathrm{Cu}}}=\frac{1}{56\mathrm{S}\cdot m/m{\mathrm{m}}^2}=0.01786\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2/m$$

Resistance equation for the measurement at the main distributor:

The measured resistance $R_1$ corresponds to the resistance of the loop of length $2l_1$:

$$R_1=\rho \frac{2l_1}{A}$$

Rearranged for $l_1$:

$$l_1=\frac{R_1\cdot A}{2\rho }$$

Substituting:

$$l_1=\frac{6.1\Omega \cdot 0.385\mathrm{m}{\mathrm{m}}^2}{2\cdot 0.01786\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2/m}=\frac{2.35\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2}{0.0357\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2/m}=65.7\mathrm{m}\approx \underline{66\mathrm{m}}$$

c) Distance between the house connection box and the main distributor

Resistance equation for the measurement at the house connection box:

The measured resistance $R_2$ corresponds to the resistance of the loop of length $2(l_{\mathrm{tot}}-l_1)$:

$$R_2=\rho \frac{2(l_{\mathrm{tot}}-l_1)}{A}$$

Rearranged:

$$l_{\mathrm{tot}}-l_1=\frac{R_2\cdot A}{2\rho }$$

Substituting:

$$l_{\mathrm{tot}}-l_1=\frac{11\Omega \cdot 0.385\mathrm{m}{\mathrm{m}}^2}{2\cdot 0.01786\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2/m}=\frac{4.24\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2}{0.0357\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2/m}=118.6\mathrm{m}$$

Total distance:

$$l_{\mathrm{tot}}=l_1+(l_{\mathrm{tot}}-l_1)=65.7\mathrm{m}+118.6\mathrm{m}=184.3\mathrm{m}\approx \underline{184\mathrm{m}}$$

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Plausibility check

Alternative calculation via the total resistance:

The sum of the two measured resistances gives the resistance of the entire two-core electric line:

$$R_1+R_2=\rho \frac{2l_{\mathrm{tot}}}{A}$$ $$l_{\mathrm{tot}}=\frac{(R_1+R_2)\cdot A}{2\rho }=\frac{(6.1+11)\Omega \cdot 0.385\mathrm{m}{\mathrm{m}}^2}{2\cdot 0.01786\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2/m}=\frac{6.58\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2}{0.0357\mathrm{\Omega }\cdot \mathrm{m}{\mathrm{m}}^2/m}=184.3\mathrm{m}✓$$

Ratio check:

The ratio of the measured resistances corresponds to the ratio of the partial distances:

$$\frac{l_1}{l_{\mathrm{tot}}-l_1}=\frac{R_1}{R_2}=\frac{6.1}{11}=0.555$$ $$\frac{65.7\mathrm{m}}{118.6\mathrm{m}}=0.554✓$$

Fault localization in practice

This method of fault localization, in which the resistance is measured from both ends, is used routinely in telecommunications engineering and on power supply lines. It works particularly reliably for low-resistance faults such as short circuits.

In practice, refined methods such as the Murray loop (German: Murray-Schleife) or the Varley loop (German: Varley-Schleife) are used. They work with bridge networks and minimize the influence of contact resistances at the fault location.