Exercise EE1-04.R1 (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-12-07)

Relates to EE1-04 Resistance

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Task

When a photovoltaic system is installed, an earthing cable made of copper is laid (${\rho }_{Cu}=1.78\cdot 10^{-8}\mathrm{\Omega }\cdot \mathrm{m}$). The electric line has a length of $l=25\mathrm{m}$ and a cross-section of $A=16\mathrm{m}{\mathrm{m}}^2$.

a) Calculate the conductor resistance of the earthing cable at $20°\mathrm{C}$.

b) What is the conductance of the earthing cable?

c) In the event of an earth fault, a current of $I=100\mathrm{A}$ flows briefly. What voltage drop occurs across the earthing cable?

Solution

a) $R=27.8\mathrm{m\Omega }$
b) $G=36.0\mathrm{S}$
c) $\Delta U=2.78\mathrm{V}$

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Solution

Given

Given explicitly:

• Cable length: $l=25\mathrm{m}$
• Conductor cross-section: $A=16\mathrm{m}{\mathrm{m}}^2$
• Material: copper
• Earth fault current: $I=100\mathrm{A}$

Known:

• Resistivity of copper (20 °C): ${\rho }_{\mathrm{Cu}}=0.0178\Omega \mathrm{m}{\mathrm{m}}^2/m$
• Conductor resistance: $R=\rho \cdot \frac{l}{A}$
• Conductance: $G=\frac{1}{R}$
• Ohm’s law: $\Delta U=R\cdot I$

where:

• $R$ the resistance in ohms ($\Omega$)
• $\rho$ the resistivity in $\Omega \mathrm{m}{\mathrm{m}}^2/m$
• $l$ the length in metres ($\mathrm{m}$)
• $A$ the cross-sectional area in square millimetres ($\mathrm{m}{\mathrm{m}}^2$)
• $G$ the conductance in siemens ($\mathrm{S}$)
• $I$ the current in amperes ($\mathrm{A}$)
• $\Delta U$ the voltage drop in volts ($\mathrm{V}$)

Required

a) Conductor resistance $R$ in $\Omega$
b) Conductance $G$ in $\mathrm{S}$
c) Voltage drop $\Delta U$ in the event of an earth fault in $\mathrm{V}$

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a) Conductor resistance

The conductor resistance results from the resistivity, the length and the cross-section:

$$R={\rho }_{\mathrm{Cu}}\cdot \frac{l}{A}=0.0178\frac{\Omega \mathrm{m}{\mathrm{m}}^2}{\mathrm{m}}\cdot \frac{25\mathrm{m}}{16\mathrm{m}{\mathrm{m}}^2}$$ $$R=0.0178\cdot 1.5625\Omega =0.0278\Omega \approx \underline{27.8\mathrm{m\Omega }}$$

Note

For earthing cables, it is important that the resistance is as low as possible so that high currents can flow in the event of a fault and the protective devices (e.g. fuses, residual-current circuit breakers) trip reliably.

b) Conductance

The conductance is the reciprocal of the resistance:

$$G=\frac{1}{R}=\frac{1}{0.0278\Omega }=\underline{36.0\mathrm{S}}$$

Check: $R\cdot G=0.0278\Omega \cdot 36.0\mathrm{S}=1.0$ ✔

c) Voltage drop in the event of an earth fault

The voltage drop across the earthing cable at an earth fault current of $100\mathrm{A}$:

$$\Delta U=R\cdot I=0.0278\Omega \cdot 100\mathrm{A}=\underline{2.78\mathrm{V}}$$

Practical significance

A voltage drop of about $2.8\mathrm{V}$ is acceptable in the event of an earth fault. The touch voltage at the earth electrode would be correspondingly low. In practice, resistances in the range below a few ohms are often aimed for in earthing systems (cf. DIN VDE 0100-540).