Collection of Solved Problems in Physics

Two-wire Cable between Electrical Wiring and Appliance

Task number: 1537

• Hint 1

  How to calculate a surface area of a cross section? Note that the wire is a conductor with a constant cross section.

• Solution of Hint 1

  We determine the surface area of a cross section of a wire with a constant cross section from the following relationship: \[R=\frac{\rho l}{S},\] where R is the resistance of the conductor, ρ is the electric resistivity, which can be found in The Handbook of Chemistry and Physics, l is the length of the conductor and S is its cross section. We evaluate the cross section S as follows: \[S=\frac{\rho l}{R}.\]

• Hint 2: Losses of two-wire cable

  For a better idea, draw a two-wire cable as a circuit with the appliance.

  

  What does it mean that the losses must not exceed 3 %? What physical quantity represents the losses? Think about how this quantity behaves in the circuit above.

• Solution of Hint 2

  Losses are represented by electric power P. In an electric circuit the power P of electric current is given by \[P=U I,\] where U is the voltage at the relevant point of the circuit and I is electric current in the circuit. The current is the same throughout the circuit, so the losses of power are represented by voltage losses. The total voltage is divided into two wires and the appliance. The total loss on both wires can be 3 % maximum of total voltage that is 1.5 % in each wire, the rest is on the appliance.

• Solution

  To determine the cross section S of the wire, we use the relation for calculation of resistance of a wire with a constant cross section \[R=\frac{\rho l}{S} \qquad \Rightarrow \qquad S={\rho l}{R},\] where R is the resistance of the conductor, ρ is the electrical resistivity of the conductor material, which can be found in The Handbook of Chemistry and Physics and l is the length of the conductor.

  We use Ohm's law to determine the resistance R \[R=\frac{U}{I},\] where U is the voltage and I is the current in the conductor.

  What is a two-wire cable? For a better idea we draw the two-wire cable as a circuit with the appliance.

  

  We evaluate the losses through the electric power P. In electric circuit the following applies \[P=U I,\] where U is voltage and I is the current in the circuit. The current in the circuit is constant. Therefore, the power losses are represented by the voltage losses. The voltage is divided into the two wires and the appliance. The losses on the wiring can be maximum 3 % of the total voltage. That is 1.5 % on each wire. The voltage on the conductor U_c is expressed as: \[U_V=0{,}015\,U.\]

  Now we put the above together and we obtain the final relation for the cross section S of the wire \[ S = \frac{l\rho}{R}=\frac{l\rho}{\frac{U_V}{I}}=\frac{Il\rho}{0.015\,U}. \]

• Given values and numerical substitution

  U = 5.6 kV = 5 600 V	Output voltage of the electrical wiring
  l = 10 km = 10 000 m	Distance between the appliance and the station
  I = 20 A	Electric current in the wiring
  UV = 0.015U V	Voltage on one conductor
  S = ? m2	Cross section of the conductor
  From The Handbook of Chemistry and Physics:
  ρ = 0.018 μΩm	Electrical resistivity of copper

  ---

  \[S=\frac{I\rho l}{0.015\,U}=\frac{20{\cdot} 0.018{\cdot} 10^{-6}\cdot 10\,000}{0.015{\cdot} 5\,600}\,\dot{=}\,43 {\cdot} 10^{-6}\, \mathrm {m^2}=43\, \mathrm {mm^2}\]

• Answer

  The cross section of a wire must be 43 mm² at minimum.