Problem — voltage drop in a copper cable

Problem

A copper cable ($\rho = 1{,}7\cdot 10^{-8}\ \Omega\cdot\text{m}$) of cross-section $S = 2{,}5\ \text{mm}^2$ feeds a load drawing $I = 10\ \text{A}$ at $230\ \text{V}$. (a) What length of cable produces a voltage drop $\Delta V = 2\ \text{V}$ (about 1%)? (b) How much power is dissipated as heat in the cable at that length? Comment on the importance of adequate cross-sections in domestic wiring.

Solution

(a) Length. The allowed drop fixes the cable’s resistance: $R = \frac{\Delta V}{I} = \frac{2\ \text{V}}{10\ \text{A}} = 0{,}2\ \Omega$ From the second law of Ohm $R = \rho\,\ell/S$ (with $S = 2{,}5\ \text{mm}^2 = 2{,}5\cdot 10^{-6}\ \text{m}^2$): $\ell = \frac{R\,S}{\rho} = \frac{0{,}2 \cdot 2{,}5\cdot 10^{-6}}{1{,}7\cdot 10^{-8}} = \frac{5\cdot 10^{-7}}{1{,}7\cdot 10^{-8}} \approx 29\ \text{m}$

(b) Power dissipated in the cable (Joule heating): $\Delta P = R\,I^2 = 0{,}2 \cdot 10^2 = 20\ \text{W}$ equivalent to $\Delta V \cdot I = 2 \cdot 10 = 20\ \text{W}$.

Comment. $20\ \text{W}$ dissipated along the cable is energy wasted as heat, and on a long or thin cable it can overheat the insulation. Since $R \propto 1/S$, doubling the cross-section halves both the voltage drop and the power lost: this is why domestic wiring uses larger cross-sections the higher the line current (e.g. $2{,}5\ \text{mm}^2$ for sockets, $1{,}5\ \text{mm}^2$ for lighting, larger cross-sections for ovens and water heaters).

$\ev{\ell \approx 29\ \text{m} \qquad \Delta P \approx 20\ \text{W}}$

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Topics: Electric circuits Concepts: Ohm’s law · Joule heating