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The maximum allowable resistance for an underwater cable is one hundredth of an ohm per
meter and the resistivity of copper is 1.54 x 10-80m.
a) Calculate the smallest cross sectional area of copper cable that could be used.​


Sagot :

Answer:

A = 1.54 x 10⁻⁵ m² = 15.4 mm²

Explanation:

The resistance of a wire can be given by the following formula:

[tex]R = \frac{\rho L}{A}\\\\A = \frac{\rho L}{R} = \frac{\rho}{\frac{R}{L} }[/tex]

where,

A = smallest cross-sectional area = ?

ρ = resistivity of copper = 1.54 x 10⁻⁸ Ωm

[tex]\frac{R}{L}[/tex] = resistance per unit length of wire = 0.001 Ω/m

Therefore,

[tex]A = \frac{1.54\ x\ 10^{-8}\ \Omega m}{0.001\ \Omega/m}[/tex]

A = 1.54 x 10⁻⁵ m² = 15.4 mm²