Question

A copper cable needs to carry a current of 160 A with a power loss of only 2.0 W/m. What is the required radius of the copper cable? (The resistivity of copper is 1.7 × 10 −8 Ω⋅m).

Answer 1

Answer:

The radius of the cable is 0.0083 m or 8.3 mm.

Explanation:

The resistance of copper cable of 1 meter length will be given by

$R_{cable} = \frac{\rho \times l}{a}$    ....     (i)

where the resistivity of copper is $\rho = 1.7 \times 10^{-8} \Omega.m$ , and l is the length of the wire which is considered to be 1m, and a is the cross sectional area of the wire in $m^{2}$.

From the formula of power we know that, $P = I^2 R$    ....    (ii)

Therefore 2 W/m  = $(160)^2 \times R$     ....     (iii)  

where  the resistance,R, actually means the resistance of the cable per meter.

Therefore R ( resistance of cable per meter)

= $\frac{2}{160^2} = 7.812 \times 10^{-5}$ ohms / meter.       ....    (iv)

Therefore from (i)

$7.812 \times 10^{-5}$ = $\frac{1.7 \times 10^{-8} \times 1}{a} = \frac{1.7 \times 10^{-8} \times 1}{\pi r^{2} }$       .....     (v)

where cross sectional area of the cable, a  = $\pi r^2$,

where r is the radius of the cable, and length of cable,l = 1m.

Therefore r  = $\sqrt{\frac{ 1.7 \times 10^{-8}}{\pi \times 7.812 \times 10^{-5} } } = 0.0083m = 8.3 mm$