Answer to Question #146299 in Electric Circuits for Jacob

Question

A power station delivers 1 MW of power at 230 V, and delivers the electricity to a town 50 km away through copper cables. Assuming that 10% Ohmic losses are acceptable, calculate the cross sectional area of the cable required. You may assume that the resistivity of copper, ρcopper = 1.7 x 10−8 Ωm.

Answer 1

Current delivered$(I) =\frac{P}{V}=\frac{1000000}{230}=4347.826A$

10 percent ohmic loss $\implies ohmic loss=\frac{10}{100}\times10^6=10^5W=I^2R\implies R=\frac{10^5}{4347.826^2}=5.29 m\varOmega$

We have

$R =\rho \frac{l}{A}$

Therefore,

$A=\rho \frac{l}{R}=1.7\times10^{-8} \frac{50\times10^3}{5.29\times10^{-3}}=0.161m^2$