Question

Imagine a copper wire 10 km long that needs to supply a city with 10 Mwatts of power--say from a waterfall. If the city ran on DC power, with voltages of 100 V, how much current would run through that wire? If you were willing to lose 1% of your power as heat in the wire, how thick would such a wire have to be? The resistivity of copper is 1.7 x 10^-8 ohm-m.

Answer 1

The amount of power dissipated is:

$$ \begin{aligned} P &=\frac{V^{2}}{R} \\ &=\frac{V^{2} A}{\rho L} \\ A &=\frac{P \rho L}{V^{2}} \\ r &=\sqrt{\frac{P \rho L}{\pi V^{2}}} \\ t &=2 r=\sqrt{\frac{4 P \rho L}{\pi V^{2}}}=\sqrt{\frac{4\left(0.01\left(10 \times 10^{6}\right)\right)\left(1.7 \times 10^{-8}\right)\left(10 \times 10^{3}\right)}{\pi(100)^{2}}}=4.8 \times 10^{-2} \mathrm{~m}=4.8 \mathrm{~cm} \end{aligned} $$

Answer 2

Current that would run through the wire,

$$ \begin{aligned} I &=\frac{P}{V} \\ &=\frac{10 \times 10^{6} \mathrm{~W}}{100 \mathrm{~V}} \\ &=10^{5} \mathrm{~A} \end{aligned} $$

Net power,

$$ \begin{aligned} P_{n s t} &=P-\frac{1}{100} P \\ &=\left(10 \times 10^{6} \mathrm{~W}\right)-\frac{\left(10 \times 10^{6} \mathrm{~W}\right)}{100} \mathrm{Current} \\ &=9.9 \times 10^{6} \mathrm{~W} \end{aligned} $$

Current

$$ I=\frac{P}{V} $$

$$ \begin{array}{l} =\frac{9.9 \times 10^{6} \mathrm{~W}}{100 \mathrm{~V}} \\ =9.9 \times 10^{4} \mathrm{~A} \end{array} $$

Resistance

$$ \begin{aligned} R &=\frac{V}{I} \\ &=\frac{100 \mathrm{~V}}{9.9 \times 10^{4} \mathrm{~A}} \\ &=1 \times 10^{-3} \Omega \end{aligned} $$

Also,

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

$$ \begin{aligned} A &=\frac{\rho l}{R} \\ &=\frac{\left(1.7 \times 10^{-8} \mathrm{~W} \cdot \mathrm{m}\right)\left(9.9 \times 10^{4} \mathrm{~A}\right)}{1 \times 10^{-3} \Omega} \\ &=168.3 \times 10^{-3} \mathrm{~m}^{2} \end{aligned} $$

Area,

$$ \begin{aligned} A &=\pi r^{2} \\ r &=\sqrt{\frac{168.3 \times 10^{-3} \mathrm{~m}^{2}}{\pi}} \\ &=0.23 \mathrm{~m} \end{aligned} $$

Thickness of the wire, \(d=2 r\)

\(=2(0.23 \mathrm{~m}) \mid\)

\(=0.46 \mathrm{~m}\)