A 22 -kV power line that is \(10.0 \mathrm{km}\) long supplies the electric energy to a small town at an average rate of \(6.0 \mathrm{MW} .\) (a) If a pair of aluminum cables of diameter \(9.2 \mathrm{cm}\) are used, what is the average power dissipated in the transmission line? (b) Why is aluminum used rather than a better conductor such as copper or silver?

First, we need to find the cross-sectional area A of the aluminum cables. This can be calculated using the formula for the area of a circle, which is \(A = \pi r^2\), where r is the radius of the cable. Since the diameter is given, we can find the radius by dividing the diameter by 2. Diameter, d = 9.2 cm = 0.092 m Radius, r = d/2 = 0.092/2 = 0.046 m Cross-sectional area, A = \(\pi r^2 = \pi(0.046)^2=\pi(0.002116) \mathrm{m^2}\).

Now that we have the cross-sectional area, we can find the resistance of the aluminum cables using the formula for resistance: \(R = \frac{\rho L}{A}\), where \(\rho\) is the resistivity of aluminum, \(L\) is the length of the cables, and \(A\) is the cross-sectional area. The resistivity of aluminum is \(\rho = 2.82 \times 10^{-8} \, \Omega \cdot \mathrm{m}\). Length of the cables, L = 10.0 km = 10000 m Resistance, R = \(\frac{2.82 \times 10^{-8}\cdot 10000}{\pi(0.002116)} \, \Omega\).

Next, we will calculate the current I in the cables using the formula for power: \(P = IV\), where P is the power and V is the voltage. In this case, the power is given as 6.0 MW and the voltage is given as 22 kV. Power, P = 6.0 MW = \(6.0 \times 10^6 \mathrm{W}\) Voltage, V = 22 kV = \(22 \times 10^3 \mathrm{V}\) Current, I = \(\frac{P}{V} = \frac{6.0 \times 10^6}{22 \times 10^3} \, \mathrm{A}\).

Now that we have the resistance and the current, we can calculate the power dissipated in the transmission line using the formula for power in terms of current and resistance: \(P_{dissipated} = I^2R\). Power dissipated, \(P_{dissipated} = I^2R\).

Although copper and silver are better conductors than aluminum, there are several reasons for using aluminum in power lines. Aluminum is lighter and cheaper than copper, which makes it more economical for use in power transmission lines. Additionally, aluminum oxide, an insulator, forms on the surface of the aluminum wire, which provides corrosion resistance and helps protect the conductor from environmental factors. Despite having a higher resistance, the benefits of aluminum make it a suitable choice for power transmission lines. To summarize, we have now: 1. Calculated the cross-sectional area of the aluminum cables. 2. Found the resistance of the aluminum cables. 3. Calculated the current flowing in the cables. 4. Determined the power dissipated in the transmission line. 5. Discussed the reason for using aluminum instead of copper or silver.