Chapter 5, Problem 32PP

Electric wire, like that in the photo, is a cylindrical conductor covered by an insulating material. The resistance of a piece of wire is given by the formula

$R=\frac{\rho L}{A}=\frac{4\rho L}{\pi d^2}$

where ρ is the resistivity of the conductor, and L, A, and d are the length, crosssectional area, and diameter of the wire. The resistivity of copper is 1.678 × 10⁻⁸ Ω m.

The wire diameter, d, is commonly specified by the American wire gauge (AWG), which is an integer, n. The diameter of an AWG n wire is given by the formula

$d=0.127\times {92}^{\frac{36-n}{39}}mm$

Write a method

public static double diameter(int wireGauge)

that accepts the wire gauge and returns the corresponding wire diameter. Write another method

public static double copperWireResistance(double length, int wireGauge)

that accepts the length and gauge of a piece of copper wire and returns the resistance of that wire. The resistivity of aluminum is 2.82 × 10⁻⁸ Ω m. Write a third method

public static double aluminumWireResistance(double length, int wireGauge)

that accepts the length and gauge of a piece of aluminum wire and returns the resistance of that wire.

Write a program to test these methods.