Answer:
a
Solid Wire [tex]I = 0.01237 \ A [/tex]
Stranded Wire [tex]I_2 = 0.00978 \ A [/tex]
b
Solid Wire [tex]R = 0.0149 \ \Omega [/tex]
Stranded Wire [tex]R_1 = 0.0189 \ \Omega [/tex]
Explanation:
Considering the first question
From the question we are told that
The radius of the first wire is [tex]r_1 = 1.53 mm = 0.0015 \ m[/tex]
The radius of each strand is [tex]r_0 = 0.306 \ mm = 0.000306 \ m[/tex]
The current density in both wires is [tex]J = 1750 \ A/m^2[/tex]
Considering the first wire
The cross-sectional area of the first wire is
[tex]A = \pi r^2[/tex]
= > [tex]A = 3.142 * (0.0015)^2[/tex]
= > [tex]A = 7.0695 *10^{-6} \ m^2 [/tex]
Generally the current in the first wire is
[tex]I = J*A[/tex]
=> [tex]I = 1750*7.0695 *10^{-6}[/tex]
=> [tex]I = 0.01237 \ A [/tex]
Considering the second wire wire
The cross-sectional area of the second wire is
[tex]A_1 = 19 * \pi r^2[/tex]
=> [tex]A_1 = 19 *3.142 * (0.000306)^2[/tex]
=> [tex]A_1 = 5.5899 *10^{-6} \ m^2[/tex]
Generally the current is
[tex]I_2 = J * A_1[/tex]
=> [tex]I_2 = 1750 * 5.5899 *10^{-6} [/tex]
=> [tex]I_2 = 0.00978 \ A [/tex]
Considering question two
From the question we are told that
Resistivity is [tex]\rho = 1.69* 10^{-8} \Omega \cdot m[/tex]
The length of each wire is [tex]l = 6.25 \ m[/tex]
Generally the resistance of the first wire is mathematically represented as
[tex]R = \frac{\rho * l }{A}[/tex]
=> [tex]R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }[/tex]
=> [tex]R = 0.0149 \ \Omega [/tex]
Generally the resistance of the first wire is mathematically represented as
[tex]R_1 = \frac{\rho * l }{A_1}[/tex]
=> [tex]R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }[/tex]
=> [tex]R_1 = 0.0189 \ \Omega [/tex]
