Question

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper, with a circular cross section of radius 1.53 mm. The other wire is composed of 19 strands of thin copper wire bundled together. Each strand has a circular cross section of radius 0.306 mm. The current density ???? in each wire is the same. ????=1750 A/m2 Two circles. One circle is solid, whereas the other contains 19 tightly packed smaller circles How much current does each wire carry? solid wire current: A stranded wire current: A The resistivity of copper is ????=1.69×10−8 Ω·m. What is the resistance of a 3.25 m length of each wire? solid wire resistance: Ω stranded wire resistance: Ω

Answer 1

Answer:

a

 Solid Wire     $I = 0.01237 \ A$      

  Stranded  Wire  $I_2 = 0.00978 \ A$

b

  Solid Wire   $R = 0.0149 \ \Omega$

   Stranded  Wire  $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

  The  radius of the first wire is  $r_1 = 1.53 mm = 0.0015 \ m$

  The radius of  each strand is  $r_0 = 0.306 \ mm = 0.000306 \ m$

  The current density in both wires is  $J = 1750 \ A/m^2$

Considering the first wire

     The  cross-sectional area of the first wire is

      $A = \pi r^2$

= >  $A = 3.142 * (0.0015)^2$

= >  $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is    

     $I = J*A$

=>  $I = 1750*7.0695 *10^{-6}$

=>  $I = 0.01237 \ A$

Considering the second wire  wire

The  cross-sectional area of the second wire is

     $A_1 = 19 * \pi r^2$

=>     $A_1 = 19 *3.142 * (0.000306)^2$

=>  $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is  

      $I_2 = J * A_1$

=>    $I_2 = 1750 * 5.5899 *10^{-6}$

=>    $I_2 = 0.00978 \ A$

Considering question two  

 From the question we are told that

     Resistivity is  $\rho = 1.69* 10^{-8} \Omega \cdot m$

     The  length of each wire  is  $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

    $R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

    $R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$