Question

Three cylindrical wires, 1, 2, and 3 are made of the same materialand have resistances R1, R2, and R3, respectively. Wires 1 and 2 have the same lengthbut diameter of wire 2 is twice that of wire 1. Wires 2 and 3 have the same diameterbut length of wire 3is twice that of wire 2.
1. Rank the wires according to their resistances, greatest first.
A. R1> R2> R3R1
B. R1> R3> R2R2
C. R2> R1> R3R3
D. R2> R3> R1
E. R3> R1> R2
F. R3> R2> R1Q2.
2. If same voltage is applied across each of the wires, which one will dissipate heat at the highest rate

Answer 1

Answer:

1)     R₁ > R₃ > R₂ correct B , 2) the wire that dissipates the most is wire 2

Explanation:

1) The resistance of a wire is given by the expression

          R = $\rho \ \frac{l}{A}$

where ρ is the resistivity of the material, l the length of the wire and A the area of ​​the wire

The area is given by

          A = π r² = π d² / 4

we substitute

         R = (ρ 4 /π)  $\frac{l}{d^2}$

the amount in parentheses is constant for this case

let's analyze the situation presented, to find the resistance of each wire

* indicate l₁ = l₂ and d₂ = 2 d₁

the resistance of wire 1 is

          R₁ = (ρ 4 /π)  $\frac{l_1}{d_1^2}$  

the resistance of wire 2 is

          R₂ = (ρ 4 /π) \frac{l_2}{d_2^2}

          R₂ = (ρ 4 /π)   $\frac{l_1}{ (2 d_1)^2}$

          R₂ = (ρ 4 /π ) $\frac{l_1}{d_1^2}$   ¼

          R₂ = ¼ R₁

 

* indicate that d₂ = d₃    and  l₃ = 2 l₂

           R2 = (ρ 4 /π)  $\frac{l_2}{d_2^2}$

the resistance of wire 3 is substituting the indicated condition

           R3 = (ρ 4 /π 2)  \frac{l_3}{d_3^2}

            R3 = (ρ 4 /π)   $\frac{2 \ l_2}{d_2}$

            R3 = 2 R₂

let's write the relations obtained

          R₁ = (ρ 4 /π)  $\frac{I_1}{ d_1^2}$

          R₂ = ¼ R₁

          R₃ = 2 R₂

let's write everything as a function of R1

           R₁ =(ρ 4 /π)   $\frac{l_1}{d_1^2}$

           R₂ = ¼ R₁

           R₃ = ½ R₁

the resistance of the wire in decreasing order is

           R₁ > R₃ > R₂

2) The power dissipated by a wire is

           P = V I

the voltage is

           V = I R

            I = V / R

substituting

          P = V² / R

therefore the power dissipated by each wire is

wire 1

           P₁ = V² / R₁

wire 2

           P₂ = V² / R₂

           P₂ = $\frac{V^2}{ \frac{1}{4} R_1}$

           P₂ = 4 P₁

wire 3

           P₃ = V² / R₃

           P₃ = $\frac{V^2}{ \frac{1}{2} R_1}$

           P₃ = 2 P₁

Therefore, the wire that dissipates the most is wire 2