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

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

Step-by-step explanation:

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

R =
`\rho \ (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)
`(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)
`(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)
`(l_1)/( (2 d_1)^2)`

R₂ = (ρ 4 /π )
`(l_1)/(d_1^2)` ¼

R₂ = ¼ R₁

* indicate that d₂ = d₃ and l₃ = 2 l₂

R2 = (ρ 4 /π)
`(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)
`(2 \ l_2)/(d_2)`

R3 = 2 R₂

let's write the relations obtained

R₁ = (ρ 4 /π)
`(I_1)/( d_1^2)`

R₂ = ¼ R₁

R₃ = 2 R₂

let's write everything as a function of R1

R₁ =(ρ 4 /π)
`(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 =
`(V^2)/( (1)/(4) R_1)`

P₂ = 4 P₁

wire 3

P₃ = V² / R₃

P₃ =
`(V^2)/( (1)/(2) R_1)`

P₃ = 2 P₁

Therefore, the wire that dissipates the most is wire 2