Respuesta :
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
A 1.60 m cylindrical rod of diameter 0.450 cm is connected to a power supply that maintains a constant potential difference of 12.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.4 A , while at 92.0 ∘C it reads 17.4 A . You can ignore any thermal expansion of the rod.
a) Find the resistivity and for the material of the rod at 20° C .
b) Find the temperature coefficient of resistivity at 20° C for the material of the rod.
Given Information:
Room temperature = T₀ = 20° C
Temperature = T = 92° C
Current at 20° C = I₀ = 18.4 A
Current at 92° C = I = 17.4 A
Voltage = V = 12 V
Length = L = 1.60 m
Diameter = d = 0.450 cm = 0.0045 m
Required Information:
Resistivity of the material at 20° C = ρ = ?
Temperature coefficient of resistivity at 20° C = α = ?
Answer:
Resistivity of the material at 20° C = 2.062x10⁻⁶ Ω.m
Temperature coefficient of resistivity at 20° C = 7.986x10⁻⁴ per °C
Explanation:
a) We want to find out the resistivity of the material at 20° C
The resistivity of any material can be found using,
ρ = R₀A/L
Where R₀ is the resistance of the rod at 20° C, A is the area of rod and L is the length of the cylindrical rod.
We also know that area is given by
A = πr²
where r = d/2 = 0.0045/2 = 0.00225 m
A = π(0.00225)²
A = 5.062⁻⁶ m²
We know that resistance of the material is given by
R₀ = V/I₀
R₀ = 12/18.4
R₀ = 0.6521 Ω
Therefore, the resistivity of the material is
ρ = R₀A/L
ρ = (0.6521*5.062⁻⁶)/1.60
ρ = 2.062x10⁻⁶ Ω.m
b) We want to find out the temperature coefficient of resistivity of the rod at 20° C
The temperature coefficient of resistivity is given by
α = R/R₀ - 1/(T - T₀)
Where R is the resistance of the rod at 90° C
R = V/I
R = 12/17.4
R = 0.6896 Ω
α = R/R₀ - 1/(T - T₀)
α = (0.6896/0.6521) - 1/(92° - 20°)
α = 0.0575/72°
α = 0.000798 per °C
α = 7.986x10⁻⁴ per °C
Answer:
is incomplete:
Part A
Find the resistivity and for the material of the rod at 20 ∘C.
Part B
Find the temperature coefficient of resistivity at 20∘C for the material of the rod.
Answer:
a) p = 0.00000648 Ω*m
b) the temperature coefficient of resistivity is 0.00078/ºC
Explanation:
a) L = length of the cylindrical rod = 1.6 m
D = diameter of the cylindrical rod = 0.45 cm = 0.0045 m
the radius r = 0.00225 m
V = potential difference = 12 V
I at 20ºC = 18.4 V
I at 92ºC = 17.4 A
The area is equal to
[tex]A=\pi r^{2} =\pi *0.00225^{2} =0.0000159 m^{2}[/tex]
the resistivity at 20ºC is
[tex]p=\frac{VA}{IL} =\frac{12*0.0000159}{18.4*1.6} =0.00000648ohm*m[/tex]
b) the coefficient of resistivity at 20ºC is
R = V/I = 12/18.4 = 0.652 Ω
the coefficient of resistivity at 92ºC is
R = 12/17.4 = 0.689 Ω
the temperature coefficient of resistivity is
[tex]R(T)=R_{0} (1+\alpha (T-T_{0} )\\0.689=0.652(1+\alpha (92-20))\\\alpha =\frac{0.689-0.652}{46.94} =0.00078/C[/tex]
