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.

Answer 1

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

$A=\pi r^{2} =\pi *0.00225^{2} =0.0000159 m^{2}$

the resistivity at 20ºC is

$p=\frac{VA}{IL} =\frac{12*0.0000159}{18.4*1.6} =0.00000648ohm*m$

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

$R(T)=R_{0} (1+\alpha (T-T_{0} )\\0.689=0.652(1+\alpha (92-20))\\\alpha =\frac{0.689-0.652}{46.94} =0.00078/C$

Answer 2

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