A 1.70 m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 14.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.7 A, while at 92.0 ∘C it reads 17.2 A. You can ignore any thermal expansion of the rod.

part 1) Find the resistivity and for the material of the rod at 20 ∘C.

part 2) Find the temperature coefficient of resistivity at 20 ∘C for the material of the rod.

Question

Physics

Giovanni Rosario

1 January, 10:32

A 1.70 m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 14.0 V across its ends, while an ammeter measures the current through it. You observe that at room temperature (20.0 ∘C) the ammeter reads 18.7 A, while at 92.0 ∘C it reads 17.2 A. You can ignore any thermal expansion of the rod.

part 1) Find the resistivity and for the material of the rod at 20 ∘C.

part 2) Find the temperature coefficient of resistivity at 20 ∘C for the material of the rod.

+1

Answers (2)

Know the Answer?

Jorden Casey · Answer 1

Given Information:

Length of rod = L = 1.70 m

diameter of rod = d = 0.500 cm = 0.005 m

Voltage = V = 14 volts

Current at 20° C = I₀ = 18.7 A

Current at 92° C = I = 17.2 A

Room temperature = T₀ = 20° C

Temperature = T = 92° C

Required Information:

Resistivity at 20° C = ρ = ?

Temperature coefficient = α = ?

Answer:

Resistivity = ρ = 8.63x10⁻⁶ Ω. m

Temperature coefficient = α = 0.00121 per °C

Explanation:

Part 1) Find the resistivity and for the material of the rod at 20° C

We know that resistivity is given by

ρ = R₀A/L

Where R₀ is the resistance of the rod at 20° C, A is the area and L is the length of the cylindrical rod.

Area is given by

A = πr²

A = π (d/2) ²

A = π (0.005/2) ²

A = 0.0000196 m²

Resistance can be found using Ohm's law

R₀ = V/I₀

R₀ = 14/18.7

R₀ = 0.7486 Ω

ρ = R₀A/L

ρ = 0.7486*0.0000196/1.70

ρ = 8.63x10⁻⁶ Ω. m

Part 2) Find the temperature coefficient of resistivity at 20° C for the material of the rod

The temperature coefficient of resistivity can be found using

α = R/R₀ - 1 / (T - T₀)

Where R is the resistance of the rod at 20° C

R = V/I = 14/17.2 = 0.8139 Ω

α = R/R₀ - 1 / (T - T₀)

α = (0.8139/0.7486) - 1 / (92° - 20°)

α = 0.08722/72°

α = 0.00121 per °C

Conley · Answer 2

(1) The resistivity of the rod at 20 °C is 8.652*10^-6 ohm-meter.

(2) The temperature coefficient of resistivity at 20 °C is 0.00125/°C

Explanation:

(1) Resistance at 20 °C = V/I = 14/18.7 = 0.749 ohm

Length = 1.7 m

Diameter (d) = 0.5 cm = 0.5/100 = 0.005 m

Area = πd^2/4 = 3.142*0.005^2/4 = 1.96375*10^-5 m^2

Resistivity at 20 °C = resistance * area/length = 0.749*1.96375*10^-5/1.7 = 8.652*10^-6 ohm-meter.

(2) Resistance at 92 °C = V/I = 14/17.2 = 0.814 ohm

Temperature coefficient at 20 °C = (0.814/0.749 - 1) : (92 - 20) = (1.09 - 1) : 72 = 0.09 : 72 = 0.00125/°C