Respuesta :
Answer:
(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
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