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8 June, 08:39

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.

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Answers (2)
  1. 8 June, 11:21
    0
    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
  2. 8 June, 12:39
    0
    (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
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