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# Temperature changes from [25∘C] up to [125∘C] a tungsten wire filament [15 cm] long with a diameter of [1.7 mm] will be used.?

Temperature changes from [25∘C] up to [125∘C] a tungsten wire filament [15 cm] long with a diameter of [1.7 mm] will be used. The wire will carry [12.5 A] at all temperatures.

The resistivity of tungsten at [25∘C] is [5 × 10^-8 Ωm], the temperature coefficient of resistivity at [25∘C] is [0.005∘C^-1]

A) Calculate the maximum electric field in this filament? Unit of V/m

B) Calculate the resistance of the filament with that field? Unit of Ω

C) Calculate the maximum potential drop over the full length of the filament? Unit of V

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• hace 1 mes
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(A) By J = σE, where J is the current density at a given location in a resistive material, E is the electric field at that location, and σ (sigma) is a material-dependent parameter called the conductivity.

=>E = J/σ

=>E = (I/A)/(1/ρ)

=>E = Iρ/A

=>E(max) = I/A x ρ(max)

=>E(max) = [15/(1/4 x π x d^2)] x [ρ25*{1 + α(125-25)]

=>E(max) = [15/{0.25 x 3.14 x (1.7 x 10^-3)^2}] x [5 x 10^-8 x {1 + 0.005 x 100}]

=>E(max) = 0.50 V/m

B) As ρ125* = ρ25*{1 + α(125-25)}

=>ρ125* = 5 x 10^-8 x {1 + 0.005 x 100}

=>ρ125* = 7.5 x 10^-8 Ωm

=>R125* = ρ125* x L/A = [7.5 x 10^-8 x 15 x 10^-2]/[{0.25 x 3.14 x (1.7 x 10^-3)^2}]

=>R125* = 4.96 x 10^-3 Ω OR 4.96 mΩ

C) By V = iR

=>V(max) = i x R(max)

=>V(max) = 15 x 4.96 x 10^-3

=>V(max) = 74.38 x 10^-3 V OR 74.38 mV

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