Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
The current that will flow in the silver wire of equal length is 21.13 mA.
The given parameters;
The resistance of a wire with given length and area is given as;
[tex]R = \frac{\rho l}{A}[/tex]
where;
Let the diameter of the metal wire = d
diameter of the silver wire = 2d
[tex]A = \frac{\pi d^2}{4}[/tex]
[tex]A_s = \frac{\pi (2d)^2}{4} \\\\A_s = 4(\frac{\pi d^2}{4} )\\\\A_s = 4A_{metal}[/tex]
The current that will flow in the silver wire is calculated as;
[tex]l = \frac{RA}{\rho} \\\\\frac{R_1A_1}{\rho_1} = \frac{R_2A_2}{\rho_2} \\\\V= IR\\\\R = \frac{V}{I} \\\\\frac{V d^2}{I_1 \rho_1} = \frac{V(4d^2)}{I_2 \rho_2} \\\\\frac{1}{I_1 \rho_1} = \frac{4}{I_2\rho_2} \\\\I_2\rho_2=4 I_1\rho_1\\\\I _2 = \frac{4 I_1\rho_1}{\rho_2} \\\\I_2 = \frac{4\times (5\ mA)\times 1.68\times 10^{-8}}{1.59\times 10^{-8}} \\\\I_2 = 21.13 \ mA[/tex]
Thus, the current that will flow in the silver wire of equal length is 21.13 mA.
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