When a voltage difference is applied to a piece of metal wire, a 5.0 ma current flows through it. if this metal wire is now replaced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire? the lengths of both wires are the same, and the voltage difference remains unchanged. (the resistivity of the original metal is 1.68 � 10-8 ? ? m, and the resistivity of silver is 1.59 � 10-8 ? ? m.)
answer
5.3 ma
11 ma
19 ma
21 ma

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

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