When a voltage difference is applied to a piece of metal wire, a 1.6-mA current

Question

When a voltage difference is applied to a piece of metal wire, a 1.6-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 1.9 × 10^-8 m, and the resistivity of silver is 1.59 × 10^-8 m.) (Give your answer to the nearest 0.1 mA).

The correct answer is 7.6. I need an explanation as to how this is the answer. Thanks in advance!

Explanation / Answer

resistance = resistivity * length / area

resistance for first wire = 1.9 * 10^-8 * length / pi * (diameter / 2)^2

resistance of the silver wire = 1.59 * 10^-8 * length / pi * (2 * diameter / 2)^2

resistance of the silver wire = 1.59 * 10^-8 * length / pi * (diameter)^2

by ohm's law

V = IR

since voltage is same in both the cases so,

1.6 * 10^-3 * 1.9 * 10^-8 * length / pi * (diameter / 2)^2 = I * 1.59 * 10^-8 * length / pi * (diameter)^2

4 * 1.6 * 10^-3 * 1.9 = I * 1.59

current I = 0.007647 A or 7.647 mA

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