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Question

Please show all work and specify the correct answer at the bottom. Please do not put down an answer that is not one of the choices, I will thumbs you down. Correct answer will get five stars and thumbs up.

When a voltage difference is applied to a piece of metal wire, a 5-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.)

Explanation / Answer

R = rho*L/A

here rho = resistivity of metal

R1 = rho1*L/(pi*r1^2)

and R2 = rho2*L/(pi*r2^2)

here r2 = 2*r1 so

R2 = rho2*L/(4*pi*r1^2)

R1/R2 = (rho1/rho2)*(4) = (1.68*10^(-8)/(1.59*10^(-8)))*4 = 4.226

and R1/R2 = I2/I1

I2 = 4.226*I1 = 4.226*5 = 21 mA (Answer)

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