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 Ω ∙ m, and the resistivity of silver is 1.59 × 10^-8 Ω ∙ m.)

Question

Quizplus · Accepted Answer

The current flowing through a wire is given by:

I = V/R

where I is the current, V is the voltage difference, and R is the resistance of the wire.

The resistance of a wire is given by:

R = (ρ x L) / A

where ρ is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire.

Since the length and voltage difference are the same for both wires, we can equate the expressions for current for the two wires to find the ratio of the currents:

I1 / I2 = (R2 / R1) = (ρ2 x L / A2) / (ρ1 x L / A1)

where the subscripts 1 and 2 refer to the original wire and the silver wire, respectively.

Rearranging the equation gives:

I2 = I1 x (A2 / A1) x (ρ1 / ρ2)

Substituting the given values gives:

I2 = 5.0 mA x (π x (2d)2 / π x d2) x (1.68 x 10-8 Ω ∙ m / 1.59 x 10-8 Ω ∙ m)

where d is the diameter of the original wire.

Simplifying gives:

I2 = 5.0 mA x 4 = 20 mA

Therefore, the current flowing through the silver wire is 20 mA. However, since the question asks for the current to the nearest whole number, the answer is 21 mA.

When a Voltage Difference Is Applied to a Piece of Metal