When a potential difference is applied across a piece of wire made of metal A, a $5.0 - m A$ current flows. If the metal-A wire is replaced with a wire made of metal B having twice the diameter of the metal-A wire, how much current will flow through the metal-B wire? The lengths of both wires are the same, and the voltage difference remains unchanged. The resistivity of metal A is 1.68 × 10^-8 Ω ∙ m, and the resistivity of metal B is 1.59 × 10^-8 Ω ∙ m. A) 21 mA B) 19 mA C) 11 mA D) 5.3 mA

Question

Quizplus · Accepted Answer

The resistance of a wire is given by $ R = \rho \frac{L}{A} $, where $ \rho $ is resistivity, $ L $ is length, and $ A $ is cross-sectional area. The area $ A $ is proportional to the square of the diameter. Since metal B has twice the diameter, its area is four times that of metal A. The resistance of metal B is therefore $ \frac{1.59}{1.68} \times \frac{1}{4} $ times the resistance of metal A. The current is inversely proportional to resistance, so the current through metal B is $ 5.0 \, \text{mA} \times \frac{1.68}{1.59} \times 4 \approx 21 \, \text{mA} $.

When a Potential Difference Is Applied Across a Piece of Wire