Basics of Current: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Basics of Current. Understanding the underlying formula is key to arriving at the correct answer below:

Using \(R = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} \propto \frac{1}{d^2}\). Since diameter \(d_B = 3d_A\), the resistance \(R_B = \frac{R_A}{3^2} = \frac{81}{9} = 9 \Omega\).