Solution:
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 \(R_B = \frac{R_A}{3^2} = \frac{81}{9} = 9 \Omega\).
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 \(R_B = \frac{R_A}{3^2} = \frac{81}{9} = 9 \Omega\).
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