Answer
a. $3.8\times10^{-4}\Omega$
b. $1.5\times10^{-3}\Omega$
c. $6.0\times10^{-3}\Omega$
Work Step by Step
a. Use equation 18–3 to find the resistance.
$$R_x=\rho\frac{\mathcal{l}}{A}$$
$$=(3.0\times10^{-5}\Omega \cdot m)\frac{1.0\times10^{-2}m}{(2.0\times10^{-2}m)(4.0\times10^{-2}m)}$$
$$=3.8\times10^{-4}\Omega$$
b. Use equation 18–3 to find the resistance.
$$R_y=\rho\frac{\mathcal{l}}{A}$$
$$=(3.0\times10^{-5}\Omega \cdot m)\frac{2.0\times10^{-2}m}{(1.0\times10^{-2}m)(4.0\times10^{-2}m)}$$
$$=1.5\times10^{-3}\Omega$$
c. Use equation 18–3 to find the resistance.
$$R_z=\rho\frac{\mathcal{l}}{A}$$
$$=(3.0\times10^{-5}\Omega \cdot m)\frac{4.0\times10^{-2}m}{(1.0\times10^{-2}m)(2.0\times10^{-2}m)}$$
$$=6.0\times10^{-3}\Omega$$
