# Chapter 1 - 1.6 - Problems - Introduction to Circuit Elements - Page 40: P1.56

The diameter must be increased by a factor of 1.26

#### Work Step by Step

We can write an expression for the resistance in a wire. $R = \frac{\rho~L}{A} = \frac{\rho~L}{\pi~r^2}$ We can write an expression for the resistance in the copper wire. $R_c = \frac{\rho_c~L}{\pi~r_c^2}$ We can write an expression for the resistance in the aluminum wire. $R_a = \frac{\rho_a~L}{\pi~r_a^2}$ Since the resistance is the same in both wires, we can equate the two expressions to find the factor that the radius must be increased. $R_c = R_a$ $\frac{\rho_c~L}{\pi~r_c^2} = \frac{\rho_a~L}{\pi~r_a^2}$ $r_a^2 = \frac{\rho_a~r_c^2}{\rho_c}$ $r_a = \sqrt{\frac{\rho_a}{\rho_c}}~r_c$ $r_a = \sqrt{\frac{2.73\times 10^{-8}~\Omega~m}{1.72\times 10^{-8}~\Omega~m}}~r_c$ $r_a = 1.26~r_c$ Since the radius must be increased by a factor of 1.26, the diameter must be increased by a factor of 1.26

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