Question

A circuit consists of a copper wire of length \(10.0 \mathrm{~m}\) and radius \(1.00 \mathrm{~mm}\) connected to a \(10.0-\mathrm{V}\) battery. An aluminum wire of length \(5.00 \mathrm{~m}\) is connected to the same battery and dissipates the same amount of power. What is the radius of the aluminum wire?

Short answer

Answer: To find the radius of the aluminum wire, we can use the formula \(r_{Al} = \sqrt{\frac{\rho_{Al}L_{Al}}{R_C \pi}}\), where \(\rho_{Al}\) is the resistivity of aluminum, \(L_{Al}\) is the length of the aluminum wire, and \(R_C\) is the resistance of the copper wire calculated in step 1. Plugging in the values, we can solve for the radius \(r_{Al}\) of the aluminum wire.

Step-by-step solution

Calculate the resistance of the copper wire.

To find the resistance of the copper wire, we can use the formula for resistance: \(R = \frac{\rho L}{A}\), where \(R\) is the resistance, \(\rho\) is the resistivity of the material, \(L\) is the length of the wire, and \(A\) is the cross-sectional area of the wire. For a cylindrical wire, the cross-sectional area can be calculated as \(A = \pi r^2\), where \(r\) is the radius of the wire. The resistivity of copper is \(\rho_C = 1.68 \times 10^{-8} \mathrm{\Omega m}\). Given the length \(L_C = 10.0 \mathrm{m}\) and radius \(r_C = 1.00\mathrm{mm} = 1.00 \times 10^{-3} \mathrm{m}\), we can calculate the resistance of the copper wire: \(R_C = \frac{\rho_C L_C}{\pi r_C^2} \)

Calculate the current through the copper wire.

To obtain the current passing through the copper wire, we can use Ohm's Law: \(V = IR\), where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance. We are given a \(10.0 \mathrm{V}\) battery, so we can solve for the current: \(I_C = \frac{V}{R_C}\)

Calculate the power dissipation in the copper wire.

The power dissipation in the wire can be calculated using the formula \(P = I^2R\). \(P_C = I_C^2 R_C\)

Calculate the resistance of the aluminum wire.

Now that we have the power dissipation of the copper wire, we can find the resistance of the aluminum wire such that it dissipates the same power. \(\rho_{Al} = 2.82 \times 10^{-8} \mathrm{\Omega m}\) We are given the length \(L_{Al} =5.00 \mathrm{m}\) and need to find the radius \(r_{Al}\). The cross-sectional area for the aluminum wire is \(A_{Al} = \pi r_{Al}^2\). We set the power dissipation equal to that of the copper wire: \(P_{Al} = P_C = I_C^2 R_{Al}\) Substituting the formula for resistance, we get: \(I_C^2 \frac{\rho_{Al} L_{Al}}{A_{Al}} = I_C^2 R_C\)

Solve for the radius of the aluminum wire.

Now, we want to isolate \(r_{Al}\) on one side of the equation: \(\frac{I_C^2 \rho_{Al}L_{Al}}{I_C^2 R_C} = \frac{\rho_{Al}L_{Al}}{R_C} = A_{Al} = \pi r_{Al}^2\) Solve for \(r_{Al}\): \(r_{Al} = \sqrt{\frac{\rho_{Al}L_{Al}}{R_C \pi}}\) Plug in the resistivity of aluminum, length of the aluminum wire, and the resistance of the copper wire calculated in step 1: \(r_{Al} = \sqrt{\frac{(2.82 \times 10^{-8} \mathrm{\Omega m})(5.00 \mathrm{m})}{R_C \pi}}\) Finally, we plug in the resistance of the copper wire \(R_C\) found in step 1 and calculate the radius of the aluminum wire: \(r_{Al}\).