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Chapter 18: Problem 23

A copper wire and an aluminum wire of the same length have the same resistance. What is the ratio of the diameter of the copper wire to that of the aluminum wire?

Short Answer

Expert verified
Answer: The approximate ratio of the diameter of the copper wire to that of the aluminum wire is 1.30.

Step by step solution

01

Find the formula for resistance of a wire

To find the resistance of a wire, we use the formula: Resistance (R) = (Resistivity (ρ) × Length (L)) / Cross-sectional Area (A) Since both wires have the same length and resistance, we can write the equation as: R_copper = R_aluminum (ρ_copper × L) / A_copper = (ρ_aluminum × L) / A_aluminum
02

Use the resistivity values for copper and aluminum

The standard resistivity values for copper and aluminum at room temperature are: ρ_copper = 1.68 × 10^{-8} ohm·m ρ_aluminum = 2.82 × 10^{-8} ohm·m Using the formula from step 1, we get: (1.68 × 10^{-8}) / A_copper = (2.82 × 10^{-8}) / A_aluminum
03

Solve for the ratio of cross-sectional areas

To find the ratio of cross-sectional areas, we can rearrange the equation from step 2: A_copper / A_aluminum = (2.82 × 10^{-8}) / (1.68 × 10^{-8}) = 1.67857 A_copper / A_aluminum ≈ 1.68
04

Convert the cross-sectional area ratio to a diameter ratio

Since both wires are cylindrical, we can use the formula for the cross-sectional area of a circle (A = π * (diameter^2) / 4) and set up a proportion for their diameters: (A_copper / A_aluminum) = (d_copper^2 / d_aluminum^2) 1.68 = (d_copper^2 / d_aluminum^2) Now, take the square root of both sides: sqrt(1.68) = (d_copper / d_aluminum) d_copper / d_aluminum ≈ 1.30 The ratio of the diameter of the copper wire to that of the aluminum wire is approximately 1.30.

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Most popular questions from this chapter

A galvanometer has a coil resistance of \(34.0 \Omega .\) It is to be made into a voltmeter with a full-scale deflection equal to \(100.0 \mathrm{V}\). If the galvanometer deflects full scale for a current of \(0.120 \mathrm{mA},\) what size resistor should be placed in series with the galvanometer?
A gold wire of 0.50 mm diameter has \(5.90 \times 10^{28} \mathrm{con}\) duction electrons/m". If the drift speed is \(6.5 \mu \mathrm{m} / \mathrm{s}\), what is the current in the wire?
A gold wire and an aluminum wire have the same dimensions and carry the same current. The electron density (in electrons/cm \(^{3}\) ) in aluminum is three times larger than the density in gold. How do the drift speeds of the electrons in the two wires, \(v_{\mathrm{Au}}\) and \(v_{\mathrm{Al}},\) compare?
Poiseuille's law [Eq. (9-15)] gives the volume flow rate of a viscous fluid through a pipe. (a) Show that Poiseuille's law can be written in the form \(\Delta P=I R\) where \(I=\Delta V / \Delta t\) represents the volume flow rate and \(R\) is a constant of proportionality called the fluid flow resistance. (b) Find \(R\) in terms of the viscosity of the fluid and the length and radius of the pipe. (c) If two or more pipes are connected in series so that the volume flow rate through them is the same, do the resistances of the pipes add as for electrical resistors \(\left(R_{e q}=R_{1}+R_{2}+\cdots\right) ?\) Explain. (d) If two or more pipes are connected in parallel, so the pressure drop across them is the same, do the reciprocals of the resistances add as for electrical resistors \(\left(1 / R_{\mathrm{eq}}=\right.\) $\left.1 / R_{1}+1 / R_{2}+\cdots\right) ?$ Explain.
Chelsea inadvertently bumps into a set of batteries with an emf of $100.0 \mathrm{V}\( that can supply a maximum power of \)5.0 \mathrm{W} .$ If the resistance between the points where she contacts the batteries is $1.0 \mathrm{k} \Omega,$ how much current passes through her?
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