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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 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?
Near Earth's surface the air contains both negative and positive ions, due to radioactivity in the soil and cosmic rays from space. As a simplified model, assume there are 600.0 singly charged positive ions per \(\mathrm{cm}^{3}\) and 500.0 singly charged negative ions per \(\mathrm{cm}^{3} ;\) ignore the presence of multiply charged ions. The electric field is $100.0 \mathrm{V} / \mathrm{m},$ directed downward. (a) In which direction do the positive ions move? The negative ions? (b) What is the direction of the current due to these ions? (c) The measured resistivity of the air in the region is $4.0 \times 10^{13} \Omega \cdot \mathrm{m} .$ Calculate the drift speed of the ions, assuming it to be the same for positive and negative ions. [Hint: Consider a vertical tube of air of length \(L\) and cross-sectional area \(A\) How is the potential difference across the tube related to the electric field strength?] (d) If these conditions existed over the entire surface of the Earth, what is the total current due to the movement of ions in the air?
The current in the electron beam of a computer monitor is \(320 \mu\) A. How many electrons per second hit the screen?
Consider a \(60.0-\mathrm{W}\) lightbulb and a \(100.0-\mathrm{W}\) lightbulb designed for use in a household lamp socket at \(120 \mathrm{V}\) (a) What are the resistances of these two bulbs? (b) If they are wired together in a series circuit, which bulb shines brighter (dissipates more power)? Explain. (c) If they are connected in parallel in a circuit, which bulb shines brighter? Explain.
Copper and aluminum are being considered for the cables in a high-voltage transmission line where each must carry a current of \(50 \mathrm{A}\). The resistance of each cable is to be \(0.15 \Omega\) per kilometer. (a) If this line carries power from Niagara Falls to New York City (approximately $500 \mathrm{km}),$ how much power is lost along the way in the cable? Compute for each choice of cable material (b) the necessary cable diameter and (c) the mass per meter of the cable. The electrical resistivities for copper and aluminum are given in Table \(18.1 ;\) the mass density of copper is $8920 \mathrm{kg} / \mathrm{m}^{3}\( and that of aluminum is \)2702 \mathrm{kg} / \mathrm{m}^{3}$
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