/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 Two cylindrical wires of identic... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 66

Two cylindrical wires of identical length are made of copper and aluminum. If they carry the same current and have the same potential difference across their length, what is the ratio of their radii?

Short Answer

Expert verified
Answer: The approximate ratio of the radii of the copper and aluminum wires is 1.296.

Step by step solution

01

1. Write the resistance formula for each wire

For the copper wire, the resistance is given by: R_Cu = 蟻_Cu * L / A_Cu For the aluminum wire, the resistance is given by: R_Al = 蟻_Al * L / A_Al
02

2. Express the area in terms of radii

Both wires are cylindrical, so the cross-sectional area is given by A = 蟺r^2. A_Cu = 蟺 * r_Cu^2 A_Al = 蟺 * r_Al^2 Thus, we can rewrite the resistance formulas as follows: R_Cu = 蟻_Cu * L / (蟺 * r_Cu^2) R_Al = 蟻_Al * L / (蟺 * r_Al^2)
03

3. Set the resistances equal to each other

Since both wires have the same current and potential difference, their resistances are equal. We can equate the above formulas: 蟻_Cu * L / (蟺 * r_Cu^2) = 蟻_Al * L / (蟺 * r_Al^2)
04

4. Simplify and solve for the ratio of radii

We can cancel out the L and 蟺 terms from both sides and rearrange the equation to find the ratio of their radii: r_Cu^2 / r_Al^2 = 蟻_Al / 蟻_Cu Take the square root of both sides: r_Cu / r_Al = sqrt(蟻_Al / 蟻_Cu)
05

5. Calculate the ratio using known resistivity values

Now, we can plug in the known resistivity values for copper (蟻_Cu = 1.68 脳 10^-8 惟鈰卪) and aluminum (蟻_Al = 2.82 脳 10^-8 惟鈰卪) to find the ratio of their radii: r_Cu / r_Al = sqrt((2.82 脳 10^-8 惟鈰卪) / (1.68 脳 10^-8 惟鈰卪)) r_Cu / r_Al 鈮� sqrt(1.679) 鈮� 1.296 So, the ratio of the radii of the copper and aluminum wires is approximately 1.296.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three resistors are connected to a power supply with \(V=110 . \mathrm{V}\) as shown in the figure a) Find the potential drop across \(R_{3}\) b) Find the current in \(R_{1}\). c) Find the rate at which thermal energy is dissipated from \(R_{2}\).

A hair dryer consumes \(1600 .\) W of power and operates at \(110 .\) V. (Assume that the current is \(D C .\) In fact, these are root-mean-square values of AC quantities, but the calculation is not affected. Chapter 30 covers AC circuits in detail.) a) Will the hair dryer trip a circuit breaker designed to interrupt the circuit if the current exceeds \(15.0 \mathrm{~A} ?\) b) What is the resistance of the hair dryer when it is operating?

A material is said to be ohmic if an electric field, \(\vec{E}\), in the material gives rise to current density \(\vec{J}=\sigma \vec{E},\) where the conductivity, \(\sigma\), is a constant independent of \(\vec{E}\) or \(\vec{J}\). (This is the precise form of Ohm's Law.) Suppose in some material an electric field, \(\vec{E}\), produces current density, \(\vec{J},\) not necessarily related by Ohm's Law; that is, the material may or may not be ohmic. a) Calculate the rate of energy dissipation (sometimes called ohmic heating or joule heating) per unit volume in this material, in terms of \(\vec{E}\) and \(\vec{J}\). b) Express the result of part (a) in terms of \(\vec{E}\) alone and \(\vec{J}\) alone, for \(\vec{E}\) and \(\vec{J}\) related via Ohm's Law, that is, in an ohmic material with conductivity \(\sigma\) or resistivity \(\rho .\)

A constant electric field is maintained inside a semiconductor. As the temperature is lowered, the magnitude of the current density inside the semiconductor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

Before bendable tungsten filaments were developed, Thomas Edison used carbon filaments in his light bulbs. Though carbon has a very high melting temperature \(\left(3599^{\circ} \mathrm{C}\right)\) its sublimation rate is high at high temperatures. So carbonfilament bulbs were kept at lower temperatures, thereby rendering them dimmer than later tungsten-based bulbs. A typical carbon-filament bulb requires an average power of \(40 \mathrm{~W}\), when 110 volts is applied across it, and has a filament temperature of \(1800^{\circ} \mathrm{C}\). Carbon, unlike copper, has a negative temperature coefficient of resistivity: \(\alpha=-0.0005^{\circ} \mathrm{C}^{-1}\) Calculate the resistance at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) of this carbon filament.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Measurements

Read Explanation

Waves Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Physical Quantities And Units

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.