/*! 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 21 (II) Two aluminum wires have the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) Two aluminum wires have the same resistance. If one has twice the length of the other, what is the ratio of the diameter of the longer wire to the diameter of the shorter wire?

Short Answer

Expert verified
The diameter ratio of the longer to the shorter wire is \( \sqrt{2} \).

Step by step solution

01

Understand The Resistance Formula

The resistance of a wire is given by the formula \( R = \frac{\rho L}{A} \), 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. For a circular wire, \( A = \frac{\pi d^2}{4} \), where \( d \) is the diameter of the wire.
02

Set Up The Equations

We know that the two wires have the same resistance, so \( R_1 = R_2 \). Let \( L_1 \) and \( L_2 \) be the lengths of the short and long wires, and \( d_1 \) and \( d_2 \) be their diameters, respectively.Since \( L_2 = 2L_1 \), we have:\[ \frac{\rho L_1}{\frac{\pi d_1^2}{4}} = \frac{\rho (2L_1)}{\frac{\pi d_2^2}{4}} \]
03

Simplify The Equation

Cancel \( \rho \) and \( \frac{\pi}{4} \) from both sides and simplify the equation:\[ \frac{L_1}{d_1^2} = \frac{2L_1}{d_2^2} \]
04

Solve For Diameter Ratio

Rearrange the simplified equation to solve for the diameter ratio:\[ \frac{1}{d_1^2} = \frac{2}{d_2^2} \]Cross multiply to get:\[ d_2^2 = 2d_1^2 \]Finally, take the square root of both sides to find the ratio \( \frac{d_2}{d_1} \):\[ \frac{d_2}{d_1} = \sqrt{2} \]
05

Conclusion

The ratio of the diameter of the longer wire to the diameter of the shorter wire is \( \sqrt{2} \).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistivity
Resistivity is a fundamental property of materials that affects how easily electric current can pass through them. It is denoted by the Greek letter \( \rho \) and is measured in ohm-meters (\( \Omega \cdot m \)). The concept of resistivity is crucial for understanding electrical resistance, as it indicates how much a material opposes the flow of electric current. High resistivity means that the material does not conduct electricity well, while low resistivity indicates that it allows electric current to flow more freely.
In practical terms, resistivity helps us determine the resistance of a wire using the formula:
  • \( R = \frac{\rho L}{A} \)
where \( R \) is the resistance, \( L \) is the length of the wire, and \( A \) is the cross-sectional area. Each material has a different resistivity, which affects how we choose materials for wiring based on their ability to conduct electricity efficiently and their ability to withstand various electrical and environmental conditions.
Cross-sectional Area
The cross-sectional area of a wire plays a significant role in determining its resistance. In simple terms, the cross-sectional area is the surface area of the wire's end, which can be thought of as a slice of the wire. For a circular wire, the cross-sectional area \( A \) is calculated using the formula:
  • \( A = \frac{\pi d^2}{4} \)
where \( d \) is the diameter of the wire. A larger cross-sectional area allows more electric current to pass through, resulting in lower resistance. Conversely, a smaller cross-sectional area restricts the flow of current, increasing resistance.
Understanding this concept is important when designing circuits, as the wire’s cross-section must be appropriate for the intended current flow to ensure efficient operation without overheating. Thus, the choice of wire diameter is directly influenced by the desired electrical characteristics.
Diameter Ratio Calculation
Understanding the diameter ratio is essential in exercises where wires of varying lengths have the same resistance, like the one given. In this specific problem, both wires have equal resistance, but their lengths differ, with one being twice as long as the other. Therefore, we need to calculate the ratio of their diameters.
We start by using the resistance formula \( R = \frac{\rho L}{A} \). Given that the resistance \( R \) is the same for both wires, we set up the equation:
  • \( \frac{L_1}{d_1^2} = \frac{2L_1}{d_2^2} \)
Cross-multiplying yields: \( d_2^2 = 2d_1^2 \). By taking the square root of both sides, we find that:
  • \( \frac{d_2}{d_1} = \sqrt{2} \)
This calculation shows that the diameter of the longer wire is \( \sqrt{2} \) times that of the shorter wire. This ratio ensures that despite differing lengths, both wires possess the same resistance. This principle is crucial in applications where maintaining constant electrical characteristics is important, even when physical dimensions change.

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

What is the average current drawn by a 1.0 -hp \(120-\mathrm{V}\) motor? \((1 \mathrm{hp}=746 \mathrm{~W} .)\)

(II) A length of wire is cut in half and the two lengths are wrapped together side by side to make a thicker wire. How does the resistance of this new combination compare to the resistance of the original wire?

The Tevatron accelerator at Fermilab (Illinois) is designed to carry an 11 -mA beam of protons traveling at very nearly the speed of light \(\left(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\) around a ring \(6300 \mathrm{~m}\) in circumference. How many protons are in the beam?

(II) An electric clothes dryer has a heating element with a resistance of \(8.6 \Omega .\) (a) What is the current in the element when it is connected to \(240 \mathrm{~V} ?(b)\) How much charge passes through the element in \(50 \mathrm{~min}\) ? (Assume direct current.)

A proposed electric vehicle makes use of storage batteries as its source of energy. Its mass is \(1560 \mathrm{~kg}\) and it is powered by 24 batteries, each \(12 \mathrm{~V}, 95 \mathrm{~A} \cdot \mathrm{h}\). Assume that the car is driven on level roads at an average speed of \(45 \mathrm{~km} / \mathrm{h},\) and the average friction force is \(240 \mathrm{~N}\). Assume \(100 \%\) efficiency and neglect energy used for acceleration. No energy is consumed when the vehicle is stopped, since the engine doesn't need to idle. (a) Determine the horsepower required. (b) After approximately how many kilometers must the batteries be recharged?

See all solutions

Recommended explanations on Physics Textbooks

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.