/*! 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 18 What diameter must a copper wire... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 18

What diameter must a copper wire have if its resistance is to be the same as that of an equal length of aluminum wire with diameter 3.26 \(\mathrm{mm}\) ?

Short Answer

Expert verified
The copper wire must have a diameter of approximately 2.59 mm.

Step by step solution

01

Understand the Formula for Resistance

The resistance of a wire is given by the formula: \[ R = \frac{\rho L}{A} \]where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. The cross-sectional area of a wire is \( A = \pi \left(\frac{d}{2}\right)^2 \), where \( d \) is the diameter.
02

Calculate the Resistance of Aluminum Wire

For aluminum wire, use the formula: \[ R_{Al} = \frac{\rho_{Al} L}{\pi \left(\frac{d_{Al}}{2}\right)^2} \]Given \( d_{Al} = 3.26 \ \text{mm} = 0.00326 \ \text{m} \) and known resistivity of aluminum, \( \rho_{Al} \approx 2.65 \times 10^{-8} \ \Omega \cdot \text{m} \). The resistance simplifies to a product of known quantity since the exact length cancels out in both equations as we're equating resistances.
03

Set Up the Equality for Resistances

Since both wires have the same resistance, set up the equation:\[ \frac{\rho_{Al} L}{\pi \left(\frac{d_{Al}}{2}\right)^2} = \frac{\rho_{Cu} L}{\pi \left(\frac{d_{Cu}}{2}\right)^2} \] Cancel \( L \) and \( \pi \) from both sides.
04

Solve for Diameter of Copper Wire

Rearrange the equation to solve for \(d_{Cu}^2\):\[ \left(\frac{d_{Cu}}{2}\right)^2 = \frac{\rho_{Cu}}{\rho_{Al}} \left(\frac{d_{Al}}{2}\right)^2 \]Given \( \rho_{Cu} \approx 1.68 \times 10^{-8} \ \Omega \cdot \text{m} \), substitute values and solve for \(d_{Cu}\):\[ d_{Cu} = 2 \sqrt{\frac{\rho_{Cu}}{\rho_{Al}}} \times \frac{d_{Al}}{2} \approx 2 \sqrt{\frac{1.68}{2.65}} \times \frac{0.00326}{2} \]
05

Calculate Final Result

Do the calculations to find the diameter of the copper wire:\[ \frac{1.68}{2.65} \approx 0.634 \]\[ \sqrt{0.634} \approx 0.796 \]\[ d_{Cu} \approx 2 \times 0.796 \times 0.00163 \approx 0.00259 \ \text{m} \approx 2.59 \ \text{mm} \]

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!

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 defines how strongly a material opposes the flow of electric current. It is represented by the symbol \( \rho \) and measured in ohm-meters (\( \Omega \cdot \text{m} \)). A lower resistivity indicates that a material easily allows the passage of electric current.
For wires, knowing the resistivity helps us understand how various materials behave when used to conduct electricity.
  • Copper and aluminum are common conductive materials, but they have different resistivities.
  • Copper typically has a resistivity of about \( 1.68 \times 10^{-8} \, \Omega \cdot \text{m} \), making it an excellent conductor.
  • Aluminum has a higher resistivity of around \( 2.65 \times 10^{-8} \, \Omega \cdot \text{m} \), making it slightly less efficient for conducting electricity.
Understanding resistivity is essential when designing circuits or electrical systems to ensure optimal performance and efficiency.
Cross-sectional Area
The cross-sectional area of a wire is crucial in determining its resistance, with larger areas reducing resistance. The area is calculated using the formula \( A = \pi \left(\frac{d}{2}\right)^2 \), where \( d \) is the diameter of the wire. This formula shows how the resistance is inversely proportional to the area.
  • A larger diameter results in a larger cross-sectional area, reducing resistance.
  • The decrease in resistance with a larger area is due to more pathways for electrons to travel without colliding.
Thus, when comparing wires like copper and aluminum, adjusting the diameter affects how they conduct electricity. For a given length, the wire with a larger diameter will have less resistance, assuming the material resistivity remains constant.
Copper and Aluminum Wire Resistance Comparison
Comparing copper and aluminum wire requires considering both resistivity and diameter to achieve the same resistance. Since copper has a lower resistivity, it naturally provides lower resistance per unit length if the diameter is constant.
When equating resistance of copper and aluminum wires of the same length:
  • Copper's lower resistivity allows for a smaller diameter to achieve the same resistance.
  • In our example, to equalize resistance with aluminum (diameter \(3.26\) mm), the copper wire must have a smaller diameter of approximately \(2.59\) mm.
  • This difference highlights copper's superior conductivity, allowing it to transmit electricity effectively even with reduced size.
Choosing between copper and aluminum wires involves balancing cost, weight, and electrical efficiency, with copper often being favored in applications where space and performance are priorities.

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

The current in a wire varies with time according to the relationship \(I=55 \mathrm{A}-\left(0.65 \mathrm{A} / \mathrm{s}^{2}\right) t^{2}\) . (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t=0\) and \(t=8.0 \mathrm{s} ?(\mathrm{b})\) What constant current would transport the same charge in the same time interval?

A typical cost for electric power is 12.0\(\phi\) per kilowatt-hour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a \(75-\mathrm{W}\) bulb burning day and night? (b) Suppose your refrigerator uses 400 \(\mathrm{W}\) of power when it's running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?

A resistor with a \(15.0 .\) V potential difference across its ends develops thermal energy at a rate of 327 \(\mathrm{W}\) (a) What is its resistance? (b) What is the current in the resistor?

The following measurements were made on a Thyrite resistor: $$ \begin{array}{l|llll}{I(A)} & {0.50} & {1.00} & {2.00} & {4.00} \\\ {V_{\omega}(V)} & {2.55} & {3.11} & {3.77} & {4.58}\end{array} $$ (a) Graph \(V_{a b}\) as a function of \(I .\) (b) Does Thyrite obey Ohm's law? How can you tell?(c) Graph the resistance \(R=V_{a b} / I\) as a function of \(I .\)

A strand of wire has resistance 5.60\(\mu \Omega\) . Find the net resistance of 120 such strands if they are (a) placed side by side to form a cable of the same length as a single strand, and (b) connected end to end to form a wire 120 times as long as a single strand.
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Wave Optics

Read Explanation

Scientific Method Physics

Read Explanation

Classical Mechanics

Read Explanation

Nuclear Physics

Read Explanation

Physics of Motion

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.