/*! 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 25 A wire with a resistance of \(6.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A wire with a resistance of \(6.0 \Omega\) is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

Short Answer

Expert verified
The resistance of the longer wire is \(54.0\,\Omega\).

Step by step solution

01

Understand the relationship between resistance and wire dimensions

The resistance of a wire is given by the equation \( R = \rho \frac{L}{A} \), where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area. As the length of the wire increases, the cross-sectional area decreases if the volume of the wire remains constant.
02

Use the volume conservation principle

Since the density of the material remains unchanged, the volume before and after drawing the wire remains the same. For a wire with initial length \( L_1 \) and cross-sectional area \( A_1 \) and final length \( L_2 = 3L_1 \), the initial and final volumes \( V \) are \( V_1 = L_1 A_1 \) and \( V_2 = L_2 A_2 \). Therefore, \( L_1 A_1 = L_2 A_2 \).
03

Substitute the given relationship and simplify

By substituting \( L_2 = 3L_1 \) into the volume conservation equation, you get \( L_1 A_1 = 3L_1 A_2 \). Simplifying, this gives \( A_2 = \frac{A_1}{3} \).
04

Find the new resistance

Using the resistance formula \( R_2 = \rho \frac{L_2}{A_2} \) and substituting \( L_2 = 3L_1 \) and \( A_2 = \frac{A_1}{3} \), we get \( R_2 = \rho \frac{3L_1}{A_1/3} = \rho \frac{9L_1}{A_1} \). Since the original resistance \( R_1 = \rho \frac{L_1}{A_1} = 6.0\,\Omega \), the new resistance is \( R_2 = 9 \times 6.0\,\Omega = 54.0\,\Omega \).

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 quantifies how strongly a given material opposes the flow of electric current. It is denoted by the symbol \( \rho \) and is expressed in ohm-meters (\( \Omega \cdot m \)).
In simpler terms, resistivity tells us how much resistance a material will provide against electrical conduction.
This property depends on the material's intrinsic nature and remains unchanged unless there is a change in the material itself, such as its temperature.
For example, the resistivity of copper is lower than that of rubber, which is why copper is commonly used in electrical wires.
  • Materials with low resistivity are good conductors (e.g., metals).
  • Materials with high resistivity are insulators (e.g., rubber).
Wire Length
The length of a wire is one of the key factors that affect its resistance. According to the formula \( R = \rho \frac{L}{A} \), resistance \( R \) is directly proportional to the length \( L \) of the wire.
This means that when the length of the wire increases, so does its resistance, assuming that other factors like resistivity and cross-sectional area remain constant.
In the given exercise, the wire's length was tripled, leading to an increase in resistance.
Thus, we must consider length changes when calculating electrical resistance for wires of different sizes.
  • Longer wires have higher resistance because electrons have to travel a greater distance.
  • Shorter wires provide less resistance as the path for electron flow is reduced.
Cross-Sectional Area
The cross-sectional area (A) of a wire is also crucial in determining its resistance. It appears in the denominator of the resistance formula \( R = \rho \frac{L}{A} \), indicating that resistance is inversely proportional to the cross-sectional area.
Simply put, if you increase the area, the resistance decreases, and vice versa. When a wire is stretched and its length increased, its cross-sectional area decreases if the volume is conserved.
For the exercise, as the wire was lengthened to three times its original size, its cross-sectional area decreased to one-third, affecting the overall resistance.
  • A larger cross-sectional area allows more current to pass through with less resistance.
  • Smaller cross-sectional areas restrict the flow, increasing resistance.
Volume Conservation
Volume conservation is a principle often applied when dealing with changes in the dimensions of objects, like wires.
The principle states that when an object is deformed, like being stretched, its volume remains constant if the material density is unchanged.
For example, if we extend a wire, making it longer, the cross-sectional area must decrease for the volume to remain constant.
In the exercise, the wire's length increased, which caused the cross-sectional area to decrease, all while retaining its original volume.
This was crucial for calculating the new resistance because it guided us to adjust the area calculation accordingly. Volume conservation helps in maintaining a balance of properties when an object's shape changes.
  • If a wire's length triples, its cross-sectional area becomes a third due to constant volume.
  • Volume conservation aligns with keeping the material's mass constant.

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

A \(30 \mu \mathrm{F}\) capacitor is connected across a programmed power supply. During the interval from \(t=0\) to \(t=3.00 \mathrm{~s}\) the output voltage of the supply is given by \(V(t)=6.00+4.00 t-2.00 t^{2}\) volts. At \(t=0.500 \mathrm{~s}\) find (a) the charge on the capacitor, (b) the current into the capacitor, and (c) the power output from the power supply.

Current is set up through a truncated right circular cone of resistivity \(731 \Omega \cdot \mathrm{m}\), left radius \(a=2.00 \mathrm{~mm}\), right radius \(b=2.30 \mathrm{~mm}\), and length \(L=1.94 \mathrm{~cm}\). Assume that the current density is uniform across any cross section taken perpendicular to the length. What is the resistance of the cone?

A \(100 \mathrm{~W}\) lightbulb is plugged into a standard \(120 \mathrm{~V}\) outlet. (a) How much does it cost per 31 -day month to leave the light turned on continuously? Assume electrical energy costs US \(\$ 0.06 / \mathrm{kW} \cdot \mathrm{h} .\) (b) What is the resistance of the bulb? (c) What is the current in the bulb?

A \(120 \mathrm{~V}\) potential difference is applied to a space heater whose resistance is \(14 \Omega\) when hot. (a) At what rate is electrical energy transferred to thermal energy? (b) What is the cost for \(5.0 \mathrm{~h}\) at \(\mathrm{US} \$ 0.05 / \mathrm{kW} \cdot \mathrm{h} ?\)

A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to \(440 \mathrm{~A} / \mathrm{cm}^{2} .\) What diameter of cylindrical wire should be used to make a fuse that will limit the current to \(0.50 \mathrm{~A} ?\)

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.