/*! 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 119 A piece of gold wire of length \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 119

A piece of gold wire of length \(L\) has a resistance \(R_{0}\) Suppose the wire is drawn out so that its length increases by a factor of three. What is the new resistance \(R\) in terms of the original resistance?

Short Answer

Expert verified
Answer: The new resistance is 9 times the original resistance.

Step by step solution

01

Define the variables and given information

We are given: - Original length of the wire: \(L\) - Original resistance of the wire: \(R_{0}\) - New length of the wire: \(3L\) We need to find the new resistance, \(R\), when the length of the wire is increased by a factor of three.
02

Use the formula for resistance

The formula for resistance is: $$R=\rho\frac{L}{A}$$ where: - \(R\) is the resistance, - \(\rho\) is the resistivity of the material (constant for a given material), - \(L\) is the length of the wire, - \(A\) is the cross-sectional area of the wire.
03

Use the fact that volume remains constant

When the wire is drawn out, its volume remains constant. Therefore, we can write the equation for the volume of the wire before and after being drawn out as: $$A \cdot L = A' \cdot 3L$$ where \(A'\) is the new cross-sectional area of the wire. Now, we can solve for \(A'\): $$A'=\frac{A}{3}$$
04

Calculate the new resistance

Using the formula for resistance, we can find the new resistance \(R\) as follows: $$R=\rho\frac{3L}{\frac{A}{3}}=\rho\frac{9L}{A}$$
05

Relate the new resistance to the original resistance

Now let's express the new resistance in terms of the original resistance \(R_{0}\). We know that: $$R_{0}=\rho\frac{L}{A}$$ Now we can solve for \(\rho\frac{L}{A}\): $$\rho\frac{L}{A} = R_{0}$$
06

Substitute the expression for the original resistance into the new resistance formula

Now, using our expression for \(\rho\frac{L}{A}\) from Step 5, we can write the new resistance as: $$R=9(\rho\frac{L}{A})=9R_{0}$$
07

Conclusion

The new resistance \(R\) is 9 times the original resistance \(R_{0}\). So, we have found that: $$R=9R_{0}$$

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

Suppose a collection of five batteries is connected as shown. (a) What is the equivalent emf of the collection? Treat them as ideal sources of emf. (b) What is the current through the resistor if its value is \(3.2 \Omega ?\)
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?

A common flashlight bulb is rated at \(0.300 \mathrm{A}\) and \(2.90 \mathrm{V}\) (the values of current and voltage under operating conditions). If the resistance of the bulb's tungsten filament at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) is \(1.10 \Omega,\) estimate the temperature of the tungsten filament when the bulb is turned on.
A 1.5 -horsepower motor operates on \(120 \mathrm{V}\). Ignoring \(l^{2} R\) losses, how much current does it draw?
A 2.00 - \(\mu\) F capacitor is charged using a \(5.00-\mathrm{V}\) battery and a 3.00 - \(\mu\) F capacitor is charged using a \(10.0-\mathrm{V}\) battery. (a) What is the total energy stored in the two capacitors? (b) The batteries are disconnected and the two capacitors are connected together \((+\text { to }+\) and \- to \(-\) ). Find the charge on each capacitor and the total energy in the two capacitors after they are connected. (c) Explain what happened to the "missing" energy. [Hint: The wires that connect the two have some resistance.]
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Fundamentals of Physics

Read Explanation

Oscillations

Read Explanation

Nuclear Physics

Read Explanation

Rotational Dynamics

Read Explanation

Translational Dynamics

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.