Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 25: Problem 38

A potential difference of \(12.0 \mathrm{~V}\) is applied across a wire of cross-sectional area \(4.50 \mathrm{~mm}^{2}\) and length \(1000 . \mathrm{km} .\) The current passing through the wire is \(3.20 \cdot 10^{-3} \mathrm{~A}\). a) What is the resistance of the wire? b) What type of wire is this?

Short Answer

Expert verified
Answer: The wire is likely made of copper.

Step by step solution

01

Find the resistance using Ohm's Law

Ohm's Law states that \(V = IR\), where \(V\) is the voltage, \(I\) is the current, and \(R\) is the resistance. We have \(V = 12.0 \mathrm{~V}\) and \(I = 3.20 \cdot 10^{-3} \mathrm{~A}\). To find the resistance, we can rearrange the formula to get \(R = V/I\). So, \(R = \frac{12.0 \mathrm{~V}}{3.20 \cdot 10^{-3} \mathrm{~A}}\).
02

Calculate the resistance

Now, we can plug in the values to find the resistance: \(R = \frac{12.0 \mathrm{~V}}{3.20 \cdot 10^{-3} \mathrm{~A}} = 3750 \Omega\).
03

Rearrange the resistance equation to find resistivity

We can use the formula \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length of the wire, and \(A\) is the cross-sectional area. We want to find the resistivity, so we can rearrange the formula to get \(\rho = R \frac{A}{L}\).
04

Convert the length and area to the proper units

The length is given in kilometers, which we need to convert to meters. We know that 1 km = 1000 m, so \(1000 \mathrm{~km} = 1000 \cdot 1000 \mathrm{~m} = 1 \cdot 10^{6} \mathrm{~m}\). The cross-sectional area is given in mm², which we need to convert to m². We know that 1 m² = 1,000,000 mm², so \(4.50 \mathrm{~mm}^{2} = \frac{4.50}{1,000,000} \mathrm{~m}^{2} = 4.50 \cdot 10^{-6} \mathrm{~m}^{2}\).
05

Calculate the resistivity

Now we can plug in the values to find the resistivity: \(\rho = 3750 \Omega \cdot \frac{4.50 \cdot 10^{-6} \mathrm{~m}^{2}}{1 \cdot 10^{6} \mathrm{~m}} = 1.69 \cdot 10^{-8} \Omega \cdot \mathrm{m}\).
06

Identify the type of wire

The resistivity value we found, \(1.69 \cdot 10^{-8} \Omega \cdot \mathrm{m}\), is close to the resistivity of copper, which is approximately \(1.68 \cdot 10^{-8} \Omega \cdot \mathrm{m}\). Therefore, the wire is likely made of copper.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

If the current through a resistor is increased by a factor of \(2,\) how does this affect the power that is dissipated? a) It decreases by a factor of 4 . b) It increases by a factor of 2 . c) It decreases by a factor of 8 . d) It increases by a factor of 4 .

Which of the following wires has the largest current flowing through it? a) a 1 -m-long copper wire of diameter \(1 \mathrm{~mm}\) connected to a \(10-V\) battery b) a \(0.5-\mathrm{m}\) -long copper wire of diameter \(0.5 \mathrm{~mm}\) connected to a 5 -V battery c) a 2 -m-long copper wire of diameter \(2 \mathrm{~mm}\) connected to a \(20-V\) battery d) a \(1-\mathrm{m}\) -long copper wire of diameter \(0.5 \mathrm{~mm}\) connected to a 5 -V battery e) All of the wires have the same current flowing through them.

An infinite number of resistors are connected in parallel. If \(R_{1}=10 \Omega, R_{2}=10^{2} \Omega, R_{3}=10^{3} \Omega,\) and so on, show that \(R_{e q}=9 \Omega\).

A charged-particle beam is used to inject a charge, \(Q_{0}\), into a small, irregularly shaped region (not a cavity, just some region within the solid block) in the interior of a block of ohmic material with conductivity \(\sigma\) and permittivity \(\epsilon\) at time \(t=0\). Eventually, all the injected charge will move to the outer surface of the block, but how quickly? a) Derive a differential equation for the charge, \(Q(t)\), in the injection region as a function of time. b) Solve the equation from part (a) to find \(Q(t)\) for all \(t \geq 0\). c) For copper, a good conductor, and for quartz (crystalline \(\mathrm{SiO}_{2}\) ), an insulator, calculate the time for the charge in the injection region to decrease by half. Look up the necessary values. Assume that the effective "dielectric constant" of copper is \(1.00000 .\)

One brand of \(12.0-\mathrm{V}\) automotive battery used to be advertised as providing " 600 cold-cranking amps." Assuming that this is the current the battery supplies if its terminals are shorted, that is, connected to negligible resistance, determine the internal resistance of the battery.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Translational Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Physical Quantities And Units

Read Explanation

Torque and Rotational 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.

Sign-up for free