Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 27: Problem 25

A 220 -mH inductor carries 350 mA. How much energy must be supplied to the inductor in raising the current to 800 mA?

Short Answer

Expert verified
The energy required to raise the current to 800mA is the difference between the final energy and the initial energy, which can be calculated by plugging in the given values in the equations in step 3.

Step by step solution

01

Calculate the initial energy

Calculate the energy of the inductor when the current is 350mA using the formula \(E1 = \frac{1}{2}*L*I1^2\), where L=220mH and \(I1=350mA =0.35A\). Convert the inductance to Henry by dividing by 1000, to get L=0.22H.
02

Calculate the final energy

Calculate the energy of the inductor when the current is 800mA using the formula \(E2 = \frac{1}{2}*L*I2^2\), where \(I2=800mA =0.8A\).
03

Calculate the energy difference

Calculate the difference between the final energy \(E2\) and the initial energy \(E1\). That is the energy needed to raise the current to 800mA.

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

During lab, you're given a circular wire loop of resistance \(R\) and radius \(a\) with its plane perpendicular to a uniform magnetic field. You're supposed to increase the field strength from \(B_{1}\) to \(B_{2}\) and measure the total charge that moves around the loop. Your lab partner claims that the details of how you vary the field will make a difference in the total charge; your hunch is that it won't. By integrating the loop current over time, determine who's right.

A 2.0 -A current is flowing in a \(20-\mathrm{H}\) inductor. A switch opens, interrupting the current in 1.0 ms. Find the induced emf in the inductor.

A single-turn loop of radius \(R\) carries current \(I\). How does the magnetic- energy density at the loop center compare with that of a long solenoid of the same radius, carrying the same current, and consisting of \(n\) turns per unit length?

A conducting disk with radius \(a,\) thickness \(h,\) and resistivity \(\rho\) is inside a solenoid of circular cross section, its axis coinciding with the solenoid axis. The magnetic field in the solenoid is given by \(B=b t,\) where \(b\) is a constant. Find expressions for (a) the current density in the disk as a function of the distance \(r\) from the disk center and (b) the power dissipation in the entire disk. (Hint: Consider the disk as consisting of infinitesimal conducting loops.)

You have a fixed length of wire to wind into an inductor. Will you get more inductance if you wind a short coil with large diameter, or a long coil with small diameter?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Torque and Rotational Motion

Read Explanation

Scientific Method Physics

Read Explanation

Dynamics

Read Explanation

Oscillations

Read Explanation

Quantum Physics

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