Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 30: Q50P (page 899)

The current in an RL circuit builds up to one-third of its steady-state value in 5.00 s . Find the inductive time constant.

Short Answer

Expert verified

τL=12.3s

Step by step solution

01

Given

02

Understanding the concept

im-currentatsteadyvalueIf a constant emf εis introduced into a single-loop circuit containing a resistance Rand an inductance L, the current rises to an equilibrium value ofε/Ras

i=εR(1-e-tτL)

Formulae:

i=εR(1-e-tτL)

Here,ε-emf;R-resistance;t-time;τL-inductivetimeconstantofcircuit

Given:

At t=5.0s,i=im/3

03

Calculate the inductive time constant

We have current in the inductive circuit as\

i=εR(1-e-tτL)

We know that steady state current of RL circuit is

im=ε/R

Now plug in this value in equation:

im3=εR1-e-tτLε3R=εR1-e-tτLe-tτL=1-13=2/3

Hence,

-tτL=In23--0.5τL=-0.4054

Therefore,

τL=12.3s

The inductive time constant is 12.3s

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

How long would it take, following the removal of the battery, for the potential difference across the resistor in an RL circuit (with L = 2.00H, R = 3.00) to decay to 10.0% of its initial value?

Figure 30-24 shows two circuits in which a conducting bar is slid at the same speed vthrough the same uniform magnetic field and along a U-shaped wire. The parallel lengths of the wire are separated by 2Lin circuit 1 and by Lin circuit 2. The current induced in circuit 1 is counter clockwise. (a) Is the magnetic field into or out of the page? (b) Is the current induced in circuit 2 clockwise or counter clockwise? (c) Is the emf induced in circuit 1 larger than, smaller than, or the same as that in circuit 2?

A long cylindrical solenoid with 100 turns/cmhas a radius of 1.6 cm. Assume that the magnetic field it produces is parallel to its axis and is uniform in its interior. (a) What is its inductance per meter of length? (b) If the current changes at the rate of 13A/s, what emf is induced per meter?

A long solenoid has a diameter of 12.0 cm.When a current i exists in its windings, a uniform magnetic field of magnitude B = 30.0 mTis produced in its interior. By decreasing i, the field is caused to decrease at the rate of 6.50mTs.(a) Calculate the magnitude of the induced electric field 2.20 cmfrom the axis of the solenoid.

(b) Calculate the magnitude of the induced electric field 8.20 cmfrom the axis of the solenoid.

A circular loop of wire 50 mmin radius carries a current of 100 A. (a) Find the magnetic field strength. (b) Find the energy density at the center of the loop
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Modern Physics

Read Explanation

Radiation

Read Explanation

Physical Quantities And Units

Read Explanation

Thermodynamics

Read Explanation

Nuclear 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