/*! 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} Q37E An LC聽circuit containing an 80.... [FREE SOLUTION] | 91影视

91影视

An LCcircuit containing an 80.0mH inductor and a 1.25nF capacitor oscillates with a maximum current of 0.750 A. Assuming the capacitor had its maximum charge at time t= 0, calculate the energy stored in the inductor after 2.50 ms of oscillation.

Short Answer

Expert verified

The energy stored in the inductor after 2.50 ms of oscillation is 0.0212J.

Step by step solution

01

Energy conservation in LC circuit

The stored energy in capacitor in addition to the stored energy in inductor at any instant equals to the maximum stored energy in the capacitor.

Mathematically it can be given by

UL+UC=UCmax12Li2+q22C=Q22C

Where , UCmaxis the maximum energy stored in capacitor and ULand UCare the energy stored in inductor and capacitor respectively.

The definition of oscillating frequency of the LC circuit

It is frequency at which current or charge move in whole LC circuit.

Mathematically it is given

f=12LC=1LC

Where, Lis the inductance of inductor, Cis the capacitance of capacitor , fis frequency of oscillation and is the angular frequency .

The current at a particular time in LC circuit

The current at particular time is given by

i=-Qsint

Where, iis the current in the circuit at time , Qis the maximum charge over capacitor .

02

Step 2:  Calculation of maximum charge on the capacitor in LC circuit

Using

12Li2+q22C=Q22C

For maximum current in LC circuit stored charge on capacitor must be zero .

So,

12Liimax2=Q22CQ=imaxLC

Put the values of constants in above equation

Q=0.750A0.0800H1.2510-9FQ=7.5010-6C

7.5010-6C

Thus, the maximum charge on the capacitor is .

Calculation of angular frequency of oscillation

Using

=1LC

Put the values of constants in above equation

=10.0800H1.2510-9F=1.00105rad/s

1.00105rad/s

Thus, the angular frequency of oscillation .

t=2.50ms

03

Calculation of current in the LC circuit at a time

Using

i=-Qsint

Now put the value of constant in above equation

i=-1.00105rad/s7.5010-6Csin1.00105rad/s2.5010-3si=0.7279A

Thus, the current in the LC circuit at time 2.50ms is 0.7279A .

04

Calculation of the energy stored in the inductor after 2.50 ms of oscillation

Using

UL=12Li

Now put the values of constants in above equation

UL=120.0800H0.7279A2UL=0.0212J

Hence ,the energy stored in the inductor after 2.50 ms of oscillation is 0.0212J.

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

We have seen that a coulomb is an enormous amount of charge; it is virtually impossible to place a charge of 1 C on an object. Yet, a current of 10A,10C/sis quite reasonable. Explain this apparent discrepancy.

A battery-powered global positioning system (GPS) receiver operating 9.0 V on draws a current of 0.13 A. How much electrical energy does it consume during 30 minutes?

Two copper wires with different diameter are joined end to end. If a current flow in the wire combination, what happens to electrons when they move from the large diameter wire into the smaller diameter wire? Does their drift speed increase, decrease, or stay the same? If the drift speed change, what is the role the force that causes the change? Explain your reasoning.

A 140-g ball containing excess electrons is dropped into a 110-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.300 T and direction from east to west. If air resistance is negligibly small, find the magnitude ond direction of the force that this magnetic field exerts on the ball just as it enters the field.

A particle with charge-5.60nCis moving in a uniform magnetic fieldrole="math" localid="1655717557369" B=-(1.25T)k^

The magnetic force on the particle is measured to berole="math" localid="1655717706597" F鈬赌=-(3.4010-7N)i^-(7.4010-7N)j^ (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there
components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar productv鈬赌F鈬赌. What is the angle between velocity and force?

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.