/*! 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} Q88P C=1.4 μ¹óIn an oscillating LC... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

C=1.4 μFIn an oscillating LCcircuit, L=8.0″¾±áand . At the time , t=0the current is maximum at 12.0mA. (a) What is the maximum charge on the capacitor during the oscillations? (b) At what earliest time t>0is the rate of change of energy in the capacitor maximum? (c) What is that maximum rate of change?

Short Answer

Expert verified

a) The maximum charge on the capacitor during the oscillation is 1.27×10−6 C.

b) The earliest time when the rate of change of energy in the capacitor is maximum is .8.31×10−5 s

c) The maximum rate of change in energy is .5.44×10−3 J/²õ

Step by step solution

01

Step 1: Given

  • Inductance in the circuit, .L=8.0″¾±á
  • Capacitance,°ä=1.40 μF
  • At ,t=0 maximum current .I=12 mA
02

Determining the concept 

By using the formulae for angular frequency and amplitude of current, find the maximum charge on the capacitor. By differentiating the total energy of the capacitor with respect to time, find the time corresponding to the maximum energy and the maximum rate of change of energy.

Formulae:

  • Angular frequency of oscillation,.Ó¬=1LC
  • The amplitude of varying current,.I=Ó¬Q
  • The total energy in the capacitor,.UE=q22C=Q22C(²õ¾±²Ô Ӭt)2
03

(a) Determining the maximum charge on the capacitor during the oscillation

The angular frequency of oscillation is .Ó¬=1LC

Also, the amplitude of varying current, .I=Ó¬Q

Using these two equations, it can be written as,

Q=ILCQ=(0.012)(1.40×10-6)(0.008)Q=1.27×10-6 C.

Hence, the maximum charge on the capacitor during the oscillation is.1.27×10-6 C

04

(b) Determining the earliest time, when the rate of change of energy in the capacitor is maximum

The total energy in the capacitor is,

.UE=q22C=Q22C(²õ¾±²Ô Ӭt)2

Differentiating it with time and use,sin(2θ)=2sin θ cos θ

dUEdt=Q22CӬ sin(2Ӭt)

The maximum value occurs when .sin(2Ó¬t)=1

Therefore,

.2Ӭt=π2t=12Ӭ(π2)t=π4LCt=3.144(1.40×10-6)(0.008)t=8.31×10-5 s

Hence, the earliest time when the rate of change of energy in the capacitor is maximum is .8.31×10−5 s

05

(c) Determining the maximum rate of change of energy

Since,

dUEdt=Q22CÓ¬sin(2Ó¬t).

The maximum value occurs when .sin(2Ó¬t)=1

(dUEdt)max=Q22CÓ¬.

Substituting the values of ,Q â¶Ä‰andÓ¬

(dUEdt)max=((ILC)22C)(ILC)(dUEdt)max=I22LC(dUEdt)max=(0.012)22(0.008)(1.40×10-6)(dUEdt)max=5.44×10-3J/s

Hence,the maximum rate of change in energy is .5.44×10−3 J/²õ

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 91Ó°ÊÓ!

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

In an oscillating series RCL circuit, show thatΔU/U the fraction of the energy lost per cycle of oscillation is given to a close approximation by 2πR/ӬL. The quantity is often called the Qof the circuit (for quality). A high-Qcircuit has low resistance and a low fractional energy loss(=2π/Q) per cycle.

In an oscillating LCcircuit withL=50mHandC=40μ¹ó, the current is initially a maximum. How long will it take before the capacitor is fully charged for the first time?

In Fig. 31-7, R=15.0Ω, C=4.70μF, and L=25.0mH. The generator provides an emf with rms voltageVRCL=75.0Vand frequency f=550Hz. (a) What is the rms current? What is the rms voltage across (b) R, (c) C, (d) L, (e) C and L together, and (f) R, C, and L together? At what average rate is energy dissipated by (g) R , (h) C, and (i) L?

An electric motor has an effective resistance of 32.0Ωand an inductive reactance of45.0Ωwhen working under load. The rms voltage across the alternating source is420V.Calculate the rms current.

A variable capacitor with a range from 10 to 365pF is used with a coil to form a variable-frequency LCcircuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor? If this circuit is to obtain frequencies from 0.54 MHz to 1.6 0MHz, the ratio computed in (a) is too large. By adding a capacitor in parallel to the variable capacitor, this range can be adjusted. To obtain the desired frequency range,(b) What capacitance should be added ? (c) What inductance should the coil have?

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.