   Chapter 17.R, Problem 19E

Chapter
Section
Textbook Problem

A Series circuit contains a resistor with R = 40   Ω , an inductor with L = 2  H , a capacitor with C = 0.0025  F , and a 12-V battery. The initial charge is Q = 0.01  C and the initial current is 0. Find the charge at time t.

To determine

To find:

The charge at time t

Explanation

1) Concept:

i. The general second order differential equation for L-C-R circuit is

Ld2Qdt2+RdQdt+1CQ=Et

ii. The general solution of nonhomogeneous differential equation is ay''+by'+cy=Gx can be written as yx=yc+yP

where yc is the general solution of complementary equation; ay''+by'+cy=0 and yP is a particular solution of ay''+by'+cy=Gx

Find yc&yP

2) General solution of ay''+by'+cy=0:

 Roots of ar2+br+c=0 General solution r1, r2 are real and distinct y=c1er1x+c2er2x r1=r2=r y=c1erx+c2xerx r1 , r2 complex :α±iβ y=eαx(c1cosβx+c2sinβx

3) Method of undetermined coefficient:

It is reasonable to guess that there is a particular solution yP that is a polynomial of the same degree as G because if y is polynomial, then ay''+by'+cy is also polynomial. We therefore substitute yPx= a polynomial into (Of the same degree as ) the differential equation and determine the coefficient.

4) Calculations:

We are given the circuit consist of resistor with R=40Ω, an inductor with L=2Η, a capacitor with C=0.0025F, and a 12-V battery.

Let Q be the charge in circuit at time t then dQdt be the rate of charge flowwith time t that is current at time t

The general second order differential equation for L-C-R circuit is

Ld2Qdt2+RdQdt+1CQ=Et

By using the given information the differential equation for given L-C-R circuit is

2Q''+40Q'+10.0025Q=12

Given that the initial charge is Q(0)=0.01 C and the initial current Q'(0) is 0.

Therefore, the given initial value problem is

2Q''+40Q'+400Q=12,  Q0=0.01,  Q'0=0

To find the charge means to find the solution of given initial value problem

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