BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Solutions

Chapter
Section
BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

A series circuit contains a resistor with R = 24 Ω, an inductor with L = 2 H, a capacitor with C = 0.005 F. and a 12-V battery. The initial charge is Q = 0.001 C and the initial current is 0.

(a) Find the charge and current at time t.

(b) Graph the charge and current functions.

(a)

To determine

To find: The charge of series RLC circuit at time t and the current of series RLC circuit at time t .

Explanation

Given data:

Series RLC circuit with following values.

R=24Ω , L=2H , C=0.005F , E(t)=12V , Q(0)=0.001 and Q(0)=I(0)=0 .

Formula used:

Write the expression for differential equation of electric circuit.

Ld2Qdt2+RdQdt+1CQ=E(t)

LQ+RQ+1CQ=E(t) (1)

Write the expression for general solution with complex roots.

Q(t)=eαt[c1cos(βt)+c2sin(βt)] (2)

Write the expression for r .

r=α+βi (3)

Write the expression for auxiliary equation.

ar2+br+c=0 (4)

Write the expression for differential equation.

ay+by+cy=0 (5)

Find the differential equation for electric circuit using equation (1).

Substitute 2 for L, 24 for R, 0.005 for C and 12 for E(t) in equation (1),

2Q+24Q+10.005Q=12

2Q+24Q+200Q=12 (6)

Modify equation (5) as follows.

aQ+bQ+cQ=0 (7)

Compare equation (6) and (7).

a=2b=24c=200

Substitute 2 for a, 24 for b and 200 for c in equation (4),

2r2+24r+200=0r2+12r+100=0

Find the value of r .

r=12±(12)24(1)(100)2(1)=12±1444002(1)=12±2562=12±16i2

Simplify r as follows.

r=6±8i (8)

Compare equations (3) and (8).

α=6β=8

Substitute 6 for α and 8 for β in equation (2),

Q(t)=e6t[c1cos(8t)+c2sin(8t)] (9)

Modify equation (9) for complementary equation as follows.

Qc(t)=e6t[c1cos(8t)+c2sin(8t)]

Consider the value of Qp(t) as follows.

Qp(t)=A (10)

Differentiate equation (10) with respect to t .

Qp(t)=0

Differentiate equation with respect to t .

Qp(t)=0

Modify equation (6) as follows.

2Qp(t)+24Qp(t)+200Qp(t)=12

Substitute 0 for Qp(t) , 0 for Qp(t) and A for Qp(t) ,

2(0)+24(0)+200(A)=12200(A)=12A=12200A=350

Substitute 350 for A in equation (10),

Qp(t)=350

Write the expression for general solution.

Q(t)=Qc(t)+Qp(t)

Substitute e6t[c1cos(8t)+c2sin(8t)] for Qc(t) and 350 for Qp(t) ,

Q(t)=e6t[c1cos(8t)+c2sin(8t)]+350 (11)

Substitute 0 for t ,

Q(0)=e6(0)[c1cos(8×0)+c2sin(8×0)]+350=c1+350

Substitute 0

(b)

To determine

To graph: The charge functions and the current functions.

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Chapter 17 Solutions

Show all chapter solutions add
Sect-17.1 P-11ESect-17.1 P-12ESect-17.1 P-13ESect-17.1 P-14ESect-17.1 P-15ESect-17.1 P-16ESect-17.1 P-17ESect-17.1 P-18ESect-17.1 P-19ESect-17.1 P-20ESect-17.1 P-21ESect-17.1 P-22ESect-17.1 P-23ESect-17.1 P-24ESect-17.1 P-25ESect-17.1 P-26ESect-17.1 P-27ESect-17.1 P-28ESect-17.1 P-29ESect-17.1 P-30ESect-17.1 P-31ESect-17.1 P-32ESect-17.1 P-33ESect-17.1 P-34ESect-17.2 P-1ESect-17.2 P-2ESect-17.2 P-3ESect-17.2 P-4ESect-17.2 P-5ESect-17.2 P-6ESect-17.2 P-7ESect-17.2 P-8ESect-17.2 P-9ESect-17.2 P-10ESect-17.2 P-11ESect-17.2 P-12ESect-17.2 P-13ESect-17.2 P-14ESect-17.2 P-15ESect-17.2 P-16ESect-17.2 P-17ESect-17.2 P-18ESect-17.2 P-19ESect-17.2 P-20ESect-17.2 P-21ESect-17.2 P-22ESect-17.2 P-23ESect-17.2 P-24ESect-17.2 P-25ESect-17.2 P-26ESect-17.2 P-27ESect-17.2 P-28ESect-17.3 P-1ESect-17.3 P-2ESect-17.3 P-3ESect-17.3 P-4ESect-17.3 P-5ESect-17.3 P-6ESect-17.3 P-7ESect-17.3 P-8ESect-17.3 P-9ESect-17.3 P-10ESect-17.3 P-11ESect-17.3 P-12ESect-17.3 P-13ESect-17.3 P-14ESect-17.3 P-15ESect-17.3 P-16ESect-17.3 P-17ESect-17.3 P-18ESect-17.4 P-1ESect-17.4 P-2ESect-17.4 P-3ESect-17.4 P-4ESect-17.4 P-5ESect-17.4 P-6ESect-17.4 P-7ESect-17.4 P-8ESect-17.4 P-9ESect-17.4 P-10ESect-17.4 P-11ESect-17.4 P-12ECh-17 P-1RCCCh-17 P-2RCCCh-17 P-3RCCCh-17 P-4RCCCh-17 P-5RCCCh-17 P-1RQCh-17 P-2RQCh-17 P-3RQCh-17 P-4RQCh-17 P-1RECh-17 P-2RECh-17 P-3RECh-17 P-4RECh-17 P-5RECh-17 P-6RECh-17 P-7RECh-17 P-8RECh-17 P-9RECh-17 P-10RECh-17 P-11RECh-17 P-12RECh-17 P-13RECh-17 P-14RECh-17 P-15RECh-17 P-16RECh-17 P-17RECh-17 P-18RECh-17 P-19RECh-17 P-20RECh-17 P-21RE

Additional Math Solutions

Find more solutions based on key concepts

Show solutions add

Describe several ways in which a limit can fail to exist illustrate with sketches.

Single Variable Calculus: Early Transcendentals, Volume I

Find the limit of f(x) as x approaches 0. f(x)={ x2+1,x02x+1,x0

Calculus: An Applied Approach (MindTap Course List)

For the following set of stores, find the value of each expression: a. X b. (X) c. (X 3) d. (X 3) X 4 5 2 1 3

Essentials of Statistics for The Behavioral Sciences (MindTap Course List)

In Exercises 90-98, determine whether the statement is true or false. If it is true, explain why it is true. If...

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

If f=vt,v=45,andt=4findd

Elementary Technical Mathematics

Prove the following identities. sintan+cos=sec

Trigonometry (MindTap Course List)

The polar form for the graph at the right is:

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th