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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

The battery in Exercise 14 is replaced by a generator producing a voltage of E(t) = 12 sin 10t.

(a) Find the charge at time t.

(b) Graph the charge function.

(a)

To determine

To find : The charge 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)=12sin10tV , 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 12sin10t for E(t) in equation (1),

2Q+24Q+10.005Q=12sin10t

2Q+24Q+200Q=12sin10t (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)=Acos10t+Bsin10t (10)

Differentiate equation (10) with respect to t .

Qp(t)=10Asin10t+10Bcos10t

Differentiate equation with respect to t .

Qp(t)=100Acos10t100Bsin10t

Modify equation (6) as follows.

2Qp(t)+24Qp(t)+200Qp(t)=12sin10t

Substitute 100Acos10t100Bsin10t for Qp(t) , 10Asin10t+10Bcos10t for Qp(t) and Acos10t+Bsin10t for Qp(t) ,

{2(100Acos10t100Bsin10t)+24(10Asin10t+10Bcos10t)+200(Acos10t+Bsin10t)}=12sin10t{200Acos10t200Bsin10t240Asin10t+240Bcos10t+200Acos10t+200Bsin10t}=12sin10t{(200A+240B+200A)cos10t+(200B240A+200B)sin10t}=12sin10t

Compare left hand side (LHS) and right hand side (RHS) of equation

(b)

To determine

To graph: The charge functions.

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Chapter 17 Solutions

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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

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