Chapter 17.3, Problem 16E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### 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=24â€‰Î© , L=2H , C=0.005â€‰F , E(t)=12sin10tâ€‰V , Q(0)=0.001 and Qâ€²(0)=I(0)=0 .

Formula used:

Write the expression for differential equation of electric circuit.

Ld2Qdt2+RdQdt+1CQ=E(t)

LQâ€³+RQâ€²+1CQ=E(t) (1)

Write the expression for general solution with complex roots.

Q(t)=eÎ±t[c1cos(Î²t)+c2sin(Î²t)] (2)

Write the expression for r .

r=Î±+Î²i (3)

Write the expression for auxiliary equation.

ar2+br+c=0 (4)

Write the expression for differential equation.

ayâ€³+byâ€²+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),

2Qâ€³+24Qâ€²+10.005Q=12sin10â€‰t

2Qâ€³+24Qâ€²+200Q=12sin10t (6)

Modify equation (5) as follows.

aQâ€³+bQâ€²+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=âˆ’12Â±(âˆ’12)2âˆ’4(1)(100)2(1)=âˆ’12Â±144âˆ’4002(1)=âˆ’12Â±âˆ’2562=âˆ’12Â±16i2

Simplify r as follows.

r=âˆ’6Â±8i (8)

Compare equations (3) and (8).

Î±=âˆ’6Î²=8

Substitute âˆ’6 for Î± and 8 for Î² in equation (2),

Q(t)=eâˆ’6t[c1cos(8t)+c2sin(8t)] (9)

Modify equation (9) for complementary equation as follows.

Qc(t)=eâˆ’6t[c1cos(8t)+c2sin(8t)]

Consider the value of Qp(t) as follows.

Qp(t)=Acosâ€‰10t+Bsinâ€‰10t (10)

Differentiate equation (10) with respect to t .

Qâ€²p(t)=âˆ’10Asinâ€‰10t+10Bcosâ€‰10t

Differentiate equation with respect to t .

Qâ€³p(t)=âˆ’100Acosâ€‰10tâˆ’100Bsinâ€‰10t

Modify equation (6) as follows.

2Qâ€³p(t)+24Qâ€²p(t)+200Qp(t)=12sin10â€‰t

Substitute âˆ’100Acosâ€‰10tâˆ’100Bsinâ€‰10t for Qâ€³p(t) , âˆ’10Asinâ€‰10t+10Bcosâ€‰10t for Qâ€²p(t) and Acosâ€‰10t+Bsinâ€‰10t for Qp(t) ,

{2(âˆ’100Acosâ€‰10tâˆ’100Bsinâ€‰10t)+24(âˆ’10Asinâ€‰10t+10Bcosâ€‰10t)+200(Acosâ€‰10t+Bsinâ€‰10t)}=12sin10â€‰t{âˆ’200Acosâ€‰10tâˆ’200Bsinâ€‰10tâˆ’240Asinâ€‰10t+240Bcosâ€‰10t+200Acosâ€‰10t+200Bsinâ€‰10t}=12sin10â€‰t{(âˆ’200A+240B+200A)cosâ€‰10t+(âˆ’200Bâˆ’240A+200B)sinâ€‰10t}=12sin10â€‰t

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

(b)

To determine

To graph: The charge functions.

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