Chapter 17.3, Problem 13E

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

A series circuit consists of a resistor with R = 20 Ω,  an inductor with L= 1 H, a capacitor with C = 0.002 F, and a 12-V battery. If the initial charge and current are both 0, find the charge and current at time t.

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=20â€‰Î© , L=1â€‰H , C=0.002â€‰F , E(t)=12V , Q(0)=0 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 1 for L, 20 for R, 0.002 for C and 12 for E(t) in equation (1),

1Qâ€³+20Qâ€²+10.002Q=12

Qâ€³+20Qâ€²+500Q=12 (6)

Modify equation (5) as follows.

aQâ€³+bQâ€²+cQ=0 (7)

Compare equation (6) and (7).

a=1b=20c=500

Substitute 1 for a, 20 for b and 500 for c in equation (4),

r2+20r+500=0

Find the value of r .

r=âˆ’20Â±(20)2âˆ’4(1)(500)2(1)=âˆ’20Â±400âˆ’20002=âˆ’20Â±âˆ’16002=âˆ’20Â±40i2

Simplify r as follows.

r=âˆ’10Â±20i (8)

Compare equations (3) and (8).

Î±=âˆ’10Î²=20

Substitute âˆ’10 for Î± and 20 for Î² in equation (2),

Q(t)=eâˆ’10t[c1cos(20t)+c2sin(20t)] (9)

Modify equation (9) for complementary equation as follows.

Qc(t)=eâˆ’10t[c1cos(20t)+c2sin(20t)]

Consider the value of Qp(t) as follows.

Qp(t)=A (10)

Differentiate equation (10) with respect to t .

Qâ€²p(t)=0

Differentiate equation with respect to t .

Qâ€³p(t)=0

Modify equation (6) as follows.

Qâ€³p(t)+20Qâ€²p(t)+500Qp(t)=12

Substitute 0 for Qâ€³p(t) , 0 for Qâ€²p(t) and A for Qp(t) ,

1(0)+20(0)+500(A)=12500(A)=12A=12500A=3125

Substitute 3125 for A in equation (10),

Qp(t)=3125

Write the expression for general solution.

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

Substitute eâˆ’10t[c1cos(20t)+c2sin(20t)] for Qc(t) and 3125 for Qp(t) ,

Q(t)=eâˆ’10t[c1cos(20t)+c2sin(20t)]+3125 (11)

Substitute 0 for t ,

Q(0)=eâˆ’10(0)[c1cos(20Ã—0)+c2sin(20Ã—0)]+3125=c1+3125

Substitute 0 for Q(0) ,

0=c1+3125c1=0âˆ’3125c1=âˆ’3125

Write the expression for I

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