   Chapter 17.3, Problem 15E

Chapter
Section
Textbook Problem

The battery in Exercise 13 is replaced by a generator producing a voltage of E(t) = 12 sin 10t. Find the charge at time t.

To determine

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

Explanation

Given data:

Series RLC circuit with following values.

R=20Ω , L=1H , C=0.002F , E(t)=12V , Q(0)=0 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 1 for L, 20 for R, 0.002 for C and 12sin10t for E(t) in equation (1),

1Q+20Q+10.002Q=12sin10t

Q+20Q+500Q=12sin10t (6)

Modify equation (5) as follows.

aQ+bQ+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=20±(20)24(1)(500)2(1)=20±40020002=20±16002=20±40i2

Simplify r as follows.

r=10±20i (8)

Compare equations (3) and (8).

α=10β=20

Substitute 10 for α and 20 for β in equation (2),

Q(t)=e10t[c1cos(20t)+c2sin(20t)] (9)

Modify equation (9) for complementary equation as follows.

Qc(t)=e10t[c1cos(20t)+c2sin(20t)]

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.

Qp(t)+20Qp(t)+500Qp(t)=12sin10t

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

{100Acos10t100Bsin10t+20(10Asin10t+10Bcos10t)+500(Acos10t+Bsin10t)}=12sin10t{100Acos10t100Bsin10t200Asin10t+200Bcos10t+500Acos10t+500Bsin10t}=12sin10t(400A+200B)cos10t+(200A+400B)sin10t=12sin10t

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

400A+200B=0 (11)

200A+400B=12 (12)

Solve for B using equations (11) and (12).

1000B=24B=241000B=3125

Solve for A .

Substitute 3125 for B in equation (11),

400A+200(3125)=0400A=600125A=600125×400A=3250

Substitute 3250 for A and 3125 for B in equation (10),

Qp(t)=3250cos10t+3125sin10t

Write the expression for general solution

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