EE 98: Fundamentals of Electrical Circuits - With Connect Access
EE 98: Fundamentals of Electrical Circuits - With Connect Access
6th Edition
ISBN: 9781259981807
Author: Alexander
Publisher: MCG
Question
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Chapter 16, Problem 8P
To determine

Find the expression of voltage v(t).

Expert Solution & Answer
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Answer to Problem 8P

The expression of voltage v(t) is.[1515e2t(cos(2t)+sin(2t))]u(t)V.

Explanation of Solution

Given data:

A branch voltage in an RLC circuit is given by,

d2vdt2+4dvdt+8v=120 (1)

The value of initial voltage v(0) is 0.

The value of dv(0)dt is 0.

Calculation:

Apply Laplace transform for equation (1) with initial conditions.

[s2V(s)dv(0)dtsv(0)+4(sV(s)v(0))+8V(s)]=120s{L(d2vdt2)=s2V(s)dv(0)dtsv(0)L(dvdt)=sV(s)v(0),L(u(t))=1s}

Simplify the above equation as follows.

s2V(s)dv(0)dtsv(0)+4sV(s)4v(0)+8V(s)=120s (2)

Substitute 0 for v(0), and 0 for dv(0)dt in equation (2).

s2V(s)0s(0)+4sV(s)4(0)+8V(s)=120ss2V(s)+4sV(s)+8V(s)=120s[s2+4s+8]V(s)=120s

Simplify the above equation to find V(s).

V(s)=120s(s2+4s+8) (3)

From equation (3), the characteristic equation is written as follows,

s2+4s+8=0 (4)

Write an expression to calculate the roots of characteristic equation (as2+bs+c=0).

s1,2=b±b24ac2a . (5)

Here,

a is the coefficient of second order term,

b is the coefficient of first order term, and

c is the coefficient of constant term.

Compare equation (4) with quadratic equation (as2+bs+c=0).

a=1b=4c=8

Substitute 1 for a, 4 for b, and 8 for c in equation (5) to find s1,2.

s1,2=4±(4)24(1)(8)2(1)=4±162=4±j42=2+j2,2j2

Now, the equation (3) is written as follows.

V(s)=120s(s(2+j2))(s(2j2))

V(s)=120s(s+2j2)(s+2+j2) (6)

Take partial fraction for equation (6).

V(s)=120s(s+2j2)(s+2+j2)=As+B(s+2j2)+C(s+2+j2) (7)

The equation (7) can be written as follows.

120s(s+2j2)(s+2+j2)=A(s+2j2)(s+2+j2)+Bs(s+2+j2)+Cs(s+2j2)s(s+2j2)(s+2+j2)

Simplify the above equation.

120=A(s+2j2)(s+2+j2)+Bs(s+2+j2)+Cs(s+2j2) (8)

Substitute 0 for s in equation (8) to find A.

120=A(0+2j2)(0+2+j2)+B(0)(0+2+j2)+C(0)(0+2j2)120=A(2j2)(2+j2)+0+0A(2j2)(2+j2)=120A(22(j2)2)=120{a2b2=(a+b)(ab)}

Simplify the above equation as follows.

A(4j24)=120A(4(1)4)=120{j2=1}A(4+4)=1208A=120

Simplify the above equation to find A.

A=1208=15

Substitute 2+j2 for s in equation (8) to find B.

120=[A(2+j2+2j2)(2+j2+2+j2)+B(2+j2)(2+j2+2+j2)+C(2+j2)(2+j2+2j2)]120=[A(0)(j4)+B(2+j2)(j4)+C(2+j2)(0)]120=0+B(2+j2)(j4)+0120=B(2+j2)(j4)

Simplify the above equation as follows.

120=B(2+j2)(j4)120=B(j8+j28)120=B(j8+(1)8){j2=1}120=B(j88)

Simplify the above equation to find B.

B=120(j88)=1208(1j)=15(1j)

Multiply and divide by 1+j on right hand side of above equation.

B=15(1+j)(1j)(1+j)=15(1+j)(1)2(j)2{a2b2=(a+b)(ab)}=15(1+j)(1)2(1){j2=1}=15(1+j)1+1

Simplify the above equation to find B.

B=15(1+j)2=7.5(1+j)

Substitute 2j2 for s in equation (8) to find C.

120=[A(2j2+2j2)(2j2+2+j2)+B(2j2)(2j2+2+j2)+C(2j2)(2j2+2j2)]120=[A(j4)(0)+B(2j2)(0)+C(2j2)(j4)]120=0+0+C(j8+j28)120=C(j8+(1)8){j2=1}

Simplify the above equation as follows.

120=C(j88)

Simplify the above equation to find C.

C=120(j88)=1208(1+j)=15(1+j)

Multiply and divide by 1j on right hand side of above equation.

C=15(1j)(1+j)(1j)=15(1j)(1)2(j)2{a2b2=(a+b)(ab)}=15(1j)(1)2(1){j2=1}=15(1j)1+1

Simplify the above equation to find C.

C=15(1j)2=7.5(1j)

Substitute 15 for A, 7.5(1+j) for B, and 7.5(1j) for C in equation (7) to find V(s).

V(s)=15s+7.5(1+j)(s+2j2)+7.5(1j)(s+2+j2)

V(s)=15s+(7.5+j7.5)(s+2j2)+(7.5j7.5)(s+2+j2) (9)

Apply inverse Laplace transform for equation (9) to find v(t).

v(t)=[15u(t)+(7.5+j7.5)e(2+j2)tu(t)+(7.5j7.5)e(2j2)tu(t)]{L1(1s+a)=eatu(t),L1(1s)=u(t)}=[15+(7.5+j7.5)e(2+j2)t+(7.5j7.5)e(2j2)t]u(t)V=[15+(7.5+j7.5)e(2)te(j2)t+(7.5j7.5)e(2)te(j2)t]u(t)V=[157.5e(2)te(j2)t+j7.5e(2)te(j2)t7.5e(2)te(j2)tj7.5e(2)te(j2)t]u(t)V

Simplify the above equation to find v(t).

v(t)=[67.5e2t(e(j2)t+e(j2)t)+j7.5et(e(j2)te(j2)t)]u(t)V=[157.5e2t(2cos(2t))+j7.5e2t(2jsin(2t))]u(t)V{cosθ=ejθ+ejθ2,sinθ=ejθejθ2j}=[1515e2tcos(2t)+j215e2tsin(2t)]u(t)V=[1515e2tcos(2t)+(1)15e2tsin(2t)]u(t)V{j2=1}

Simplify the above equation to find v(t).

v(t)=[1515e2tcos(2t)15e2tsin(2t)]u(t)V=[1515e2t(cos(2t)+sin(2t))]u(t)V

Conclusion:

Thus, the expression of voltage v(t) is [1515e2t(cos(2t)+sin(2t))]u(t)V.

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Chapter 16 Solutions

EE 98: Fundamentals of Electrical Circuits - With Connect Access

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