Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
6th Edition
ISBN: 9780078028229
Author: Charles K Alexander, Matthew Sadiku
Publisher: McGraw-Hill Education
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Chapter 16, Problem 17P

If is(t) = 7.5e–2t u(t) A in the circuit shown in Fig. 16.40, find the value of io(t).

Chapter 16, Problem 17P, If is(t) = 7.5e2t u(t) A in the circuit shown in Fig. 16.40, find the value of io(t).

Expert Solution & Answer
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To determine

Find the expression of current io(t) in the circuit of Figure 16.40.

Answer to Problem 17P

The expression of current io(t) in the circuit of Figure 16.40 is 7.5[e2t0.7559e0.5tsin(1.3229t)]u(t)A.

Explanation of Solution

Given data:

Refer to Figure 16.40 in the textbook.

The value of source current is,

is(t)=7.5e2tu(t)A (1)

Formula used:

Write a general expression to calculate the impedance of a resistor in s-domain.

ZR=R (2)

Here,

R is the value of resistance.

Write a general expression to calculate the impedance of an inductor in s-domain.

ZL=sL (3)

Here,

L is the value of inductance.

Write a general expression to calculate the impedance of a capacitor in s-domain.

ZC=1sC (4)

Here,

C is the value of capacitance.

Calculation:

The given circuit is redrawn as shown in Figure 1.

Fundamentals of Electric Circuits, Chapter 16, Problem 17P , additional homework tip  1

Substitute 2Ω for R in equation (2) to find ZR.

ZR=2Ω

Substitute 1H for L in equation (3) to find ZL.

ZL=s(1H)=s

Substitute 0.5F for C in equation (4) to find ZC.

ZC=1s(0.5F)=2s

Apply Laplace transform for equation (1) to find Is(s).

Is(s)=7.5s+2{L(eatu(t))=1s+a}

Convert the Figure 1 into s-domain as shown in Figure 2.

Fundamentals of Electric Circuits, Chapter 16, Problem 17P , additional homework tip  2

Apply nodal analysis at node V(s) in the circuit shown in Figure 2.

7.5s+2+V(s)s+V(s)2+V(s)(2s)=0V(s)s+V(s)2+sV(s)2=7.5s+2V(s)[1s+12+s2]=7.5s+2V(s)[2+s+s22s]=7.5s+2

Simplify the above equation to find V(s).

V(s)=(7.5s+2)(2ss2+s+2)

V(s)=15s(s+2)(s2+s+2) (5)

From the above equation , the characteristic equation is

s2+s+2=0 (6)

Write a general expression to calculate the roots of quadratic equation (as2+bs+c=0).

s1,2=b±b24ac2a (7)

Comparing the equation (6) with the equation (as2+bs+c=0).

a=1b=1c=2

Substitute 1 for a, 1 for b, and 2 for c in equation (7) to find s1,2.

s1,2=1±(1)24(1)(2)2(1)=1±182(1)=1±72=1±j2.64572

Simplify the above equation to find s1,2.

s1,2=1±j2.64572,1±j2.64572=0.5+j1.3229,0.5j1.3229

Substitute the roots of characteristic equation in equation (5) to find V(s).

V(s)=15s(s+2)(s(0.5+j1.3229))((s(0.5j1.3229)))=15s(s+2)(s(0.5+j1.3229))(s(0.5j1.3229))

Refer to Figure 2, the current through the capacitor is calculated as follows:

Io(s)=V(s)(2s)=sV(s)2

Substitute 15s(s+2)(s(0.5+j1.3229))(s(0.5j1.3229)) for V(s) in above equation to find Io(s).

Io(s)=s[15s(s+2)(s(0.5+j1.3229))(s(0.5j1.3229))]2=7.5s2(s+2)(s(0.5+j1.3229))(s(0.5j1.3229))

Take partial fraction for above equation.

Io(s)=[7.5s2[(s+2)(s(0.5+j1.3229))(s(0.5j1.3229))]]=[As+2+B(s(0.5+j1.3229))+C(s(0.5j1.3229))] (8)

The equation (8) can be written as follows:

7.5s2[(s+2)(s(0.5+j1.3229))(s(0.5j1.3229))]=[A(s(0.5+j1.3229))(s(0.5j1.3229))+B(s+2)(s(0.5j1.3229))+C(s+2)(s(0.5+j1.3229))[(s+2)(s(0.5+j1.3229))(s(0.5j1.3229))]]

Simplify the above equation as follows:

7.5s2=[A(s(0.5+j1.3229))(s(0.5j1.3229))+B(s+2)(s(0.5j1.3229))+C(s+2)(s(0.5+j1.3229))] . (9)

Substitute 2 for s in equation (9) to find A.

7.5(2)2=[A(2(0.5+j1.3229))(2(0.5j1.3229))+B(2+2)(2(0.5j1.3229))+C(2+2)(2(0.5+j1.3229))]30=[A(2(0.5+j1.3229))(2(0.5j1.3229))+B(0)(2(0.5j1.3229))+C(0)(2(0.5+j1.3229))]30=[A[4+2((0.5j1.3229))+2(0.5+j1.3229)+(0.5+j1.3229)(0.5j1.3229)]+0+0]30=[A[4+1j2.64581+j2.6458+(0.5)2(j1.3229)2]+0+0]{a2b2=(a+b)(ab)}

Simplify the above equation as follows:

30=A[2+0.25j21.75]30=A(2.25(1)1.75){j2=1}30=A(2.25+1.75)30=A(4)

Simplify the above equation to find A.

A=304=7.5

Substitute 0.5+j1.3229 for s in equation (9) to find B.

7.5(0.5+j1.3229)2=[A((0.5+j1.3229)(0.5+j1.3229))((0.5+j1.3229)(0.5j1.3229))+B(0.5+j1.3229+2)((0.5+j1.3229)(0.5j1.3229))+C((0.5+j1.3229)+2)((0.5+j1.3229)(0.5+j1.3229))]7.5(1.5j1.3229)=[A(0)+B(1.5+j1.3229)(j2.646)+C(0)]11.25j9.92175=B(3.5+j3.969)

Simplify the above equation to find B.

B=11.25j9.92175(3.5+j3.969)=2.834690°

Substitute 0.5j1.3229 for s in equation (8) to find C.

7.5(0.5j1.3229)2=[A((0.5j1.3229)(0.5+j1.3229))((0.5j1.3229)(0.5j1.3229))+B(0.5j1.3229+2)((0.5j1.3229)(0.5j1.3229))+C((0.5j1.3229)+2)((0.5j1.3229)(0.5+j1.3229))]7.5(1.5+j1.3229)=[A(0)+B(0)+C(1.5j1.3229)(j2.646)]11.25+j9.92175=C(3.5j3.969)

Simplify the above equation to find C.

C=11.25+j9.92175(3.5j3.969)=2.834690°

Substitute 7.5 for A, 2.834690° for B, and 2.834690° for C in equation (8) to find Io(s).

Io(s)=[7.5s+2+2.834690°(s(0.5+j1.3229))+2.834690°(s(0.5j1.3229))]=[7.5s+2+2.8346(cos(90°)+jsin(90°))(s(0.5+j1.3229))+2.8346(cos(90°)+jsin(90°))(s(0.5j1.3229))]=[7.5s+2+2.8346ej90°(s(0.5+j1.3229))+2.8346ej90°(s(0.5j1.3229))]{eiθ=cosθ+isinθ}

Take inverse Laplace transform for above equation to find io(t).

io(t)=[7.5e2tu(t)+2.8346ej90°e(0.5+j1.3229)tu(t)+2.8346ej90°e(0.5j1.3229)tu(t)]{L1(1sa)=eatu(t)}=[7.5e2t+2.8346e(0.5t+j1.3229t+j90°)+2.8346e(0.5tj1.3229tj90°)]u(t){eaeb=ea+b}=[7.5e2t+2.8346e0.5tej1.3229t+j90°+2.8346e0.5tej1.3229tj90°]u(t)=[7.5e2t+2.8346e0.5tej(1.3229t+90°)+2.8346e0.5tej(1.3229t+j90°)]u(t)

Simplify the above equation to find io(t).

io(t)=[7.5e2t+2.8346e0.5tej(1.3229t+90°)+2.8346e0.5tej(1.3229t+j90°)]u(t)=[7.5e2t+2.8346e0.5t[ej(1.3229t+90°)+ej(1.3229t+j90°)]]u(t)=[7.5e2t+2.8346e0.5t[2cos(1.3229t+90°)]]u(t){cosθ=eiθ+eiθ2}=[7.5e2t+5.6692e0.5tcos(1.3229t+90°)]u(t)A

Simplify the above equation to find io(t).

io(t)=[7.5e2t+5.6692e0.5t(sin(1.3229t))]u(t)A{cos(90°+θ)=sinθ}=7.5[e2t0.7559e0.5tsin(1.3229t)]u(t)A

Conclusion:

Thus, the expression of current io(t) in the circuit of Figure 16.40 is 7.5[e2t0.7559e0.5tsin(1.3229t)]u(t)A.

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Chapter 16 Solutions

Fundamentals of Electric Circuits

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