CHAPTER 16Applications of Laplace TransformsSolution:Figure 16.12(b) represents the entire circuit in the s domain with the initialconditions incorporated. We now have a straightforward nodal analysisproblem. Since the value of V₁ is also the value of the capacitor voltagein the time domain and is the only unknown node voltage, we only needto write one equation.V₁ - VsV₁-0i (0)V₁ - [v(0)/s]+10/35s= 0S1/(0.1s)(16.4.1)or31How0.1 s +3+3) vV₁ = =+ = +0.5(16.4.2)SSSwhere v (0) = 5 V and i (0) = -1 A. Simplifying we get(s² + 3s +2) V₁ = 40+5sor40+5s35V₁ =(s + 1)(s+2)s+1Taking the inverse Laplace transform yields(16.4.4)v₁ (t) = (35e¹ - 30e-2¹)u(t) VCTICE PROBLEM 16.4circuit shown in Fig. 16.12 with the same initial conditions, findent through the inductor for all time t > 0.7e¹+3e-21)u(t) A.i(t) =Fig 16.12 and the initial conditions used inMP+30s+2(16.4.3)v (1)V₂Figure 16.1.

The given equation can be easily simplified by taking the common multiplier across the terms and do…

Answered: 1/(0.1s) or 2 How 0.1 s +3+ s +3+ + v 3…

1/(0.1s) or 2 How 0.1 s +3+ s +3+ + v 3 1 ²) V₁ = V₁=+=+0.5 S S S where v (0) = 5 V and i (0) = -1 A. Simplifuin S = 0 (16.4.1) (16.4.2)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

CHAPTER 16
Applications of Laplace Transforms
Solution:
Figure 16.12(b) represents the entire circuit in the s domain with the initial
conditions incorporated. We now have a straightforward nodal analysis
problem. Since the value of V₁ is also the value of the capacitor voltage
in the time domain and is the only unknown node voltage, we only need
to write one equation.
V₁ - Vs
V₁-0
i (0)
V₁ - [v(0)/s]
+
10/3
5s
= 0
S
1/(0.1s)
(16.4.1)
or
3
1
How
0.1 s +3+
3) v
V₁ = =+ = +0.5
(16.4.2)
S
S
S
where v (0) = 5 V and i (0) = -1 A. Simplifying we get
(s² + 3s +2) V₁ = 40+5s
or
40+5s
35
V₁ =
(s + 1)(s+2)
s+1
Taking the inverse Laplace transform yields
(16.4.4)
v₁ (t) = (35e¹ - 30e-2¹)u(t) V
CTICE PROBLEM 16.4
circuit shown in Fig. 16.12 with the same initial conditions, find
ent through the inductor for all time t > 0.
7e¹+3e-21)u(t) A.
i(t) =
Fig 16.12 and the initial conditions used in
MP
+
30
s+2
(16.4.3)
v (1)
V₂
Figure 16.1.

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