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)
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
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F31669f38-de4d-405f-bafd-7e9a7b3a5c37%2Fe395e6b1-ee19-40dd-a71b-45c0df58a679%2Fgb4d3va_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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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