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)

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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.
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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