Determine vo(t) in the circuit of Fig. 16.6, assuming zero initial conditions. Answer: 40(1-21 — 2te21)u(t) V. 14 FL 1H m 10u(t) V 492 + vo(t)

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For the following question involving the Laplace transform of a simple circuit, I simply want to know why using source transformation first does not give the same result as solving the question by using current division or KVL.  When I use current division and solve the Laplace, it gives the correct answer.  The second method of source transformation which I have included in the image does not.

Determine v(t) in the circuit of Fig. 16.6, assuming zero initial conditions.
Answer: 40(1-e-21 — 2te21)u(t) V.
14
1 H
m
F:
10u(t) V
492
This methods does NOT give
the correct result: WHY?
JFS
915
+
vo(t)
15
m
t
I (s)
42 V(t)
+ I(s).
Tsys I) 4 = 0
+
(s) [+S+4]
In [4+ 5+ 45] = 40
=
40
S
40
ICs) =
5²+45+4
:.
당(t)=
4. I(s)
160
5²+45 +4
160
(5 +2)
1
Transcribed Image Text:Determine v(t) in the circuit of Fig. 16.6, assuming zero initial conditions. Answer: 40(1-e-21 — 2te21)u(t) V. 14 1 H m F: 10u(t) V 492 This methods does NOT give the correct result: WHY? JFS 915 + vo(t) 15 m t I (s) 42 V(t) + I(s). Tsys I) 4 = 0 + (s) [+S+4] In [4+ 5+ 45] = 40 = 40 S 40 ICs) = 5²+45+4 :. 당(t)= 4. I(s) 160 5²+45 +4 160 (5 +2) 1
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