how to solve Node equations (10.1.1) and (10.1.2) in detail

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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I would like to know how to solve Node equations (10.1.1) and (10.1.2) in detail.
20/0°,
jwL = j4
jwL = j2
1
jwC
Thus, the frequency-domain equivalent circuit is as shown in Fig. 10.2.
or
20 cos 4t
1 H
0.5 H
20/0° V
Figure 10.2
At node 2,
But Ix
0.1 F
1092
Applying KCL at node 1,
20 - V₁
10
V₁
By simplifying, we get
=
21x +
-j2.5 92
2V1
-j2.5
j4 92
m
Frequency-domain equivalent of the circuit in Fig. 10.1.
V₁
-j2.5
= -j2.5
(1 + j1.5)V₁ + j2.5V₂
+
+
1
V₁ - V₂
j4
21₁
j2
=
= V₁/-j2.5. Substituting this gives
V₁ - V₂
j4
@ = 4 rad/s
V₁ - V₂
j4
11V₁ + 15V₂ = 0
= 20
گرانه
еее
j2
j2 92
(10.1.1)
(10.1.2)
Transcribed Image Text:20/0°, jwL = j4 jwL = j2 1 jwC Thus, the frequency-domain equivalent circuit is as shown in Fig. 10.2. or 20 cos 4t 1 H 0.5 H 20/0° V Figure 10.2 At node 2, But Ix 0.1 F 1092 Applying KCL at node 1, 20 - V₁ 10 V₁ By simplifying, we get = 21x + -j2.5 92 2V1 -j2.5 j4 92 m Frequency-domain equivalent of the circuit in Fig. 10.1. V₁ -j2.5 = -j2.5 (1 + j1.5)V₁ + j2.5V₂ + + 1 V₁ - V₂ j4 21₁ j2 = = V₁/-j2.5. Substituting this gives V₁ - V₂ j4 @ = 4 rad/s V₁ - V₂ j4 11V₁ + 15V₂ = 0 = 20 گرانه еее j2 j2 92 (10.1.1) (10.1.2)
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