VI I₁ = 1 AZ0° IB Y₂ Y₁ IA Y3 Y₁ = 0.5 S Y₂ = -10.5 S Y3 = +1 0.25 S FIGURE 19-22 Now, substituting the values of admittance into the nodal equations, we get (0.5j0.5)V₁ - (-j0.5)V2 = 120° Node 1: Node 2: -(-j0.5)V₁ + (-j0.25)V₂ = -220° which, when solved, results in: and V₂ V₁ = 4.243 VZ135° V₂ = 6.324 VZ-161.57° Ic 1₂ = 2 AZ0° Cengage Learning 2013
VI I₁ = 1 AZ0° IB Y₂ Y₁ IA Y3 Y₁ = 0.5 S Y₂ = -10.5 S Y3 = +1 0.25 S FIGURE 19-22 Now, substituting the values of admittance into the nodal equations, we get (0.5j0.5)V₁ - (-j0.5)V2 = 120° Node 1: Node 2: -(-j0.5)V₁ + (-j0.25)V₂ = -220° which, when solved, results in: and V₂ V₁ = 4.243 VZ135° V₂ = 6.324 VZ-161.57° Ic 1₂ = 2 AZ0° Cengage Learning 2013
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Question
![IB
I₁ = 1 AZ0°
V₂
Y₁
Y₁ = 0.5 S
Y₂ = -j 0.5 S
Y3 = +j 0.25 S
FIGURE 19-22
Now, substituting the values of admittance into the nodal equations, we get
(0.5j0.5)V₁ - (-j0.5)V₂ = 120°
Node 1:
Node 2:
-(-j0.5)V₁ + (-j0.25)V₂ = -220°
which, when solved, results in:
and
V₁ = 4.243 VZ135°
V2 = 6.324 VZ-161.57°
Ic
1₂ = 2 AZ0°
ⒸCengage Learning 2013](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb6b61b09-434f-4ee7-ab7e-f6853d8fc930%2F84ea6c12-1e3c-4f22-9df1-e208c2a3103b%2F90ecsf7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:IB
I₁ = 1 AZ0°
V₂
Y₁
Y₁ = 0.5 S
Y₂ = -j 0.5 S
Y3 = +j 0.25 S
FIGURE 19-22
Now, substituting the values of admittance into the nodal equations, we get
(0.5j0.5)V₁ - (-j0.5)V₂ = 120°
Node 1:
Node 2:
-(-j0.5)V₁ + (-j0.25)V₂ = -220°
which, when solved, results in:
and
V₁ = 4.243 VZ135°
V2 = 6.324 VZ-161.57°
Ic
1₂ = 2 AZ0°
ⒸCengage Learning 2013
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