Nodal AnalysIs Application: Example 1 NTELO VALL ► Example 4.1: Determine the current flowing through the 15 N resistor and its power dissipation. Talking: Ricardo Matias 152 V2 Elin = Elout 15 2 V2 502 (1) 4 A lin lout= O 2 A (1)4A 2 A 10Ω 5Ω 3Ω %3D Ref. Ref. = 3 nodes = 2 equations and 2 unknowns (v, and v,). vi – V2 V1 + 10 = 2 15 5v1 – 2v2 = 60 Node 1: Eq.1 V1 = 20 V E System of Linear = Equations v2 - Vị + 5 V2 V2 = 20 V = 4 15 -Vi + 4v2 = 60 Node 2: Eq.2 EE330: Circuit Analysis I WKO3: Chapter 04 Ricardo Matias
Nodal AnalysIs Application: Example 1 NTELO VALL ► Example 4.1: Determine the current flowing through the 15 N resistor and its power dissipation. Talking: Ricardo Matias 152 V2 Elin = Elout 15 2 V2 502 (1) 4 A lin lout= O 2 A (1)4A 2 A 10Ω 5Ω 3Ω %3D Ref. Ref. = 3 nodes = 2 equations and 2 unknowns (v, and v,). vi – V2 V1 + 10 = 2 15 5v1 – 2v2 = 60 Node 1: Eq.1 V1 = 20 V E System of Linear = Equations v2 - Vị + 5 V2 V2 = 20 V = 4 15 -Vi + 4v2 = 60 Node 2: Eq.2 EE330: Circuit Analysis I WKO3: Chapter 04 Ricardo Matias
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter6: Power Flows
Section: Chapter Questions
Problem 6.61P
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How does node 1 simplify to Eq.1
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Nodal Analysis Application: Example 1
ANTELORE
> Example 4.1: Determine the current flowing through the 15 N resistor and its power dissipation.
Talking: Ricardo Matias
15 N
V2
Elin = Elout
15 2
V2
72
lin=
lout= O
2 A
5Ω
4 A
2 A (1)
10 Ω5Ω
4 A
3Ω
Ref.
Ref.
= 3 nodes = 2 equations and 2 unknowns (v, and v,).
Vi – v2
+
V1
Node 1:
= 2
5v1 – 2v2 = 60
Eq.1
10
15
Vz = 20 V
E System of Linear =
Equations
V2
V2 - V1
V2 = 20 V
= 4
15
-Vi + 4v2 = 60
Node 2:
+
Eq.2
5
EE330: Circuit Analysis I
WK03: Chapter 04
Ricardo Matias
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Transcribed Image Text:You are viewing Ricardo Matias' screen
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Nodal Analysis Application: Example 1
ANTELORE
> Example 4.1: Determine the current flowing through the 15 N resistor and its power dissipation.
Talking: Ricardo Matias
15 N
V2
Elin = Elout
15 2
V2
72
lin=
lout= O
2 A
5Ω
4 A
2 A (1)
10 Ω5Ω
4 A
3Ω
Ref.
Ref.
= 3 nodes = 2 equations and 2 unknowns (v, and v,).
Vi – v2
+
V1
Node 1:
= 2
5v1 – 2v2 = 60
Eq.1
10
15
Vz = 20 V
E System of Linear =
Equations
V2
V2 - V1
V2 = 20 V
= 4
15
-Vi + 4v2 = 60
Node 2:
+
Eq.2
5
EE330: Circuit Analysis I
WK03: Chapter 04
Ricardo Matias
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