Practice Problem 3.7 Use mesh analysis to determine i,, iz, and iz in Fig. 3.25. Answer: i, = 3.474 A, iz = 0.4737 A, iz = 1.1052 A. ЗА *Nodal and Mesh Analyses by Inspection 3.6 Figure 3.25 For Practice Prob. 3.7. This section presents a generalized procedure for nodal or mesh a It ic chortgut bacad or mara incnaation of ww-

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Vo) LTE
1:31
A LIEI l 38%
FundamentalsofElectricCi.
100
Chapter 3
Methods of Analysis
But I, = -i hence,
iz = iz - 3i4
(3
Applying KVL in mesh 4,
2i4 + 8(i4 - i) + 10 = 0
or
Si4 - 4iz = -5
From Eqs. (3.7.1) to (3.7.4),
i = -7.5 A,
i, = -2.5 A,
iz = 3.93 A,
i, = 2,14:
Practice Problem 3.7
Use mesh analysis to determine i, i2, and iz in Fig. 3.25.
Answer: i, = 3.474 A, iz = 0.4737 A, iz = 1.1052 A.
3 20
6 V
3 A (
† Nodal and Mesh Analyses
by Inspection
3.6
Figure 3.25
For Practice Prob. 3.7.
This section presents a generalized procedure for nodal or mesh
sis. It is a shortcut approach based on mere inspection of a circu
When all sources in a circuit are independent current source
do not need to apply KCL to each node to obtain the node-vo
equations as we did in Section 3.2. We can obtain the equation
mere inspection of the circuit. As an example, let us reexamine th
cuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience.
circuit has two nonreference nodes and the node equations
derived in Section 3.2 as
G2
G
G, + G2
-G2
-G2
Observe that each of the diagonal terms is the sum of the conduct
connected directly to node 1 or 2, while the off-diagonal terms ar
negatives of the conductances connected between the nodes. Also,
term on the right-hand side of Eq. (3.21) is the algebraic sum o
currents entering the node,
In general, if a circuit with independent current sources has
reference nodes, the node-voltage equations can be written in tern
the conductances as
(a)
R
ww
ww
V2
Gu G12 ...
GIN
ןט
(b)
G21 G22
G2N
U2
Figure 3.26
(a) The circuit in Fig. 3.2, (b) the circuit
in Fig. 3,17.
GNI GN2
GNN -
UN
IN
..
3.6
Nodal and Mesh Analyses by Inspection
or simply
Gy = i
(3.23)
where
Transcribed Image Text:Vo) LTE 1:31 A LIEI l 38% FundamentalsofElectricCi. 100 Chapter 3 Methods of Analysis But I, = -i hence, iz = iz - 3i4 (3 Applying KVL in mesh 4, 2i4 + 8(i4 - i) + 10 = 0 or Si4 - 4iz = -5 From Eqs. (3.7.1) to (3.7.4), i = -7.5 A, i, = -2.5 A, iz = 3.93 A, i, = 2,14: Practice Problem 3.7 Use mesh analysis to determine i, i2, and iz in Fig. 3.25. Answer: i, = 3.474 A, iz = 0.4737 A, iz = 1.1052 A. 3 20 6 V 3 A ( † Nodal and Mesh Analyses by Inspection 3.6 Figure 3.25 For Practice Prob. 3.7. This section presents a generalized procedure for nodal or mesh sis. It is a shortcut approach based on mere inspection of a circu When all sources in a circuit are independent current source do not need to apply KCL to each node to obtain the node-vo equations as we did in Section 3.2. We can obtain the equation mere inspection of the circuit. As an example, let us reexamine th cuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience. circuit has two nonreference nodes and the node equations derived in Section 3.2 as G2 G G, + G2 -G2 -G2 Observe that each of the diagonal terms is the sum of the conduct connected directly to node 1 or 2, while the off-diagonal terms ar negatives of the conductances connected between the nodes. Also, term on the right-hand side of Eq. (3.21) is the algebraic sum o currents entering the node, In general, if a circuit with independent current sources has reference nodes, the node-voltage equations can be written in tern the conductances as (a) R ww ww V2 Gu G12 ... GIN ןט (b) G21 G22 G2N U2 Figure 3.26 (a) The circuit in Fig. 3.2, (b) the circuit in Fig. 3,17. GNI GN2 GNN - UN IN .. 3.6 Nodal and Mesh Analyses by Inspection or simply Gy = i (3.23) where
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