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98 Chapter 3 Methods of Analysis
Using mesh analysis, find in the circuit of Fig. 3.21.
Answer: 4 A.
Practice Problem 3.6^ Io
4 Ω 8 Ω
2 Ω
6 Ω
i 1 i 2
i 3
10 io
Io
Figure 3. For Practice Prob. 3.6.
10 V − 5 A
4 Ω 3 Ω
i 1 6 Ω i 2
Figure 3. A circuit with a current source.
A supermesh results when two meshes have a (dependent or inde- pendent) current source in common.
(b)
20 V 4 Ω
6 Ω 10 Ω
6 A
20 V
6 Ω 10 Ω
2 Ω 4 Ω
i 1
i 1
i 2
i 2
0
(a)
Exclude these elements Figure 3. (a) Two meshes having a current source in common, (b) a supermesh, created by excluding the current source.
Mesh Analysis with Current Sources
Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Consider the following two possible cases.
■ CASE 1 When a current source exists only in one mesh: Consider
the circuit in Fig. 3.22, for example. We set A and write a mesh equation for the other mesh in the usual way; that is, (3.17)
■ CASE 2 When a current source exists between two meshes: Con-
sider the circuit in Fig. 3.23(a), for example. We create a supermesh by excluding the current source and any elements connected in series with it, as shown in Fig. 3.23(b). Thus,
10 4 i 1 6( i 1 i 2 ) 0 1 i 1 2 A
i 2 5
As shown in Fig. 3.23(b), we create a supermesh as the periphery of the two meshes and treat it differently. (If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh.) Why treat the supermesh differently? Because mesh analy- sis applies KVL—which requires that we know the voltage across each branch—and we do not know the voltage across a current source in advance. However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in Fig. 3.23(b) gives 20 6 i 1 10 i 2 4 i 2 0
or
(3.18)
We apply KCL to a node in the branch where the two meshes inter- sect. Applying KCL to node 0 in Fig. 3.23(a) gives
(3.19)
Solving Eqs. (3.18) and (3.19), we get
(3.20) Note the following properties of a supermesh:
- The current source in the supermesh provides the constraint equa- tion necessary to solve for the mesh currents.
- A supermesh has no current of its own.
- A supermesh requires the application of both KVL and KCL.
i 1 3.2 A, i 2 2.8 A
i 2 i 1 6
6 i 1 14 i 2 20
3.5 Mesh Analysis with Current Sources 99
For the circuit in Fig. 3.24, find i 1 to i 4 using mesh analysis. Example 3.
6 Ω 8 Ω − 10 V
4 Ω 2 Ω
i 1
i 2 i 3 i 4
2 Ω
5 A
i 1
i 2
i 2 i 3
Io
P
Q
3 I (^) o
Figure 3. For Example 3.7.
Solution: Note that meshes 1 and 2 form a supermesh since they have an independent current source in common. Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown. Applying KVL to the larger supermesh,
or
(3.7.1)
For the independent current source, we apply KCL to node P :
(3.7.2)
For the dependent current source, we apply KCL to node Q :
i 2 i 3 3 Io
i 2 i 1 5
i 1 3 i 2 6 i 3 4 i 4 0
2 i 1 4 i 3 8( i 3 i 4 ) 6 i 2 0
or simply
(3.23)
where
Sum of the conductances connected to node k Negative of the sum of the conductances directly connecting nodes k and Unknown voltage at node k Sum of all independent current sources directly connected to node k , with currents entering the node treated as positive
G is called the conductance matrix ; v is the output vector; and i is the input vector. Equation (3.22) can be solved to obtain the unknown node voltages. Keep in mind that this is valid for circuits with only inde- pendent current sources and linear resistors. Similarly, we can obtain mesh-current equations by inspection when a linear resistive circuit has only independent voltage sources. Consider the circuit in Fig. 3.17, shown again in Fig. 3.26(b) for con- venience. The circuit has two nonreference nodes and the node equa- tions were derived in Section 3.4 as
We notice that each of the diagonal terms is the sum of the resistances in the related mesh, while each of the off-diagonal terms is the nega- tive of the resistance common to meshes 1 and 2. Each term on the right-hand side of Eq. (3.24) is the algebraic sum taken clockwise of all independent voltage sources in the related mesh. In general, if the circuit has N meshes, the mesh-current equations can be expressed in terms of the resistances as
or simply
(3.26)
where
Sum of the resistances in mesh k Negative of the sum of the resistances in common with meshes k and Unknown mesh current for mesh k in the clockwise direction Sum taken clockwise of all independent voltage sources in mesh k , with voltage rise treated as positive
R is called the resistance matrix ; i is the output vector; and v is the input vector. We can solve Eq. (3.25) to obtain the unknown mesh currents.
vk
ik
j , k j
R (^) kj Rjk
Rkk
Ri v
R 11 R 12 p R 1 N R 21 R 22 p R 2 N o o o o RN 1 RN 2 p RNN
i 1 i 2 o iN
v 1 v 2 o vN
c
R 1 R 3 R 3
R 3 R 2 R 3
d c
i 1 i 2
d c
v 1 v 2
d
i (^) k
vk
j , k j
G (^) k j Gjk
G (^) kk
Gv i
3.6 Nodal and Mesh Analyses by Inspection 101
102 Chapter 3 Methods of Analysis
Write the node-voltage matrix equations for the circuit in Fig. 3.27 by inspection.
Example 3.
3 A 1 A 4 A
2 A
10 Ω
5 Ω
1 Ω
v 1 v 2 8 Ω v 3 8 Ω v 4
4 Ω 2 Ω
Figure 3. For Example 3.8.
Solution: The circuit in Fig. 3.27 has four nonreference nodes, so we need four node equations. This implies that the size of the conductance matrix G , is 4 by 4. The diagonal terms of G , in siemens, are
The off-diagonal terms are
The input current vector i has the following terms, in amperes:
Thus the node-voltage equations are
which can be solved using MATLAB to obtain the node voltages v 3 ,and v 4.
v 1 , v 2 ,
v 1 v 2 v 3 v 4
i 1 3, i 2 1 2 3, i 3 0, i 4 2 4 6
G 41 0, G 42 1, G 43 0.
G 31 0, G 32 0.125, G 34
G 21 0.2, G 23
0.125, G 24
G 12
0.2, G 13 G 14 0
G 33
0.5, G 44
G 11
0.3, G 22
Thus, the mesh-current equations are:
From this, we can use MATLAB to obtain mesh currents and i 5.
i 1 , i 2 , i 3 , i 4 ,
E
E U
i 1 i 2 i 3 i 4 i 5
E U
U
104 Chapter 3 Methods of Analysis
−
30 V
12 V
20 V
50 Ω
20 Ω
i 1
i 2 i 3
i 4 i 5
15 Ω
30 Ω
20 Ω
80 Ω 60 Ω
−
Figure 3. For Practice Prob. 3.9.
By inspection, obtain the mesh-current equations for the circuit in Fig. 3.30.
Practice Problem 3.
Answer:
E
E U
i 1 i 2 i 3 i 4 i 5
E U
U
Nodal Versus Mesh Analysis
Both nodal and mesh analyses provide a systematic way of analyzing a complex network. Someone may ask: Given a network to be ana- lyzed, how do we know which method is better or more efficient? The choice of the better method is dictated by two factors.
The first factor is the nature of the particular network. Networks that contain many series-connected elements, voltage sources, or super- meshes are more suitable for mesh analysis, whereas networks with parallel-connected elements, current sources, or supernodes are more suitable for nodal analysis. Also, a circuit with fewer nodes than meshes is better analyzed using nodal analysis, while a circuit with fewer meshes than nodes is better analyzed using mesh analysis. The key is to select the method that results in the smaller number of equations. The second factor is the information required. If node voltages are required, it may be expedient to apply nodal analysis. If branch or mesh currents are required, it may be better to use mesh analysis. It is helpful to be familiar with both methods of analysis, for at least two reasons. First, one method can be used to check the results from the other method, if possible. Second, since each method has its limitations, only one method may be suitable for a particular problem. For example, mesh analysis is the only method to use in analyzing tran- sistor circuits, as we shall see in Section 3.9. But mesh analysis can- not easily be used to solve an op amp circuit, as we shall see in Chapter 5, because there is no direct way to obtain the voltage across the op amp itself. For nonplanar networks, nodal analysis is the only option, because mesh analysis only applies to planar networks. Also, nodal analysis is more amenable to solution by computer, as it is easy to pro- gram. This allows one to analyze complicated circuits that defy hand calculation. A computer software package based on nodal analysis is introduced next.
Circuit Analysis with PSpice
PSpice is a computer software circuit analysis program that we will gradually learn to use throughout the course of this text. This section illustrates how to use PSpice for Windows to analyze the dc circuits we have studied so far. The reader is expected to review Sections D.1 through D.3 of Appendix D before proceeding in this section. It should be noted that PSpice is only helpful in determining branch voltages and currents when the numerical values of all the circuit components are known.
3.8 Circuit Analysis with PSpice 105
Appendix D provides a tutorial on using PSpice for Windows.
Use PSpice to find the node voltages in the circuit of Fig. 3.31.
Solution: The first step is to draw the given circuit using Schematics. If one follows the instructions given in Appendix sections D.2 and D.3, the schematic in Fig. 3.32 is produced. Since this is a dc analysis, we use voltage source VDC and current source IDC. The pseudocomponent VIEWPOINTS are added to display the required node voltages. Once the circuit is drawn and saved as exam310.sch , we run PSpice by selecting Analysis/Simulate. The circuit is simulated and the results
Example 3.
120 V − 3 A
20 Ω
30 Ω 40 Ω
1 2 10 Ω 3
0 Figure 3. For Example 3.10.
