3.6 Nodal and Mesh Analyses by Inspection

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3.6 Nodal and Mesh Analyses by Inspection

This section presents a generalized procedure for nodal or mesh analysis. It is a shortcut approach based on mere inspection of a circuit.

When all sources in a circuit are independent current sources, we do not need to apply KCL to each node to obtain the node-v oltage equations as we did in Section 3.2. We can obtain the equations by mere inspection of the circuit. As an e xample, let us ree xamine the circuit in Fig. 3.2, shown again in Fig. 3.26(a) for convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.2 as

$$\begin{bmatrix} G_1 + G_2 & -G_2 \ -G_2 & G_2 + G_3 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} I_1 - I_2 \ I_2 \end{bmatrix}$$

(3.21)

Observe that each of the diagonal terms is the sum of the conductances connected directly to node 1 or 2, while the off-diagonal terms are the negatives of the conductances connected between the nodes. Also, each term on the right-hand side of Eq. (3.21) is the algebraic sum of the currents entering the node.

In general, if a circuit with independent current sources has N nonreference nodes, the node-v oltage equations can be written in terms of the conductances as

$$\begin{bmatrix} G_{11} & G_{12} & \cdots & G_{1N} \\ G_{21} & G_{22} & \cdots & G_{2N} \\ \vdots & \vdots & \vdots & \vdots \\ G_{N1} & G_{N2} & \cdots & G_{NN} \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{bmatrix} = \begin{bmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{bmatrix}$$

(3.22)

or simply

$$Gv = i \tag{3.23}$$

where

Gkk = Sum of the conductances connected to node k

Gk j = Gjk = Negative of the sum of the conductances directly connecting nodes k and j, k ≠ j

vk= Unknown voltage at node k

ik = 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 v ector; and i is the input vector. Equation (3.22) can be solved to obtain the unknown node voltages. Keep in mind that this is v alid for circuits with only indepen dent 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 convenience. The circuit has two nonreference nodes and the node equations were derived in Section 3.4 as

$$\begin{bmatrix} R_1 + R_3 & -R_3 \ -R_3 & R_2 + R_3 \end{bmatrix} \begin{bmatrix} i_1 \ i_2 \end{bmatrix} = \begin{bmatrix} v_1 \ -v_2 \end{bmatrix}$$

(3.24)

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 negative 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

$$\begin{bmatrix} R_{11} & R_{12} & \cdots & R_{1N} \\ R_{21} & R_{22} & \cdots & R_{2N} \\ \vdots & \vdots & \vdots & \vdots \\ R_{N1} & R_{N2} & \cdots & R_{NN} \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_N \end{bmatrix}$$

(3.25)

or simply

$$\mathbf{Ri} = \mathbf{v} \tag{3.26}$$

where

Rkk = Sum of the resistances in mesh k

• Rkj = Rjk = Negative of the sum of the resistances in common with meshes k and j, k ≠ j
• ik = Unknown mesh current for mesh k in the clockwise direction
• vk = 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 v ector; and v is the input v ector. We can solv e Eq. (3.25) to obtain the unkno wn mesh currents.

Example 3.8 Write the node-v oltage matrix equations for the circuit in Fig. 3.27 by inspection.

Figure 3.27 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

$$G_{11} = \frac{1}{5} + \frac{1}{10} = 0.3, \qquad G_{22} = \frac{1}{5} + \frac{1}{8} + \frac{1}{1} = 1.325$$ $$G_{33} = \frac{1}{8} + \frac{1}{8} + \frac{1}{4} = 0.5, \qquad G_{44} = \frac{1}{8} + \frac{1}{2} + \frac{1}{1} = 1.625$$

The off-diagonal terms are

$$G_{12} = -\frac{1}{5} = -0.2, \qquad G_{13} = G_{14} = 0$$

\n

$$G_{21} = -0.2, \qquad G_{23} = -\frac{1}{8} = -0.125, \qquad G_{24} = -\frac{1}{1} = -1$$

\n

$$G_{31} = 0, \qquad G_{32} = -0.125, \qquad G_{34} = -\frac{1}{8} = -0.125$$

\n

$$G_{41} = 0, \qquad G_{42} = -1, \qquad G_{43} = -0.125$$

The input current vector i has the following terms, in amperes:

$$i_1 = 3$$

, $i_2 = -1 - 2 = -3$ , $i_3 = 0$ , $i_4 = 2 + 4 = 6$

Thus, the node-voltage equations are

$$\begin{bmatrix} 0.3 & -0.2 & 0 & 0 \ -0.2 & 1.325 & -0.125 & -1 \ 0 & -0.125 & 0.5 & -0.125 \ 0 & -1 & -0.125 & 1.625 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \ v_4 \end{bmatrix} = \begin{bmatrix} 3 \ -3 \ 0 \ 6 \end{bmatrix}$$

which can be solved using MATLAB to obtain the node voltages v1, v2, v3, and v4.

By inspection, obtain the node-voltage equations for the circuit in Fig. 3.28.

Answer:

1.25	−0.2	−1	0	0 v1
−0.2	0.2	0	0	5 v2
−1	0	1.25	−0.25	= −3 v3
[ 0	0	−0.25	1.25	] [ ] [ 2] v4

By inspection, write the mesh-current equations for the circuit in Fig. 3.29. Example 3.9

Figure 3.29

For Example 3.9.

Solution:

We have five meshes, so the resistance matrix is 5 by 5. The diagonal terms, in ohms, are:

$$R_{11} = 5 + 2 + 2 = 9, \qquad R_{22} = 2 + 4 + 1 + 1 + 2 = 10,$$ $$R_{33} = 2 + 3 + 4 = 9, \qquad R_{44} = 1 + 3 + 4 = 8, \qquad R_{55} = 1 + 3 = 4$$

The off-diagonal terms are:

$$R_{12} = -2, \t R_{13} = -2, \t R_{14} = 0 = R_{15},$$

\n

$$R_{21} = -2, \t R_{23} = -4, \t R_{24} = -1, \t R_{25} = -1,$$

\n

$$R_{31} = -2, \t R_{32} = -4, \t R_{34} = 0 = R_{35},$$

\n

$$R_{41} = 0, \t R_{42} = -1, \t R_{43} = 0, \t R_{45} = -3,$$

\n

$$R_{51} = 0, \t R_{52} = -1, \t R_{53} = 0, \t R_{54} = -3$$

The input voltage vector v has the following terms in volts:

$$v_1 = 4$$

, $v_2 = 10 - 4 = 6$ ,
$v_3 = -12 + 6 = -6$ , $v_4 = 0$ , $v_5 = -6$

Thus, the mesh-current equations are:

	9	−2	−2	0	0		i1		4
	−2	10	−4	−1	−1		i2		6
	−2	−4	9	0	0		i3		−6
[	0	−1	0	8	−3 ]		i4	= [	0 ]
	0	−1	0	−3	4	[	] i5		−6

From this, we can use MATLAB to obtain mesh currents i1, i2, i3, i4, and i5.

Practice Problem 3.9

By inspection, obtain the mesh-current equations for the circuit in Fig. 3.30.

Figure 3.30 For Practice Prob. 3.9.

Answer:

	150	−40	0	−80	0	i1		30
	−40	65	−30	−15	0	i2		0
	−0	−30	50	0	−20	i3		−12
[	80	−15	0	95	0 ]	i4 [	= [ ]	20 ]
	0	0	−20	0	80	i5		−20