3.3 Nodal Analysis with Voltage Sources

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3.3 Nodal Analysis with Voltage Sources

We now consider ho w voltage sources af fect nodal analysis. W e use the circuit in Fig. 3.7 for illustration. Consider the following two possibilities.

■ CASE 1 If a voltage source is connected between the reference node and a nonreference node, we simply set the voltage at the non reference node equal to the voltage of the voltage source. In Fig. 3.7, for example,

$$v_1 = 10 \text{ V} \tag{3.10}$$

Thus, our analysis is somewhat simplified by this knowledge of the voltage at this node.

■ CASE 2 If the voltage source (dependent or independent) is con nected between two nonreference nodes, the two nonreference nodes

form a generalized node or supernode; we apply both KCL and KVL to determine the node voltages.

A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.

In Fig. 3.7, nodes 2 and 3 form a supernode. (W e could have more than two nodes forming a single supernode. F or example, see the circuit in Fig. 3.14.) We analyze a circuit with supernodes using the same three steps mentioned in the pre vious section e xcept that the supernodes are treated dif ferently. Why? Because an essential component of nodal analysis is applying KCL, which requires kno wing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be satisfied at a supernode like any other node. Hence, at the super node in Fig. 3.7,

$$i_1 + i_4 = i_2 + i_3 \tag{3.11a}$$

or

$$\frac{v_1 - v_2}{2} + \frac{v_1 - v_3}{4} = \frac{v_2 - 0}{8} + \frac{v_3 - 0}{6}$$

(3.11b)

To apply Kirchhoff’s voltage law to the supernode in Fig. 3.7, we redraw the circuit as shown in Fig. 3.8. Going around the loop in the clockwise direction gives

$$-v_2 + 5 + v_3 = 0 \qquad \Rightarrow \qquad v_2 - v_3 = 5 \tag{3.12}$$

• From Eqs. (3.10), (3.11b), and (3.12), we obtain the node voltages. Note the following properties of a supernode:
  • 1. The v oltage source inside the supernode pro vides a constraint equation needed to solve for the node voltages.
  • 2. A supernode has no voltage of its own.
  • 3. A supernode requires the application of both KCL and KVL.

Figure 3.8 Applying KVL to a supernode.

A supernode may be regarded as a closed surface enclosing the voltage source and its two nodes.

For Example 3.3.

For the circuit shown in Fig. 3.9, find the node voltages.

Solution:

The supernode contains the 2-V source, nodes 1 and 2, and the 10- Ω resistor. Applying KCL to the supernode as shown in Fig. 3.10(a) gives

$$2 = i_1 + i_2 + 7$$

Expressing i1 and i2 in terms of the node voltages

$$2 = \frac{v_1 - 0}{2} + \frac{v_2 - 0}{4} + 7 \qquad \Rightarrow \qquad 8 = 2v_1 + v_2 + 28$$

or

$$v_2 = -20 - 2v_1 \tag{3.3.1}$$

To get the relationship between v1 and v2, we apply KVL to the circuit in Fig. 3.10(b). Going around the loop, we obtain

−v¹ − 2 + v² = 0 ⇒ v² = v¹ + 2 (3.3.2)

From Eqs. (3.3.1) and (3.3.2), we write

$$v_2 = v_1 + 2 = -20 - 2v_1$$

or

$$3v_1 = -22 \qquad \Rightarrow \qquad v_1 = -7.333 \text{ V}$$

and v² = v¹ + 2 = −5.333 V. Note that the 10-Ω resistor does not make any difference because it is connected across the supernode.

Applying: (a) KCL to the supernode, (b) KVL to the loop.

Find the node voltages in the circuit of Fig. 3.12. Example 3.4

Figure 3.12 For Example 3.4. 20 V 2 Ω 4 Ω 6 Ω 3 Ω 1 Ω vx 3v^x + – + – 10 A 1 4 2 3 + –

Solution:

Nodes 1 and 2 form a supernode; so do nodes 3 and 4. We apply KCL to the two supernodes as in Fig. 3.13(a). At supernode 1-2,

$$i_3 + 10 = i_1 + i_2$$

Expressing this in terms of the node voltages,

______ v³ − v² 6 + 10 = ______ v¹ − v⁴ 3 + __ v1 2

or

$$5v_1 + v_2 - v_3 - 2v_4 = 60 \tag{3.4.1}$$

At supernode 3-4,

$$i_1 = i_3 + i_4 + i_5$$

$\Rightarrow$ $\frac{v_1 - v_4}{3} = \frac{v_3 - v_2}{6} + \frac{v_4}{1} + \frac{v_3}{4}$

or

$$4v_1 + 2v_2 - 5v_3 - 16v_4 = 0 \tag{3.4.2}$$

Figure 3.13 Applying: (a) KCL to the two supernodes, (b) KVL to the loops.

We now apply KVL to the branches involving the voltage sources as shown in Fig. 3.13(b). For loop 1,

$$-v_1 + 20 + v_2 = 0 \Rightarrow v_1 - v_2 = 20 \tag{3.4.3}$$

For loop 2,

$$-v_3 + 3v_x + v_4 = 0$$

But v^x = v¹ − v4 so that

$$3v_1 - v_3 - 2v_4 = 0 \tag{3.4.4}$$

For loop 3,

$$v_x - 3v_x + 6i_3 - 20 = 0$$

But 6i³ = v³ − v2 and v^x = v¹ − v4. Hence,

$$-2v_1 - v_2 + v_3 + 2v_4 = 20 \tag{3.4.5}$$

We need four node v oltages, v1, v2, v3, and v4, and it requires only four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fifth equation is redundant, it can be used to check results. We can solve Eqs. (3.4.1) to (3.4.4) directly using MATLAB. We can eliminate one node voltage so that we solv e three simultaneous equations instead of four . From Eq. (3.4.3), v² = v¹ − 20. Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives

$$6v_1 - v_3 - 2v_4 = 80 \tag{3.4.6}$$

and

$$6v_1 - 5v_3 - 16v_4 = 40 \tag{3.4.7}$$

Equations (3.4.4), (3.4.6), and (3.4.7) can be cast in matrix form as

3 6 6 −1 −1 −5 −2 −2 ⁻16 ] [ v1 v3 v4 ] = [ 0 80 ⁴⁰]

Using Cramer’s rule gives

$$\Delta = \begin{vmatrix} 3 & -1 & -2 \\ 6 & -1 & -2 \\ 6 & -5 & -16 \end{vmatrix} = -18, \qquad \Delta_1 = \begin{vmatrix} 0 & -1 & -2 \\ 80 & -1 & -2 \\ 40 & -5 & -16 \end{vmatrix} = -480,$$ $$\Delta_3 = \begin{vmatrix} 3 & 0 & -2 \\ 6 & 80 & -2 \\ 6 & 40 & -16 \end{vmatrix} = -3120, \qquad \Delta_4 = \begin{vmatrix} 3 & -1 & 0 \\ 6 & -1 & 80 \\ 6 & -5 & 40 \end{vmatrix} = 840$$

Thus, we arrive at the node voltages as

[

$$v_1 = \frac{\Delta_1}{\Delta} = \frac{-480}{-18} = 26.67 \text{ V}, \qquad v_3 = \frac{\Delta_3}{\Delta} = \frac{-3120}{-18} = 173.33 \text{ V},$$

$v_4 = \frac{\Delta_4}{\Delta} = \frac{840}{-18} = -46.67 \text{ V}$

and v² = v¹ − 20 = 6.667 V. We have not used Eq. (3.4.5); it can be used to cross-check results.

Find v1, v2, and v3 in the circuit of Fig. 3.14 using nodal analysis. Practice Problem 3.4

Answer: v¹ = 7.608 V, v² = −17.39 V, v³ = 1.6305 V.

3.4 Mesh Analysis

Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it.

Nodal analysis applies KCL to find unknown voltages in a gi ven circuit, while mesh analysis applies KVL to find unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar. A circuit may have crossing branches and still be planar if it can be redra wn such that it has no crossing branches. F or example, the circuit in Fig. 3.15(a) has tw o crossing branches, b ut it can be redrawn as in Fig. 3.15(b). Hence, the circuit in Fig. 3.15(a) is planar. However, the circuit in Fig. 3.16 is nonplanar , because there is no w ay to redra w it and a void the branches crossing. Nonplanar circuits can be handled using nodal analysis, but they will not be considered in this text.

Figure 3.14 For Practice Prob. 3.4.

Mesh analysis is also known as loop analysis or the mesh-current method.

Figure 3.15

(a) A planar circuit with crossing branches, (b) the same circuit redrawn with no crossing branches.

To understand mesh analysis, we should first explain more about what we mean by a mesh.

A mesh is a loop that does not contain any other loops within it.

Figure 3.17 A circuit with two meshes.

In Fig. 3.17, for example, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The current through a mesh is known as mesh current. In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit.

In this section, we will apply mesh analysis to planar circuits that do not contain current sources. In the next section, we will consider circuits with current sources. In the mesh analysis of a circuit with n meshes, we take the following three steps.

Steps to Determine Mesh Currents:

• 1. Assign mesh currents i1, i2, … , in to the n meshes.
• 2. Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents.
• 3. Solve the resulting n simultaneous equations to get the mesh currents.

To illustrate the steps, consider the circuit in Fig. 3.17. The first step requires that mesh currents i1 and i2 are assigned to meshes 1 and 2. Although a mesh current may be assigned to each mesh in an arbi trary direction, it is conventional to assume that each mesh current flows clockwise.

As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain

$$-V_1 + R_1 i_1 + R_3 (i_1 - i_2) = 0$$

or

$$(R_1 + R_3)i_1 - R_3i_2 = V_1 \tag{3.13}$$

−R³ i¹ + (R² + R3) i² = −V² (3.14)

For mesh 2, applying KVL gives

$$R_2 i_2 + V_2 + R_3 (i_2 - i_1) = 0$$

or

The shortcut way will not apply if one mesh current is assumed clockwise and the other assumed counterclockwise, although this is permissible. Note in Eq. (3.13) that the coefficient of i1 is the sum of the resistances in the first mesh, while the coefficient of i2 is the negative of the resistance common to meshes 1 and 2. Now observe that the same is true in Eq. (3.14). This can serve as a shortcut way of writing the mesh equa tions. We will exploit this idea in Section 3.6.

Although path abcdefa is a loop and not a mesh, KVL still holds. This is the reason for loosely using the terms loop analysis and mesh analysis to mean the same thing.

The direction of the mesh current is arbitrary—(clockwise or counterclockwise)—and does not affect the validity of the solution.

The third step is to solve for the mesh currents. Putting Eqs. (3.13) and (3.14) in matrix form yields

$$\begin{bmatrix} R_1 + R_3 & -R_3 \ -R_3 & R_2 + R_3 \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \end{bmatrix} = \begin{bmatrix} V_1 \\ -V_2 \end{bmatrix}$$

(3.15)

which can be solved to obtain the mesh currents i1 and i2. We are at liberty to use any technique for solving the simultaneous equations. According to Eq. (2.12), if a circuit has n nodes, b branches, and l independent loops or meshes, then l = b − n + 1. Hence, l independent simultaneous equations are required to solve the circuit using mesh analysis.

Notice that the branch currents are different from the mesh currents unless the mesh is isolated. To distinguish between the two types of currents, we use i for a mesh current and I for a branch current. The current elements I1, I2, and I3 are algebraic sums of the mesh currents. It is e vident from Fig. 3.17 that

$$I_1 = i_1,$$

$I_2 = i_2,$ $I_3 = i_1 - i_2$ (3.16)

For the circuit in Fig. 3.18, find the branch currents I1, I2, and I3 using mesh analysis.

Solution:

We first obtain the mesh currents using KVL. For mesh 1,

$$-15 + 5i_1 + 10(i_1 - i_2) + 10 = 0$$

or

$$3i_1 - 2i_2 = 1 \t\t(3.5.1)$$

For mesh 2,

$$6i_2 + 4i_2 + 10(i_2 - i_1) - 10 = 0$$

or

$$i_1 = 2i_2 - 1 \tag{3.5.2}$$

■ METHOD 1 Using the substitution method, we substitute Eq. (3.5.2) into Eq. (3.5.1), and write

$$6i_2 - 3 - 2i_2 = 1 \qquad \Rightarrow \qquad i_2 = 1 \text{ A}$$

From Eq. (3.5.2), i¹ = 2i² − 1 = 2 − 1 = 1 A. Thus,

$$I_1 = i_1 = 1$$

A, $I_2 = i_2 = 1$ A, $I_3 = i_1 - i_2 = 0$

■ METHOD 2 To use Cramer’s rule, we cast Eqs. (3.5.1) and (3.5.2) in matrix form as

[ 3 −1 −2 2 ] [i1 i2 ] ⁼ [ ¹ 1]

For Example 3.5.

We obtain the determinants

$$\Delta = \begin{vmatrix} 3 & -2 \\ -1 & 2 \end{vmatrix} = 6 - 2 = 4$$ $$\Delta_1 = \begin{vmatrix} 1 & -2 \\ 1 & 2 \end{vmatrix} = 2 + 2 = 4, \qquad \Delta_2 = \begin{vmatrix} 3 & 1 \\ -1 & 1 \end{vmatrix} = 3 + 1 = 4$$

Thus,

$$i_1 = \frac{\Delta_1}{\Delta} = 1 \text{ A}, \qquad i_2 = \frac{\Delta_2}{\Delta} = 1 \text{ A}$$

as before.

Practice Problem 3.5

Calculate the mesh currents i1 and i2 of the circuit of Fig. 3.19.

Answer: i¹ = 4.6 A, i² = 200 mA.

Figure 3.19 For Practice Prob. 3.5.

Example 3.6 Use mesh analysis to find the current Io in the circuit of Fig. 3.20.

Solution:

We apply KVL to the three meshes in turn. For mesh 1,

$$-24 + 10(i_1 - i_2) + 12(i_1 - i_3) = 0$$ $$11i_1 - 5i_2 - 6i_3 = 12 \tag{3.6.1}$$

For mesh 2,

24i² + 4 (i² − i3) + 10 (i² − i1) = 0

$$-5i_1 + 19i_2 - 2i_3 = 0 \tag{3.6.2}$$

For mesh 3,

Figure 3.20 For Example 3.6.

$$4I_o + 12(i_3 - i_1) + 4(i_3 - i_2) = 0$$

But at node A, Io = i¹ − i2, so that

$$4(i1 - i2) + 12(i3 - i1) + 4(i3 - i2) = 0$$

or

$$-i_1 - i_2 + 2i_3 = 0 \tag{3.6.3}$$

In matrix form, Eqs. (3.6.1) to (3.6.3) become

$$\begin{bmatrix} 11 & -5 & -6 \ -5 & 19 & -2 \ -1 & -1 & 2 \ \end{bmatrix} \begin{bmatrix} i_1 \ i_2 \ i_3 \end{bmatrix} = \begin{bmatrix} 12 \ 0 \ 0 \end{bmatrix}$$

We obtain the determinants as

−2

$$\Delta_3 = \begin{pmatrix}\n11 & -5 & 12 \\ -5 & 19 & 0 \\ -11 & -5 & 19\n\end{pmatrix} + 60 + 228 = 288$$

We calculate the mesh currents using Cramer’s rule as

−5

− −

0

$$i_1 = \frac{\Delta_1}{\Delta} = \frac{432}{192} = 2.25 \text{ A}, \qquad i_2 = \frac{\Delta_2}{\Delta} = \frac{144}{192} = 0.75 \text{ A},$$

$i_3 = \frac{\Delta_3}{\Delta} = \frac{288}{192} = 1.5 \text{ A}$

Thus, Io = i¹ − i² = 1.5 A.

Practice Problem 3.6

Figure 3.21 For Practice Prob. 3.6.

Figure 3.22 A circuit with a current source.

Using mesh analysis, find Io in the circuit of Fig. 3.21.

Answer: −4 A.