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Solution:

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Common symbols for indicating a reference node, (a) common ground, (b) ground, (c) chassis ground.

The number of nonreference nodes is equal to the number of independent equations that we will derive.

Figure 3.2 Typical circuit for nodal analysis.

Note that this principle is in agreement with the w ay we defined resistance in Chapter 2 (see Fig. 2.1). With this in mind, we obtain from Fig. 3.2(b),

i1=v1−0R1ori1=G1v1i_1 = \frac{v_1 - 0}{R_1} \quad \text{or} \quad i_1 = G_1 v_1

\n

i2=v1−v2R2ori2=G2(v1−v2)i_2 = \frac{v_1 - v_2}{R_2} \quad \text{or} \quad i_2 = G_2 (v_1 - v_2)

\n

i3=v2−0R3ori3=G3v2i_3 = \frac{v_2 - 0}{R_3} \quad \text{or} \quad i_3 = G_3 v_2

\n(3.4)

Substituting Eq. (3.4) in Eqs. (3.1) and (3.2) results, respectively, in

I1=I2+v1R1+v1−v2R2I_1 = I_2 + \frac{v_1}{R_1} + \frac{v_1 - v_2}{R_2}

(3.5)

I2+v1−v2R2=v2R3I_2 + \frac{v_1 - v_2}{R_2} = \frac{v_2}{R_3}

(3.6)

In terms of the conductances, Eqs. (3.5) and (3.6) become

I1=I2+G1v1+G2(v1−v2)I_1 = I_2 + G_1 v_1 + G_2 (v_1 - v_2)

\n(3.7)

I2+G2(v1−v2)=G3v2(3.8)I_2 + G_2(v_1 - v_2) = G_3v_2 \tag{3.8}

The third step in nodal analysis is to solve for the node voltages. If we apply KCL to n − 1 nonreference nodes, we obtain n − 1 simulta neous equations such as Eqs. (3.5) and (3.6) or (3.7) and (3.8). F or the circuit of Fig. 3.2, we solve Eqs. (3.5) and (3.6) or (3.7) and (3.8) to obtain the node voltages v1 and v2 using any standard method, such as the substitution method, the elimination method, Cramer’ s rule, or matrix inversion. To use either of the last two methods, one must cast the simultaneous equations in matrix form. For example, Eqs. (3.7) and (3.8) can be cast in matrix form as

[G1+G2−G2 −G2G2+G3][v1 v2]=[I1−I2 I2]\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}

(3.9)

which can be solved to get v1 and v2. Equation 3.9 will be generalized in Section 3.6. The simultaneous equations may also be solv ed using calculators or with softw are packages such as MATLAB, Mathcad, Maple, and Quattro Pro.

Example 3.1 Calculate the node voltages in the circuit shown in Fig. 3.3(a).

Solution:

Consider Fig. 3.3(b), where the circuit in Fig. 3.3(a) has been prepared for nodal analysis. Notice how the currents are selected for the application of KCL. Except for the branches with current sources, the labeling of the currents is arbitrary but consistent. (By consistent, we mean that if, for example, we assume that i2 enters the 4-Ω resistor from the left-hand side, i2 must leave the resistor from the right-hand side.) The reference node is selected, and the node voltages v1 and v2 are now to be determined.

At node 1, applying KCL and Ohm’s law gives

i1=i2+i3i_1 = i_2 + i_3

⇒\Rightarrow 5=v1−v24+v1−025 = \frac{v_1 - v_2}{4} + \frac{v_1 - 0}{2}

Multiplying each term in the last equation by 4, we obtain

20=v1−v2+2v120 = v_1 - v_2 + 2v_1

Appendix A discusses how to use Cramer’s rule.

3v1−v2=20(3.1.1)3v_1 - v_2 = 20 \tag{3.1.1}

At node 2, we do the same thing and get

i2+i4=i1+i5i_2 + i_4 = i_1 + i_5

⇒\Rightarrow v1−v24+10=5+v2−06\frac{v_1 - v_2}{4} + 10 = 5 + \frac{v_2 - 0}{6}

Multiplying each term by 12 results in

3v1−3v2+120=60+2v23v_1 - 3v_2 + 120 = 60 + 2v_2

or

−3v1+5v2=60(3.1.2)-3v_1 + 5v_2 = 60 \tag{3.1.2}

Now we have two simultaneous Eqs. (3.1.1) and (3.1.2). We can solve the equations using any method and obtain the values of v1 and v2.

■ METHOD 1 Using the elimination technique, we add Eqs. (3.1.1) and (3.1.2).

4v2=80⇒v2=20 V4v_2 = 80 \qquad \Rightarrow \qquad v_2 = 20 \text{ V}

Substituting v2 = 20 in Eq. (3.1.1) gives

3v1−20=203v_1 - 20 = 20

⇒\Rightarrow v1=403=13.333v_1 = \frac{40}{3} = 13.333 V

■ METHOD 2 To use Cramer’s rule, we need to put Eqs. (3.1.1) and (3.1.2) in matrix form as

[3−1−35][v1v2]=[2060]\begin{bmatrix} 3 & -1 \\ -3 & 5 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 20 \\ 60 \end{bmatrix}

(3.1.3)

The determinant of the matrix is

Δ=[3−1−35]=15−3=12\Delta = \begin{bmatrix} 3 & -1 \\ -3 & 5 \end{bmatrix} = 15 - 3 = 12

We now obtain v1 and v2 as

v1=Δ1Δ=∣20−1605∣Δ=100+6012=13.333 Vv_1 = \frac{\Delta_1}{\Delta} = \frac{\begin{vmatrix} 20 & -1 \\ 60 & 5 \end{vmatrix}}{\Delta} = \frac{100 + 60}{12} = 13.333 \text{ V} v2=Δ2Δ=∣320−360∣Δ=180+6012=20 Vv_2 = \frac{\Delta_2}{\Delta} = \frac{\begin{vmatrix} 3 & 20 \\ -3 & 60 \end{vmatrix}}{\Delta} = \frac{180 + 60}{12} = 20 \text{ V}

giving us the same result as did the elimination method.

If we need the currents, we can easily calculate them from the values of the nodal voltages.

i1=5 A,i_1 = 5 \text{ A},

i2=v1−v24=−1.6668 A,i_2 = \frac{v_1 - v_2}{4} = -1.6668 \text{ A}, i3=v12=6.666 Ai_3 = \frac{v_1}{2} = 6.666 \text{ A}
i4=10 A,i_4 = 10 \text{ A}, i5=v26=3.333 Ai_5 = \frac{v_2}{6} = 3.333 \text{ A}

The fact that i2 is negative shows that the current flows in the direction opposite to the one assumed.

Obtain the node voltages in the circuit of Fig. 3.4. Practice Problem 3.1

2

1= 5

2 Ω 6 Ω 10 A

(a)

2 Ω 6 Ω 10 A

2 i 5

i i 4= 10 2

5 A

4 Ω

3 i

v2 v1

1= 5 i

5 A

4 Ω

1

i

i

Answer: v1 = 30 V, v2 = −2.5 V.

14 A 7 A 5 Ω 4 Ω 5 Ω 1 2

Example 3.2 Determine the voltages at the nodes in Fig. 3.5(a).

Solution:

Figure 3.4 For Practice Prob. 3.1.

The circuit in this example has three nonreference nodes, unlike the previous example which has two nonreference nodes. We assign voltages to the three nodes as shown in Fig. 3.5(b) and label the currents.

At node 1,

3=i1+ix⇒3=v1−v34+v1−v223 = i_1 + i_x \qquad \Rightarrow \qquad 3 = \frac{v_1 - v_3}{4} + \frac{v_1 - v_2}{2}

Multiplying by 4 and rearranging terms, we get

3v1−2v2−v3=12(3.2.1)3v_1 - 2v_2 - v_3 = 12 \tag{3.2.1}

At node 2,

ix=i2+i3i_x = i_2 + i_3

⇒\Rightarrow v1−v22=v2−v38+v2−04\frac{v_1 - v_2}{2} = \frac{v_2 - v_3}{8} + \frac{v_2 - 0}{4}

Multiplying by 8 and rearranging terms, we get

−4v1+7v2−v3=0(3.2.2)-4v_1 + 7v_2 - v_3 = 0 \tag{3.2.2}

At node 3,

i1+i2=2ixi_1 + i_2 = 2i_x

⇒\Rightarrow v1−v34+v2−v38=2(v1−v2)2\frac{v_1 - v_3}{4} + \frac{v_2 - v_3}{8} = \frac{2(v_1 - v_2)}{2}

Multiplying by 8, rearranging terms, and dividing by 3, we get

2v1−3v2+v3=0(3.2.3)2v_1 - 3v_2 + v_3 = 0 \tag{3.2.3}

We have three simultaneous equations to solve to get the node voltages v1, v2, and v3. We shall solve the equations in three ways.

■ METHOD 1 Using the elimination technique, we add Eqs. (3.2.1) and (3.2.3).

5v1−5v2=125v_1 - 5v_2 = 12

or

v1−v2=125=2.4v_1 - v_2 = \frac{12}{5} = 2.4

(3.2.4)

Adding Eqs. (3.2.2) and (3.2.3) gives

−2v1+4v2=0⇒v1=2v2(3.2.5)-2v_1 + 4v_2 = 0 \Rightarrow v_1 = 2v_2 \tag{3.2.5}

Substituting Eq. (3.2.5) into Eq. (3.2.4) yields

2v2−v2=2.42v_2 - v_2 = 2.4

⇒\Rightarrow v2=2.4v_2 = 2.4 , v1=2v2=4.8v_1 = 2v_2 = 4.8 V

From Eq. (3.2.3), we get

v3=3v2−2v1=3v2−4v2=−v2=−2.4v_3 = 3v_2 - 2v_1 = 3v_2 - 4v_2 = -v_2 = -2.4

V

Thus,

v1=4.8 V,v2=2.4 V,v3=−2.4 Vv_1 = 4.8 \text{ V}, \qquad v_2 = 2.4 \text{ V}, \qquad v_3 = -2.4 \text{ V}

■ METHOD 2 To use Cramer’s rule, we put Eqs. (3.2.1) to (3.2.3) in matrix form.

[3−2−1 −47−1 2−31][v1 v2 v3]=[12 0 0]\begin{bmatrix} 3 & -2 & -1 \ -4 & 7 & -1 \ 2 & -3 & 1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 12 \ 0 \ 0 \end{bmatrix}

(3.2.6)

From this, we obtain

v1=Δ1Δv_1 = \frac{\Delta_1}{\Delta}

, v2=Δ2Δv_2 = \frac{\Delta_2}{\Delta} , v3=Δ3Δv_3 = \frac{\Delta_3}{\Delta}

where Δ, Δ1, Δ2, and Δ3 are the determinants to be calculated as follows. As explained in Appendix A, to calculate the determinant of a 3 by 3 matrix, we repeat the first two rows and cross multiply.

Similarly, we obtain

Δ3 =

ǀ

− − −

2 3 −4

−3 −2 7

0 12 0

ǀ

Δ1=∣12−2−103−1−1122∣∣2−2−1−12−107−1∣+∣1−84+0+0−0−36−0=48+∣\Delta_1 = \frac{\begin{vmatrix} 12 & -2 & -1 \\ 0 & 3 & -1 \\ -1 & 12 & 2 \end{vmatrix}}{\begin{vmatrix} 2 & -2 & -1 \\ -1 & 2 & -1 \\ 0 & 7 & -1 \end{vmatrix} + \begin{vmatrix} 1 & -84 + 0 + 0 - 0 - 36 - 0 = 48 \\ + \end{vmatrix}} Δ2=∣312−1312−1−33−2∣∣3−2−12−30−1∣+∣3−2−12+∣\Delta_2 = \frac{\begin{vmatrix} 3 & 12 & -1 \\ 3 & 12 & -1 \\ -3 & 3 & -2 \end{vmatrix}}{\begin{vmatrix} 3 & -2 & -12 \\ -3 & 0 & -1 \end{vmatrix} + \begin{vmatrix} 3 & -2 & -12 \\ + \end{vmatrix}}

= 0 + 144 + 0 − 168 − 0 − 0 = −24

Thus, we find

v1=Δ1Δ=4810=4.8 V,v2=Δ2Δ=2410=2.4 Vv_1 = \frac{\Delta_1}{\Delta} = \frac{48}{10} = 4.8 \text{ V}, \qquad v_2 = \frac{\Delta_2}{\Delta} = \frac{24}{10} = 2.4 \text{ V}

v3=Δ3Δ=−2410=−2.4 Vv_3 = \frac{\Delta_3}{\Delta} = \frac{-24}{10} = -2.4 \text{ V}

as we obtained with Method 1.

■ METHOD 3 We now use MATLAB to solve the matrix. Equation (3.2.6) can be written as

AV=B⇒V=A−1BAV = B \qquad \Rightarrow \qquad V = A^{-1}B

where A is the 3 by 3 square matrix, B is the column vector, and V is a column vector comprised of v1, v2, and v3 that we want to determine. We use MATLAB to determine V as follows:

A =

[3−2−147−12−31]\begin{bmatrix} 3 & -2 & -1 \\ 4 & 7 & -1 \\ 2 & -3 & 1 \end{bmatrix}

;

B = [120048]\begin{bmatrix} 12 & 0 & 0 \\ 4 & 8 \end{bmatrix} ;
V = inv(A) * B
= 4.8000
= 2.4000
= 2.4000

Thus, v1 = 4.8 V, v2 = 2.4 V, and v3 = −2.4 V, as obtained previously.

Practice Problem 3.2 Find the voltages at the three nonreference nodes in the circuit of Fig. 3.6.

4 A 2 Ω 3 Ω 4 Ω 6 Ω i x 4i x 1 3 2

Figure 3.6 For Practice Prob. 3.2.

Answer: v1 = 32 V, v2 = −25.6 V, v3 = 62.4 V.