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

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 = i_{1} + i_{x}    ⇒        3= \frac{v_{1} – v_{2} }{4} + \frac{v_{1} – v_{2} }{2}

Multiplying by 4 and rearranging terms, we get

3v_{1} –   2v_{2}  – v_{3} – 12                                     (3.2.1)

At node 2,

i_{x} = i_{2}  + i_{3}    ⇒      \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

-4 v_{1} + 7v_{2} – v_{3} = 0              (3.2.2)

At node 3,

i_{1} + i_{2}  =2 i_{x}    ⇒      \frac{v_{1} – v_{3}}{2} = \frac{v_{2} -v_{3}}{8} + \frac{2(v_{1} – v_{2} ) }{2}

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

2v_{1} – 3v_{2} + v_{3} = 0                      (3.2.3)

We have three simultaneous equations to solve to get the node voltages v_{1} , v_{2}   and  v_{3}   We shall solve the equations in three ways.

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

5v_{1} – 5v_{2} = 12

or

v_{1} – v_{2} = \frac{12}{5} = 2.4                          (3.2.4)

Adding Eqs. (3.2.2) and (3.2.3) gives

-2v_{1} + 4v_{2} = 0        ⇒      v_{1} = 2v_{2}              (3.2.5)

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

2v_{2} – v_{2} = 2.4        ⇒         v_{2} = 2.4 ,       v_{1} = 2v_{2} = 4.8 V

From Eq. (3.2.3), we get

v_{3} = 3v_{2} – 2v_{1} = 3v_{2} – 4v_{2} = -v_{2} = -2.4 V

Thus,

v_{1} = 4.8 V , v_{2} = 2.4 V  ,       v_{3} = -2.4 V

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

\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

v_{1} =\frac{Δ_{1}}{Δ} ,        v_{2} =\frac{Δ_{2}}{Δ} ,      v_{3} =\frac{Δ_{3}}{Δ}

where  Δ, Δ_{1}, Δ_{2}   and Δ_{3}   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.

Δ = \begin {vmatrix} 3 & -2 & -1 \\ -4 & 7 & -1 \\ 2 & -3 & 1 \end {vmatrix} =

=21 – 12 +  4 +  14 – 9 – 8  = 10

Similarly, we obtain

= 84 +  0 + 0 – 0 – 36 – 0 = 48

= 0 + 0 – 24 – 0 – 0  + 48 = 24

= 0 +  144 + 0 – 168 – 0 – 0 =  -24

Thus, we find

as we obtained with Method 1.