Question 3.4: Find the node voltages in the circuit of Fig. 3.12.

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,

\frac{v_{3}-v_{2}}{6}+10=\frac{v_{1}-v_{4}}{3}+\frac{v_{1}}{2}

or

5 v_{1}+v_{2}-v_{3}-2 v_{4}=60                (3.4.1)

At supernode 3-4,

i_{1}=i_{3}+i_{4}+i_{5} \quad \Longrightarrow \quad \frac{v_{1}-v_{4}}{3}=\frac{v_{3}-v_{2}}{6}+\frac{v_{4}}{1}+\frac{v_{3}}{4}

or

4 v_{1}+2 v_{2}-5 v_{3}-16 v_{4}=0                   (3.4.2)

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 \quad \Longrightarrow \quad v_{1}-v_{2}=20                     (3.4.3)

For loop 2,

-v_{3}+3 v_{x}+v_{4}=0

But v_{x}=v_{1}-v_{4} so that

3 v_{1}-v_{3}-2 v_{4}=0                   (3.4.4)

For loop 3,

v_{x}-3 v_{x}+6 i_{3}-20=0

But 6 i_{3}=v_{3}-v_{2} \text { and } v_{x}=v_{1}-v_{4} . Hence

-2 v_{1}-v_{2}+v_{3}+2 v_{4}=20                     (3.4.5)

We need four node voltages, v_{1}, \quad v_{2}, v_{3}, \text { and } v_{4}, 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 eliminate one node voltage so that we solve three simultaneous equations instead of four. From Eq. (3.4.3), v_{2}=v_{1}-20 . Substituting this into Eqs. (3.4.1) and (3.4.2), respectively, gives

6 v_{1}-v_{3}-2 v_{4}=80                     (3.4.6)

and

6 v_{1}-5 v_{3}-16 v_{4}=40                       (3.4.7)

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

\left[\begin{array}{rrr}3 & -1 & -2 \\6 & -1 & -2 \\6 & -5 & -16\end{array}\right]\left[\begin{array}{l}v_{1} \\v_{3} \\v_{4}\end{array}\right]=\left[\begin{array}{r}0 \\80 \\40\end{array}\right]

Using Cramer’s rule,

\Delta=\left|\begin{array}{rrr}3 & -1 & -2 \\6 & -1 & -2 \\6 & -5 & -16\end{array}\right|=-18, \quad \Delta_{1}=\left|\begin{array}{rrr}0 & -1 & -2 \\80 & -1 & -2 \\40 & -5 & -16\end{array}\right|=-480

\Delta_{3}=\left|\begin{array}{rrr}3 & 0 & -2 \\6 & 80 & -2 \\6 & 40 & -16\end{array}\right|=-3120, \quad \Delta_{4}=\left|\begin{array}{rrr}3 & -1 & 0 \\6 & -1 & 80 \\6 & -5 & 40\end{array}\right|=840

Thus, we arrive at the node voltages as

and v_{2}=v_{1}-20=6.667  V . We have not used Eq. (3.4.5); it can be used to cross check results.