Question 3.8: Write the node-voltage matrix equations for the circuit in F...
Write the node-voltage matrix equations for the circuit in Fig. 3.27 by inspection.
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
\begin{array}{cc}G_{11}=\frac{1}{5}+\frac{1}{10}=0.3, & 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, & G_{44}=\frac{1}{8}+\frac{1}{2}+\frac{1}{1}=1.625\end{array}
The off-diagonal terms are
\begin{gathered}G_{12}=-\frac{1}{5}=-0.2, \quad G_{13}=G_{14}=0 \\G_{21}=-0.2, \quad G_{23}=-\frac{1}{8}=-0.125, \quad G_{24}=-\frac{1}{1}=-1 \\G_{31}=0, \quad G_{32}=-0.125, \quad G_{34}=-\frac{1}{8}=-0.125 \\G_{41}=0, \quad G_{42}=-1, \quad G_{43}=-0.125\end{gathered}
The input current vector i has the following terms, in amperes:
i_{1}=3, \quad i_{2}=-1-2=-3, \quad i_{3}=0, \quad i_{4}=2+4=6
Thus the node-voltage equations are
\left[\begin{array}{cccc}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{array}\right]\left[\begin{array}{l}v_{1} \\v_{2} \\v_{3} \\v_{4}\end{array}\right]=\left[\begin{array}{r}3 \\-3 \\0 \\6\end{array}\right]
which can be solved to obtain the node voltages v_{1}, v_{2}, v_{3}, \text { and } v_{4} .