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
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
The off-diagonal terms are
G_{12} = -\frac{1}{5} = -0.2 , G_{13} = G_{14} = 0
G_{21} = -0.2 , G_{23} = -\frac{1}{8}= -0.125 , G_{42} = – \frac{1}{1}= – 1
G_{31} = 0 , G_{32} = -0.125 , G_{34} = – \frac{1}{8}= – 0.125
G_{41} = 0 , G_{42} = -1 , G_{34} = – 0.125
The input current vector i has the following terms, in amperes:
i_{1} = 3 , i_{2} = -1 -2 = -3 , i_{3} = 0 , i_{4} = 2+ 4 = 6
Thus the node-voltage equati
\begin {bmatrix} 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 {bmatrix} \begin {bmatrix} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end {bmatrix} = \begin {bmatrix} 3 \\ -3\\ 0\\ 6 \end {bmatrix}
which can be solved using MATLAB to obtain the node voltages v_{1} , v_{2} ,v_{3} , and v{4} .