Question 2.7: Write the node-voltage equations in matrix form for the circ...
Write the node-voltage equations in matrix form for the circuit of Figure 2.20.
Writing KCL at each node, we have
\frac{v_1}{5} + \frac{v_1-v_2}{4} + 3.5 = 0
\frac{v_2-v_1}{4}+\frac{v_2}{2.5}+\frac{v_2-v_3}{5} = 3.5
\frac{v_3-v_2}{5}+ \frac{v_3}{10} = 2
Manipulating the equations into standard form, we have
0.45v_1 − 0.25v_2 = −3.5
− 0.25v_1 + 0.85v_2 − 0.2v_3 = 3.5
− 0.2v_2 + 0.35v_3 = 2
Then, in matrix form, we obtain
\left[\begin{matrix} 0.45 & -0.25 & 0 \\ -0.25 & 0.85 & -0.20 \\ 0 & -0.20 & 0.30 \end{matrix} \right] \left[\begin{matrix} v_1 \\ v_2 \\ v_3 \end{matrix} \right] = \left[\begin{matrix} -3.5 \\ 3.5 \\ 2 \end{matrix} \right] (2.36)
Because the circuit contains no voltage sources or controlled sources, we could have used the shortcut method to write the matrix form directly. For example, g_{11} = 0.45 is the sum of the conductances connected to node 1, g_{12} = −0.25 is the negative of the conductance connected between nodes 1 and 2, i_{3} = 2 is the current pushed into node 3 by the 2-A current source, and so forth.