Question 3.9: By inspection, write the mesh-current equations for the circ...
By inspection, write the mesh-current equations for the circuit in Fig. 3.29.
We have five meshes, so the resistance matrix is 5 by 5. The diagonal terms, in ohms, are:
\begin{gathered}R_{11}=5+2+2=9, \quad R_{22}=2+4+1+1+2=10 \\R_{33}=2+3+4=9, \quad R_{44}=1+3+4=8, \quad R_{55}=1+3=4\end{gathered}
The off-diagonal terms are:
\begin{gathered}R_{12}=-2, \quad R_{13}=-2, \quad R_{14}=0=R_{15} \\R_{21}=-2, \quad R_{23}=-4, \quad R_{24}=-1, \quad R_{25}=-1 \\R_{31}=-2, \quad R_{32}=-4, \quad R_{34}=0=R_{35} \\R_{41}=0, \quad R_{42}=-1, \quad R_{43}=0, \quad R_{45}=-3 \\R_{51}=0, \quad R_{52}=-1, \quad R_{53}=0, \quad R_{54}=-3\end{gathered}
The input voltage vector v has the following terms in volts:
\begin{gathered}v_{1}=4, \quad v_{2}=10-4=6 \\v_{3}=-12+6=-6, \quad v_{4}=0, \quad v_{5}=-6\end{gathered}
Thus the mesh-current equations are:
\left[\begin{array}{rrrrr}9 & -2 & -2 & 0 & 0 \\-2 & 10 & -4 & -1 & -1 \\-2 & -4 & 9 & 0 & 0 \\0 & -1 & 0 & 8 & -3 \\0 & -1 & 0 & -3 & 4\end{array}\right]\left[\begin{array}{l}i_{1} \\i_{2} \\i_{3} \\i_{4} \\i_{5}\end{array}\right]=\left[\begin{array}{r}4 \\6 \\-6 \\0 \\-6\end{array}\right]
From this, we can obtain mesh currents i_{1}, i_{2}, i_{3}, i_{4}, \text { and } i_{5}.