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## Q. 2.13

Mesh Equations via Inspection

PROBLEM: Write, but do not solve, the mesh equations for the network shown in Figure 2.9.

## Verified Solution

Each of the previous problems has illustrated that the mesh equations and
nodal equations have a predictable form. We use that knowledge to solve this three-loop problem. The equation for Mesh 1 will have the following form:

$\left[\begin{matrix} Sum of \\ impedances \\ around Mesh 1 \end{matrix} \right] I_{1}(s)-\left[\begin{matrix} Sum of \\ impedances \\ common to \\ Mesh 1 and\\ Mesh 2 \end{matrix} \right] I_{2}(s)$

$-\left[\begin{matrix} Sum of \\impedances \\ common to\\ Mesh 1 and\\ Mesh3 \end{matrix} \right] I_{3}(s)=\left[\begin{matrix}Sum of applied \\ voltages around \\ Mesh 1 \end{matrix} \right]$

Similarly, Meshes 2 and 3, respectively, are

$-\left[\begin{matrix} Sum of \\ impedances \\ common to \\ Mesh 1 and\\ Mesh 2 \end{matrix} \right] I_{1}(s)+\left[\begin{matrix} Sum of \\ impedances \\ around Mesh 2 \end{matrix} \right] I_{2}(s)-\left[\begin{matrix} Sum of \\ impedances \\ common to\\ Mesh 2 and\\ Mesh3 \end{matrix} \right] I_{3}(s)=\left[\begin{matrix}Sum of applied \\ voltages around \\ Mesh 2 \end{matrix} \right]$

$-\left[\begin{matrix} Sum of \\ impedances \\ common to\\ Mesh 1 and\\ Mesh3 \end{matrix} \right] I_{1}(s)-\left[\begin{matrix} Sum of \\ impedances \\ common to\\ Mesh 2 and\\ Mesh3 \end{matrix} \right] I_{2}(s)$

$+\left[\begin{matrix} Sum of \\ impedances \\ around Mesh 3 \end{matrix} \right] I_{3}(s)=\left[\begin{matrix}Sum of applied \\ voltages around \\ Mesh 3 \end{matrix} \right]$

Substituting the values from Figure 2.9 into Eqs. (2.91) through (2.93) yields

$+(2s+2)I_{1}(s)-+(2s+1)I_{2}(s)$         $-I_{3}(s)=V(s)$

$-(2s+1)I_{1}(s)+(9s+1)I_{2}(s)$           $-4sI_{3}(s)=0$

$-I_{1}(s)$      $-4sI_{2}(s)+\left(4s+1+\frac{1}{s} \right) I_{3} (s)=0$

which can be solved simultaneously for any desired transfer function, for example, $I_{3}(s)/V(s)$